CS-60 IGNOU Study Material (Part-II) by IGNOU MCA

UNIT 1

APPLICATIONS OF DIFFERENTIAL CALCULUS

Structure 1.1

Introduction Objectives

Monotonic Functions Inequalities Approximate Values

12 13 1.4 1.5 1.6

Surmnary

Solutions and Answers

In this unit we see some immediate applications of the concepts which we have studied in Blocks 1 and2. Dflerential Calculus have varied applications. You have already seen some applicationsto geometrical, physical and practical problems in Block 2. In this unit, we shall study some applications to the theory of real functions. There are some questions in mathematics, that can be asked even before knowing differentiation, but can l y solved with ease by using the theorems on differentiation. In. other words, differentiation is a useful tool in solving problems that arise independent of this notion, for exarnplt , a)

Is the function sin x + cos x increasing in [O,d4]?

Do we have the inequdity ex> 1+ x for all x?

What is the limit of

What is the approximate value of cos lo?

1-ex

when x tends to 07

Do you notice that these questions do not involve any concept that you have not studied earlier? You could have asked them earlier. You could have even answered some of them. Here we shall see how the theorems of Unit 7 can be systematicallyapplied to yield solutions to such questions. Keep Block 2 ready with you since you will need to read the relevant portions from the units in that block.

Objectives After studying this unit you should be able to ; recognise the equivalence of some properti s of functions (like monotonicity and positiveness or negativeness of ib derivativ ), prove some inequalities using the mean valde theorbnu, apply Taylor's series to obtain approximate values of certain functions at certain points.

1.2

MONOTONIC FUNCTIONS

In this section we employ differentiation to decide whether a given function is monotonic or not. ,

First we recall some terms from Unit 1. We say that a function from an interval I to R is monotonic if it is either an increasing function on I or a decreasing function on I. It is said to be an increasing function on I, if x S y implies f(x) 5 f(y). Increasing functionsmay also be thoughtof as order-preserving functions. .

Appllcatlons pf CalcuIus

Let us also recall that every constant function is an increasing function, the identity function is an increasing function, the function f(x)= 2x + 3 is an increasing function, the function f(x) = sin x is not an increasing function on R.

This is because even though 0 c 3d2, we have sin 0 = 0 > - 1 = sin 3d2. However, on the interval [o, d21, sin x is an increasing function.

Further, we know that a function f from I to R is said to be a decreasing function if x S y implies fix) 2 fly). Decreasingfunctionsare the order-reversing functions. Here are some examples: The function f(x) =4 - 2x is decreasing. The function qx) * sin x is decreasing in the interval [d2, x]. Do.you agree that each constant function is both increasing and decreasing? Warlng: It is incorrect to say that if a function is not increasing, then it is decreasing. It may happen that a function is neither increasing nor decreasing. For instance, ifwe consider the interval [0, x], the function sin x is neither increasing nor decreasing. It is increasing on [0, d 2 ] and decreasing on [d2, x]. There are other functions that are even worse. They are not monotonic on any sub-interval also. But most of the functions that we consider are not so bad. Usually, by looking at the graph of the function one can say whether the function is increasing or decreasing or neither. The graph of an increasing function does not fall as we go from left to right, wbile the graph of a decreasing function does not rise as we go from left to right. But if we are not given the graph, how do we decide whether a given function is monotonic or not? Theorem 1 gives us a criterion to do just that.

f is decreasing if and only iff' (x) S 0 for all x in I.

Proof: a) Let fbe increasing. Let x E I. Then.

f(x + h) f(h) h-+O h

f ' (x) = lim

Since f is increasing, ifh > 0, then x + h > x and q x + h) 2 qx). Hence q x + h) qx) 2 0. Ifh <0, x +h<x, andf(x+h)Sf(x).Hencef(x+h)-f(x)SO.

So either f(x + h) f(x) and h are both non-negative or they are both non-positive. Therefore,

f(x + h) f(x) is non-negative for all non-zero valuesof h. h Therefore,

!cof(x+ h)h - f(x) must also be non-negative.

Thus,f '(x) 2 0. Conversely, let f '(x) 2 0, for all x in I. To show fis m0110tonically increasing. Let a <b in I. We shall prove that f(a) 5 0). By mean vake theorem (Theorem 3, of Block 2), f(b) - f(a) =f'(c)forsomec~]a,b[cI. b-a Since f '(c) 2 0, we h e

f(b) b-a

2 0. Abo b a > 0. It follows that a ) - f(a) 2 0, or a ) 2 f(a).

Thus,a <b implies f(a) S flb). Therefore, f is increasing.

Theorem 1: Let I be an open interval. Let f :I + R be differentiable. Then

The part b) of the theorem can be proved similarly. We leave it as an exercise. It can also be deduced by applying part a) to the function, - f.

E E 1) Prove Part b) of Theorem 1.

From the class of increasing functions we can separate out functions which are strictly increasing. The following definition gives the precise meaning of the term "strictly increasing function".

Definition 1: f : 1 +R is said to be strictly Increasing if a <b implies that f(a) <f@). 1

We can similarly say that s.-hction &fined on I is strictly decreasing if a <b implies fla) > fCb). For example, a constant function is not strictly increasing, nor is it strictly &creasing. The function f(x) = [x] too, is increasing, but not strictly increasing, whereas the h c t i o n flx) = x is strictly increasing. Fig. 1 shows the graphs of these three functions. In Fig. 1 (a) the graph is horizontal. In Fig. l(b) there are parts of the graph which are horizontal. But the graph in Fig. l(c) has no horizontal portions, and rises as we go from left to right.

Flg. 1: (a) graph of f(x).- 1 @) graph of f(x)

1x1 (c) graph of f(x)

Now let us see whether strict monotonicity of a h c t i o n is reflected by its derivative. We have the followingtheorem.

Theorem 2 : a) Let f be positive on I. Then f is strictly increasing on I. b) Let f be negative on I. Then f is strictly decreasing on I.

Prooi: a) By Theorem 1, we know that since f '> 0 on 1, f must be increasing on I. That is, x <y f(x)Ifly). We have only to prove that if x < y, then f(x)cannot be equal to f(y). Let, ifpossible, Qx) = fly), where x <y in I. I

Applieationa of Differentlal Caleulua

Applications of Calculus

Then, by Rolle's Theorem (Theorem 2, Unit 7) applied to the function f on the interval [x, y],wehavef'(c)=Oforsomex~c<y. But this contradicts the assumption that f is strictly positive on I. Hence f(x) cannot be equal to fly) for my x < y in 1. Thus, x < y 3 f(x) < f(y). That is, f is strictly increasing. b) We indicate two different proofs for this part. One way is to imitate the proof of part a) by changing the symbol '<' to the symbol '>', wherever it occurs. Another way is to consider -f, apply Theorem 2 a) to it and use the facts: (-0' = - f '. Therefore f is negative If and only if (-0' is positive. Also f is strictly decreasing if and only if - f is strictly increasing. Combining the results for inckeasing and decreasing functions we get the following corollary. Corollary 1: f is strictly monotonic on the interval I iff ' is of the same sign throughout I, assuming zero is neither positive nor negative.

You may have noticed that there is a difference between the statements of Theorem 1 and Theorem 2. We said: "f is increasing if and only iff ' is non-negative". "Iff '> 0, then f ' is strictly increasing" It is natural to ask: Can we have "If and only if' in Theorem 2 also? That is, iff is strictly increasing, does it follow that f '> O? Unfortunately, we have a negative answer as shown in Example 2 below. But before that, our first example shows how among many methods available to prove the monotonicity, the one using differentiationis the simplest (provided the function is differentiable, of course). Example 1: Let f(x) = x3 for all x in R. We shall prove that the function f is increasing. First Method: Let x < y. We want to prove that x3< 9.Consider two cases. Case 1: x and y are of the same sign. (Either both are positive or both are negative). In this case xy > 0. Now,

9 - x3= (y - x) ( 9 + xy + x2)2 0 since both y - x and y2+ yx + xZare non-negative. Case 2: Let x and y be not of the same sign. Since x < y, this means that x < 0 < y. (Note that if either x or y is zero, it comes under Case 1.) Therefore, x3 < 0 < y3. Because, the cube of a negative number is negative, and the cube of a positive number is positive.

Thus, in both case x3 S y3. In fact, we have the strict inequality (x3 < g ) , indicating that x + x3is a strictly increasing function. Second Method: Let x < y. We want to prove that x3I y3.

Here y - x > 0 (since x < y). Also, y2 + yx + x21 0 because

since the square of any number is non-negative. Thus y3 - x3 is a product of two non-negative numbers and, hence is non-negative. Third Method: (Using Differentiation): Let f(x) = x3. Then f(x) = 3x2.This is always non'legative. Therefore, using Theorem 1 we can say that f is an increasing function.

Example 2: Here we give an example of a strictly increasing function whose derivative is not strictly positive.

Applications o f Differential Calculus

Let f :R + R be the function defined by fix) = x3(see Fig. 2 1).It is strictly increasing because, x<y 3 y-x>Oandx2+y2>0

fy 1

Its derivative is not strictly positive because f ' (0)= 0. Our next example describes two different ways idwhich non-monotonicity of a function can be proved.

Example 3: To prove that the function f :x + sin x + cos 2x is not monotonic on the interval [O, d4], we can proceed as follows: i

First Method: We shall consider three points, 0, d l 0 and d 6 belonging to [0, d4].

'IT

f(x 1 10) = sin -+ cos - = 03090 + 08090 > 1 10 5

We have 0 < d l 0 < d 6 and f(0) < f(ldl0) > f(x/6). Therefore, f is neither increasing, nor decreasing on [0, d4]. Or, We can say that f is not monotonic on [0, d41.

Second Method: (Using Differentiation) Let fix) = sin x + cos 2x Then f '(x) = cos x - 2 sin 2x

i t

Thus, f ' is of different signs at 0 and d 4 . Therefore f is not monotonic on [0, d4]. Our next example warns us in dealing with functions not defmed at some points.

Example 4: If the function f :x - tan x is defined on an interval, we can prove that it is an increasingfunction there. See Fig. 3.

Fig. 2

Applications of Calculus

Now consider the interval [0, x]. Can we prove that tan x is increasing on this interval? Suppose we argue as follows: f '(x) = sec2x, and sec2x 2 0 y x since the square of any quantity is non-negative. Hence by , Theorem 1, f is an increasing function on [0, x]. But if we take two points

2n -n4 and 3 in [0, x].

n 2n x 2n Thus- <-, but tan - & tan -. 4 3 4 3 This indicates that tan x is not an increasing function on [0, x]. So where did we go wrong? We can explain it as follows: n

In the interval [0, x], there is a point, namely 7 ,where tan is not defmed. Hence, its derivative does not exist at that point, and therefore we can not apply Theorem 1. What we proved is that this function is increasing in an interval, provided it is defined throughout this interval. It is only when it is defined, that we can differentiate it and apply our theorems.

In the next example we use an additionalproperty of continuous functions in the first method, and repeated differentiation in the second method. Example 5: Let us prove that the function x + sin x + cos x is increasing on [0, d 4 ] and decreasing on [d4, d2]. First Method: Recall that for all x in the k t quadrant (i.e., if 0 I x I d2), sin x; cos x and tan x

are non-negative.

LetfCx)=sinx+cosx,thenf8(x)=cosx-sinx Wenotethatf'(0)- 1, f8(7d4)=0, f '(d)=-1. x = d 4 is the only point in [0, d 2 ] at which f '(x) = 0. Because f ' (x) = 0 cos x = sin x tan x = 1. But tan x is strictly increasing on [0, x/23, by Example 4. So, it cannot take the value 1 at any point other than d 4 .

Also, f ' is a continuous function because the sine and cosine functions art continuous. Therefore, f cannot take negative values in [0, d4]. Explanation: If a continuous function takes a positive value at 0 and a negative value at some x, then it must be zero somewhere in between 0 and x. In this problem, the continuous function f 'cannot take the value zero between 0 and x if x < d 4 . Therefore, since f 'is non-negativeon [0, d4], f is increasing on, [O, d4]. Similarly, f 'cannot take positive values in [d4, d2], because its value in [d4, d2] is negative. It follows that f is decreasing on [d4, d2]. Second Method: Let fix)= sin x + cos x, and f '(x) = cos x - sin x. To prove that fisincreasing on [O, d41,we have to prove that f is non-negative on [O, d4]. We first note that f ' (0) = 1 and f ' (d4) = 0. It is enough to prove that f ' is decreasing on [0, d 4 ] (for then, all values off 'in this interval will be between 0 and 1). For thispurpose we consider f "(x) = sin x - cos x. Also see Fig. 4(a) and (b).

We note that f " (x) S 0 for all x in the first quadrant, and in particular for all x in [0, d4]. Therefore, f ' is decreasing on [0, d4]. Therefore, (since f ' (0) = 1 and f ' (d4) = 0), f 'is non-negative on [O, d4]. Therefore f is increasing on [0, d43. . Next, we shall prove that f ' is non-positive on [d4, d2]. First, we note that f '(d4) = 0 and f '(62) =-1. Also, f ' is decreasing on Id4, d2], since f "(x) S 0 on [d4, d2]. Therefore all the values o f f on this interval are between 0 and 1 and hence cannot be positive. Therefore

f i 9 rlecreasinrr nn Td4 d21.

Appllcatlons of DJfferentlaI Calculus

(8)

Fig. 4: (a) Graph of qx)

l a x + cos I, @) Graph of f(x)

cos x

- sln x

If you have followed the arguments in these examples, you should not have any difficulty in solving these exercises.

E E 2) Assuming that exnever takes negative valued, prove that it is an increasing function on R

/ E E 3) Prove that lnx is an increasing function on lo,= [. .

Applications of Calculus

E E4) Using the fact that sin x and cos x are never negative in the fust quadrantl prove by differentiationthat sin x is an increasing function and cos x is a decreasing function on [O, d21.

E E5) Which of the fdowing functions are inmaking on the interval given? Which of them are decreasing?

e) x sin x on [O, d2] f) tan x + cot x on [0, d4]

E E6) Prove that the following functions are not monotonic in the intervals given a) 2x2+3x-5on[-1,0] b) x(x-l)(x+l)on[O,2]

C) xsinxon[O,n] d) tan x + cot x on [0, d 2 ]

E E 7) Give an example of a cubic polynomial that decreases on ] --, 21, increases on [2,3] and again decreases on 13, -1.

(Hint: The derivative should change sign while passing through 2, and again-while passing through 3).

1.3

INEQUALITIES

Another application on differential calculus is to prove certain inequalities. In the three examples below we illustrate how some inequalities can be deduced from Taylor's series and the mean value theorem.

Appll~ationsof Dlfferentlal Calculus

Applications of Calculus

Example 6: Suppose we want to prove ex2 1 + x for all x in R. Letf(x)=eX-x,Thenf (x)=eX- l,f"(x)=eX We know that ex> 0 for all x. Therefore, f ' (x) is a strictly increasing function. Also f '(0) = 0.

Therefore there is no other point where f vanishes. Also,f'(x)>Oifx>Oand f'(x)<Oifx<O. Therefore flx) is increasing on ] 0, m [. So x >0 implies f(x) 5. flO).Thls means ex- x > e0- 0 = 1. This proves ex> 1+ x, if x >0. It remains to prove this for negative values of x also. For this purpose we let g(x) = ex- xeX. Then g'

(x) =ex-(xeX + ex)=-xeX <0 wherever x >0.

* g is strictly decrestsing on ] 0, [ m

a g(x) < g(0) whenever x >0. 3 ex- xeX< e0- O.eO= 1 for all x.> 0.

1 In other words, ex< - for all x > 0. I-x

Putting y = - x, we get e-y < -,or I+Y

ey> 1 +yfory<O.Inotherwords,eX>1 +xforallx<O. Whenx=O,eX=l=l+x. Thus, the inequality ex 2 1 + x is true for all values of x. In the next example we give an inequality that is still better.

x2 2

x3 6

Example 7: We prove that ex> 1+ x + -+ - for all x > 0.

We have seen in Unit 6 that the Taylor's series expansion. x2 x3 e x = l + x + -+ - + .............isvalidforallx. 2! 3! This proves the inequakty x2 x3 eX> 1+ x + -+ - wheneverx>O. 2 6 Fig. 5 represents the results of Examples 6 and 7.

Example 8: To prove bn- anC n b"' (b - a), wherever 0 C a C b and n > 1, we consider the function f : x + xnon the interval [a, b]. It is continuous there. It is also differentiablein ] a, b [. Therefore, by the mean value theorem, there is some c, a < c < b such that

Cross-multiplying we get, bn an= ncP1 (b - a). Therefore, it suffices to prove that e"' < bP'.

ThisistruebecauseO<c<b,andn> 1. You can try these exercises now.

EE8) Prove the following inequalities using the methods indicated alongside in brackets. a) In(1 +x)~xforallpositivex (first prove that x - In (1 + x) is increasing). x <x for all positive x b) tantan1 (by mean value theorem on [0, x] for tan-' x). i

c) ex+ e-X> 2 t/ x. (writing ex+ e-X- 2 as a perfect square). d) ex+ e-X> 2 t/ x. (using the already proved result eX 2 1 + x). e) ex- e-X12 x t/ x > 0. (using the inequality in d) and differentiation).

Appllcatlonr of Dlfferentlnl Calculus

1.4

APPROXIMATE VALUES

In the previous two sections we have seen how the concept of derivatives can be used in proving the monotonicity of a given differentiable-function; and how this knowledge can be applied to prove some inequalities. In this section we shall see how Taylor's series can be used to find approximate values of some functions at some points. We use the symbol = to mean approximately equal to. Example 9: Taking the first two non-zero terms in Maclaurin's series for sin x, we shall prove that sin 20" is approximatelyequal to 0.342 (in symbols, sin 20" ~0.342).Remember that in the formula -~

the angle x is always measured in radians. The same holds for cos x, tan x, and so on.

Now, 20" = - radians. Therefore, 9

Taking ~c= 3.142, we get this quantity to be 3.142 .0425 -.= 0.349 - 0.007 = 0342. 9 6

If you look into a table of sines, you will find sin 20" = 0.342. This shows that our approximation is really a good approximation. In fact, the tables are written by using precisely these methods. Example 10: Let us find the approximate value of (0.99)512by taking three terms of Maclaurin's series for (1 - x)5n

Maclaurin's series for (1 x)~"is

We can write (0.99)512as (1 - 0 . 0 1 ) ~ ~ . So when x = 0.01, tak@g the first three terms ofMaclaurin's series, we get 5 15 (1 - 0 . 0 1 ) ~=~ 1- - (0.01) + -(0.001) 2 8 That is, ( 0 . 9 9 ) ~= 0.975. X

2 . If the fust two non-zero terms of Maclaurin's Example 11: We know that cos - = 6 series for cos x are taken to approximate it, let us calculate the error, rounded off to two decimal places.

The error is the difference betwtm the actual value and the approximate value.

Maclaurin's series for cos x is I

If we take the first two terms alone and put x = d 6 , then

The actual value is cos - = 6 decimal places. -

We have found that 1

1 2 . We know that JS 1 2 = 0.866 when rounded off to three

( ~ / 6 ) ~ -= 0.863 when rounded off to three decimal places. 2

The error is 0.866- 0.863= 0.003. When rounded off to two decimal places, the error is 0.00. (This means that the error is so negligible that there is no error at all when rounded off to two decimal places.) See if you can frnd the approximate values in the following exercises.

E E 9) Find the approximate value of sin 3lo by taking the fmt two non-zero terms of its Maclaurin's series.

E E10) If the fmt three non-zero terms of Maclaurin's series for cos x are used to approximate n

cos - ,show that the error is less than 1/50. 2

E E 11) Find the value of cos 5g0,rounded off to one decimal place.

Appllcatlonr of

Differential Calculur

Appllcatlons of C ~ I C U I U S

E E 12) Find (l.01)"2 upto two decimal places.

That brings us to the end of this unit. Let us summarise what we have studied in it.

1.5

SUMMARY

In this unit we have studied the following results. 1)

I 1ffis

increasing decreasing constant monotonic 2)

, then f ' is non-negative non-positive identically zero

of same Hign throughout

Iff'is

then f is

non-negative non-positive (strictly) positive (strictly) negative identically zero of same sign throughout

increasing decreasing strict1y increasing strict1y decreasing constant monotonic

Differentiation can be used to test whether a fimction is monotonic or not, to prove some inequalities, and to find some approximate values.

1.6

SOLUTIONS AND ANSWERS

E 1) Let f be a decreasing function. If x E 1, then f ' (x) exists, and f(x + h) - f(x) f' (x) = lim

Hence, f ' (x) = Iim h-0

f (x + h) - f (x) h

f(b) - f(a) = f . (c) 5 0, b-a

SO.

Appllcatlons of Differential C a l c ~ l u s

b-a>O*f(b)-qa)<Oorf@)Sf(a).

f is a decreasing function.

qx) = eX is en increasing function on R (by Theorem 1). 1

E3) Iff(x)=lnx,f8(x)=;. Now >Oy x ~ ] O , . o [ . X

Hence In x is an increasing function on ] 0, w [.

E4) If'qx) = sin x, then f '(x) = cos x. C O S X L OX~E [O,W2].

Hence sin x is an increasing function on [0, W2]. Similarly for cos x.

*f is decreasing on [0, 11

C) f (x) =- e-x S 0 on [0, 11 d) increasing

e) increasing f ) f'(x)=sec2x-cosec2x=

- cos 2x sin2 x cds2 x

5Osincecos2x20. v X E [O,W4].

.-.f is decreasing. E6) a) fr(x)=4x+3 fr(x)50ifx€ [- 1,-3/4]andff(x)20ifx~[-3/4,0].

d) similar argument E7) ~ e t f ( x ) = a x ~ + b x ~ + c x + d f '(x) = 3axZ+ 2bx + c, where a, b and c are to be determined so that the desired cubic polynomial is obtained. f has extreme at x = 2 and x = 3.

Suppose a = -2 and d = 0.

Then f(x) = -2x3 + 1Sx2- 36x satisfies the given conditions.

Applications of Calculus

b) f(x)=tan-' x. Let x >0. By the mean value theorem 3 y E 10, x [ s.t.

tan-' x - tan-' 0 = f ' (Y) x-0 tan-'x 1 o r X P 2 < I f m y €]O,X[. 1+ Y

e) Letf(x)=eX-e~x-2x.Thenf'(x)=ex+e~x-2>0. a f is an increasing function aflx)2flO)forallx>O aeX-e-x-2x20forallx>0. aeX-e-x22xforallx>0.

n We know that cos 2 = 0

UNIT 2

AREA CINDER A CURVE

Structure Introduction

2.1

Objectives

Area Under a C u m

2.2.1 Cartesian Equation 2.2.2 Polar Equations 2.2.3 Area Bounded by a Closed Curve

Numerical Integration 2.3.1 Trapezoidal Ruk 2 3.2 Simpson's Rule

2.4 2.5

Summary

Solutions and Answers

INTRODUCTION

2.1

When we introduced you to integration, we mentioned that the origin of the method of integration lies in the attempt to estimate the areas of regions bounded by plane curves. In this unit we shall see how to calculate the area under a given curve, when the equation of the c y e is given in the Cartesian or polar or parametric form. This process is also called quadrature. We shall also study two methods of numerical integration. These are helpful when the antiderivativeof the integrand cannot be expressed in terms of known functions, and the given definite integral cannot be exactly evaluated.

Objectives After reading this unit you should be able to : use your knowledge of integration to find the area under a given curve whose equation is given in the Cartesian or polar or parametric form, , recognise the role of numerical integration in solving some practical problems when some values of the' function are known, but the function, as a whole, is not know'n. use trapezoidal and Simpson's rules to find approximatevalues of some definite integrals, compare the two rules of numerical integration.

2.2

AREA UNDER A CURVE

In this section we shall see how the area under a curve can be calculated when the equation of the curve is given in the i) Cartesian f o m ii) polarfom iii) parametric form. Some curves may have a simple equation in one form, but complicated ones in others. So, once we have considered all these forms, we can choose an appropriate form for a given curve, and then integrate it accordingly. Let us consider these forms of equations one by one.

22.1 Cartesian Equation We shall quickly recall what we studied in earlier. Let y = f(x) define a continuous function of x on the closed interval [a, b]. For simplicity,we make the assumption that f(x) is positive for x E [a, b]. Let R be the plane region in Fig. 1 (a) bounded by the graphs of the four equations:

We divide the region R into n thin strips by lines perpendicular to the x-axis through the end points x = a and x =b, and through many intermediate points which we indicate by x,, x2, .... k,. Such a subdivision, as you have already seen in Sec. 2 of Block 3, is referred to as a partition P, of the interval [a, b] is indicated briefly by writing

We write

and take the set of n points on x-axis.

such that x,, Iti 5 xi for i = 1,2, ....,n. We now construct the n rectangles (Fig. 1 (b)) whose bases are the n sub-intervals [xi ,xi], i = 1,2, ......,n induced by the partition P,, and whose altitudes are f(t,), f($), ...,f(tl), ....,f(t,,), f(tn).The

sum C f ( t i ) A xi of the areas of these n rectangles will be an approximation to the "area of R . Notice (Fig. 2(a) and (b)) that if we increase the number of sub-intmals, and decrease the length of each sub-interval, we obtain a closer approximation to the "area of R".

Area Under a Curve

Applications o f Calculus

Thus, we have Definition 1: Let f be a real valued function continuous on [a, b], and let

t-l

Sub-inte~als,A xi + 0, then that limit is the area A of the region R n

That is. A =

lim

Z f ( t i ) A xi

A xl+O i = 1.2.

....n

Compare this definitionwith that of a definite integral given in Block 3. Over there we had seen that the definite integral, If(x) dx is the common limit of

i-1

mi A xi and A M i A xi as the A xi's +0. i=l

~ow'since m,4 f(ti) 5 Mi v i,we have

Hence if the limit of each of these as A xi's +0 exists, then by the Sandwich Theorem in Unit 2, n

Now, if Sf(x) dx exists, than the first and the third limits here are equal, and therefore a

we get A = If(x) dx

The equaltity in (1) is a consequence of the defdtions of the area of R and the d e f ~ t e dx .Since f(x) is assumed to be continuous on the interval [a, b], the integral

in (1) exists, and hence yields the area of the region R under consideration. From the Interval Union Property (Sec. 3, Unit 10)of definite integrals, we have b

~ ( xdx) = ]f(x) dx + jf(x) dx, a s c s b. a

~ i g 3.

...(2 )

This means if A:, A:, A: denote the areas under the graph of y = f(x) above the x-axis from a to c, fiom c to b and from a to b, respectively, (Fig. 3) then, if c is in between a and b, then we have AC,+A: =A;

...(3 )

If we defme At = 0, A: = 0 ,then the above equation is true for c = a and c = b too. Till now, we have assumed the function qx) to be positive in the interval [a, b]. In general, a function f(x) may assume both positive and negative values in the interval [a, b]. To cover such a case, we introduce (theconvention about signed areas. The area is taken to be positive above the x-axis as we go from left to right, and negative if we go from right to left. The function f(x) may be defined beyond the interval [a,b] also. .In

that case (3) is true even if c is beyond b, since according to our convention of signed areas,

Are8 Under 8 Curve

A: will turn out to be a negative quantity (Fig. 4).

Or, A: + A; = A:. Now, if f(x)S 0for all x in some interval [a, b], then by applying the defmition of6'areaof R" b

to the h c t i o n - f(x), we get the area A = - If(x) dx . a

If we do not take the negative sign, the value of the area will come>outto be negative, since f(x) is negative for all x E [a, b]. To avoid a "negative" area, we follok this convention. Thus, if f(x) 5 0 for x E [a, b] (Fig. 5), then the area between the ordinates x = a and x = b will be

The following examples will illustrate how our lcnowledge of evaluating definte integralscan be used to calculate certain areas. Example 1: Suppose we want to find the area of the region bounded by the curve y = 16- x2, the x-axis and the ordinates x = 3, x = -3. The ngion R, whose area is to be found, is shown in Fig. 6.

The arq A of the ngion R is given by

Fig. 4

Appllc8tle.s

r f Calculus

Example 2: Considerthe shaded region R in Fig. 7.

R is composed oftwo parts, the region R,and the region R,.We have Area R = Area R,+ Area R, The region R, is bounded above the x-axis by the graph of y=x3+x2-2X,x=-2BILdx=0.

The region R, is bounded below the x-axis by the graph of y=x3+x2-2x,x=Oandx-l. Hence, 1

In this example we had to calculate area R, and area R, separately, since the region R, was below the x-axis. Therefore, according to our convention I

Area R,= - If(x) dx. 0 I

If we calculate [f(x) dx ,it will amount to calculating -2 0

P ( X )dx + If(x) dx = area R,- area 4,which wouldbe a wrong -2

estimate of area R.

Example 3: Let us find the area of the smaller region lying above the x-axis and included between the circle x2 + y2= 2x and the parabola y2= x. On solving the equation x2+y2= 2x and y2= x simultaneously,we get (0, O), (1, I), (1, -1 ) as the points of intersection of the given curves. We have to find the area of the region R bounded by OAPBO (Fig. 8).

Fig. 8

From the figure we see that area of region OAPBO = area of region OCPBO - area of region OCPAO

Now,

),/G

dx = ) , / q d x

bms

0 (- cos 0) d0, on putting 1 - x = sin 0

2 Also. I&dx = 3 0

Therefore, the required area = I

Try to solve these exercises now.

E E 1) Find the area under the curve y =sin x between x = 0 and x =n.

Area Under a Curve

Appllcatlonr of Calculus

E E2) Find the area bounded by the x-axis, the c w e y = ex,and the ordinates x = 1 and x = 2.

E E3) Find the area of the regienboundedby the curve y = 5x - x2,x = 0, x = 5 and lying above the x-axis.

E E4) Findthe area cut off fiomthe parabola 9 = 4ax by its latus rectum, x = a.

Area Under a Curve

E E 5) Flnd the area between the parabola 9 = 4ax and the chord y = mx.

In this sub-section we have derived a formula (Formula (I)) to find the area under a curve when the equation of the curve is given in the cartesian form. With slight modifications we can use this formula to find the area when the curve is described by a pair of parametric equations. We shall take a look at curves given by parametric equations a little later. But first, let us consider the curves given by a polar equation.

2.2.2

Polar Equations

Sometimes the Cartesian equation of a curve is very complicated, but its polar equation is not so. Cardioids and spirals which you have encountered in Unit 9 are examples of such curves. For these curves it is much simpler to work with their polar quation rather than with the Cartesian ones. In this sub-section we shall see how to find the area under a curve if the equation of the curve is given in the polar form. Hen we shall try to approximatethe given area through the areas of a series of circular sectors. These circular sectors will perform the same function here as rectangles did in Cartesian coordinates. Let r = f(8) determine a continuous curve between the rays 8 = a and 8 = g, (

- a 5 2n).

We want to find the area A(R) of the shaded region R in Fig. 9 (a). A

(b)

Appllcatlons of Calculus

Imagine that the angle AOB is divided into n equal parts, each measuring AH.

R,, 5, .......R,,, as shown in Fig. 9 (b). Then clearly A(R) = A(R,) + A(%) + ......+ A(&), ill

Now let us take the i&slice R,, and try to approximate its area. Look at Fig. 10. Suppose f attains its minimum and maximumvalues on [ek,, Oil at ui and vi. The area of a sector of a circle of radius r and sectorial angle A0 is 10 iA0.

From this we get

The first and the third sums in this inequality are the lower and upper Riemann sums Block 3 for the same dcflaite integral, namely,

ITtf(0)12 do. a

Therefore, by applying the d w i c h theorem as A 8 + 0, we get

We shall illustrate the use of this formula through some examples. Study them carefully, so that vou can do the exercises that follow later. Example 4: Suppose we wmt to fmd the area enclosed by the cardioid r = a ( l -case).

Since cos 8 =cos (- 0). the cardioid is symmetrical about the initial lines AOX (Fig. 11).

Hence the required area A, which is twice the area of the shaded region in Fig. 1I, is given by

= 4a2 ]sin4 0

22 dB, since cos 0 = cos2 22 - sin2 -Q2

Area Under a Curve

= 8a2 'rsin4 4 d), where ) = 2 0

315c

= 8a2-

- - by applying the reduction formula fiom Section 3 of Block 3.

4 2 4

In the case of some Cartesian equations of hi@er degree it is o h convenient to change the equation into polar form. The following example gives one such situation. Example 5: To find the area of the loop of the curve. we change the given equation into a polar equation by the transformation x = r cos 8 and y = r sin 8. Thus,we obtain

which yields r = 0 for 8 = 0 and 8 = d 2 . Hence, area A of the loop is that of a sectorial area bounded by the curve and radius vectors 8 = 0 and 8 = d 2 , that is, the area swept out by the radius vector as it moves from 8 = 0 to 8 = d 2 . See Fin. 12.

Try to do these exercises now.

Applications o f Calculus

E6) Find the area of a loop of the curve r = a sin 30.

E 7) Find the area enclosed by the 8=0,0=x/2.

E8) Find the area of the region outside the circle r = 2 and inside the lemniscate 9 = 8 cos 28.

[Hint:First find the points of intersection. Then the required area = the area under the lemniscate - the area under the circle.]

2.2.3 Area Bounded by a Closed Curve

Area Under a Curve

Now we shall turn our attention to closed curves whose equations are given in the parametric form Let the parametric equations

x = $ (0, Y = (t), t E [a. PI, where $(a) = MP). and ( a ) =

w (P), represent a plane closed curve (Fig. 13).

Here we shall considcr curves which do not intersect themselves, e.g. of the form:

This means that as the parameter t increases from a value a to a value P, the point P(x, y) describes the curve completely in the counter clockwise sense. Since the curve is closed, the polnts on it corresponding to the value $ is the same as the point' corresponding to the value a. This is reflected by the conditions $(a) = $(P) and ~ ( a=)I+@). Suppose further that the curve is cut at most in two points by every line drawn parallel to the x or y-axis. We also assume that the functions $ and w are differentiable, and that the derivatives $' and \I/ do not vanish simultaneously. Let the point R on the closed curve correspond to the values a and P i.e., at R we have $(a) = $(P) and ~ ( a = ~) ( p ) . Now suppose A is a point on the curve which has the least x-coordinate, say a. Similarly, suppose B is a point on the curve which has the greatest x-coordinate, say b. Thus the lines x = a and x = b touch the curve in points A and B, respectively. Further let t, and t, be the values o f t that correspond to A and 9,respectively. Then,

Let a point Q correspond to t = t, such that t? < t, < t,. The area of the region enclosed is S = S, - S,, where S, and S, are the areas under the arcs AQB and ARB, respectively. The minu; is because on; is clockwise and other is anti-clockwise(see Fig. 13). Hence.

(AQB)

(ARB)

Now, as a point ?(x. y) moves from B to A along BQA. the value of the parameter increases from t, to t, . Therefore,

i 1:; ydx=

y-dt

Hence S2= -

Note how we have mod~tied formula ( I ) for a pal! of panmetric equation,

Appllcatlons of Calculus

Remember the point R corresponds to t = a and also to t = 8.

Now the movement of P from A to B along ARB, can be viewed in two parts: From A to R and from R to B. As P moves from A to R, the value of the parameter increases from t, to $, and as P moves from R to B, t increases from u to t2.

Thus, we have

Note that the negative sign is due to the direction in whlch we go round the curve as marked by arrows in Fig. 13.

Similarly, by drawing tangents to the curve that are parallel to the x-axis, it can be shown that

...(ii) a

From (i) and (ii),we get

Hence, the area enclosed is

s'=1 2

(xdy ydx)

We can use any of the fonnulas (i), (ii) and (5) above for calculating S. But in many cases you will find that formula (5) is more convenient because of its symmetry.

Example 6: Let us find the area of the astroid

The region bounded by the astroid is shown in Fig. 14. The area A of the region ip given by

a cos3t (3b sin2t cos t) - b sin3t (- 3a cos2t sin t) dt

Area Under r Curve

Fig. 14

We have seen in Section 3 of Unit 11 that

= 2 i f (r) dx, if f(2a

- x) = f (x).

Here cos2(2x - t) sin2(2x - t) =cos%sin2t. 2x

Hence Icos2t sin2tdt = 2 icos2t sin%dt.

Therefore, A = 3ab cos2t sin2tdt. 0

I t

Now, by a similar argument we can say that

e, by using the reduction formula fkom Section 4 of Unit 12. 8

You can solve these exercises now.

E9) Findtheareaofthecurve x=a(3sin8-sid0),y=acos38,018~2x.

Calculus

E 10) Find the area enclosed by the curve x=acos8+bsin8+c y=a'cos8+b'sin8+c',whereOS8S21c

E 11) Find the area of one ofthe loops of the curve x = a sin 2t, y = a sin t. (Hint: first find two values oft which give the same values of x and y, and take these as the limits of integratim.)

2.3

NUMERICAL INTEGRATION

In many practical problems, the value of the integrand, that is, the function whose integral is required, are known only at some chosen points. For example,

Area Under a Curve

In some cases, no simple integral is known for the given integrand. For example, the function

sin x does not have a simple indefmite integral. In such a situation the integral X

If(x) dx cannot be evaluated exactly. But we can fmd an approximate value of the integral a

by considering the sum of the areas of inscribed (inner) or circumscribed (outer) rectangles, as we have seen at the beginning of this unit. In this section, we shall describe two more methods for approximatingthe value of an integral: i) TrapezoidalRule, and ii) Simpson's Rule.

2.3.1 Trapezoidal Rule We know that a given definite integral can be approximated by inner and outer rectangles. A better method is given by the trapezoidal rule in which we approximate the area of each strip by the area of a trapezium. Such an approximation is also called a linear approximation since the potions of the curve in each strip are approximatedby line segments. As before, we divide the interval [a, b] into n subintervals, each of length

x,-, = a + ( n - l)Axbetweenxo=aandx,,=b. Then

Now, we approximate the first integral on the RHS by the area of the trapezium aP$,x, (Fig. 15), the second by x,P,P,x, and so on, thus getting

...(6) where yo=f(xo),YI =f7xl), ........,Y,,=f(\)

Fig. 15

I 'The area of a trapezium = 2 (the sum of its parallel sides) r height.

Appl~cationsor CPICUIUS

The formula (6) is known as the trapezoidal rule for approximating the value of a definite integral. See if you can solve this exercise now.

E E 12) Use the trapezoidal rule to estimate the following integrals with the given value of n.

2.3.2

Simpson's Rule

In this method, instead of approximating the given curve by line segments, we approximate it by segments of a simple curve such as a parabola.

The area under the arc of the parabola

~=AX~+BX+C between x =-h and x = h (Fig. 16)is given by

Since the curve passes through the points, ( - h, yJ, (0, y,) and (h, y2),substituting the coordinates of these points in the equation of the parabola, we get yo = A ~ ~ - B ~ + c

Y, =c y2 = Ah2+Bh+C,

from which we get

c =Y, ~ h -Bh=yo-y, ' Ah2+Bh=y2-y, *2Ah2=yo+y2-2y, Hence, on substitution we get

To obtain Simpson's rule, we apply the above result to successive pieces of the curve y = f(x)between x = a and x = b. For this we divide [a, b] ioto n sub-intervals, each of width h. Then we approximate the portion of the curve in each pair of sub-intervals by an arc of a parabola passing through the end points of that portion of the curve and the point corresponding to the common point (Fig. 17) of those sub-intervals. Now consider the first two sub-intervals [a, x,] and [x,, %I. The points on the curve corresponding to a, x, and 3 are Po, PI and P,, respectively. Let us draw a parabola passing through Po, PI and P2. We will assume that the portion of the curve passing through P, PI and P2coincides with this parabola. See Fig. 17.

Similarly we can approximate the potion of the curve in the interval [%, x,] by a parabola passing through the points P2,P, and P,. We shall repeat this process for the remaining pairs of intervals. Now the area under the parabola Po PI P, is given by

h 3

A, = - (yo+ 4y, + y2).(On using formula (7)).

Similarly, the area under the parabola passing through the points P,, P,, P, is given by

Next, we use formula (7) for the parabola passing through the points P,, PS,P, and get the area

h A~= - [y, + 4y, + yJ, and so on.

Area Under r Curve

Applications of Calculus

Note, that to approximate the whole area under the given curve in this manner, the number n of the subdivisionsof the interval [a, b] must be even. Summing all the areas we obtain

as the total area. The above formula can also be written in the form

where we have taken n =2m.Note that yo, y,, y,, ......,y2,, are the values of the function, f, for x =a, a +h, a + 2h, ......, a + 2mh = b, respectively. Hence by Simpson's rule, we have

The following example will help us compare the accuracy of the trapezoidal and Simpson's rules. You will find that Simpson's rule is more accurate than the trapezoidal rule. Example 7: Taking four subdwisions of the interval [I, 31, let us find the approximate value of

i I

x2& by the trapezoidal rule and also by Simpson's rule.

Division by 4 gives 0.5 as the width of each sub-interval of [l, 31. The values of the integrand at these points of subdivisiod are given in the following table:

Using the formula for the trapezoidal rule, we obtain

(k + + + + 2.25 4

6.25

Using the formula for Simpson's rule, we get

the actual value of the defmite integral

On comparing (a) with the actual value in (c), we observe that the value given by the trapezoidal rule has an e m r of + 0.08. Comparison of (b) with (c) shows that the error in the value given by Simpson's rule is zero in this case. Example 8: Let us use the trapezoidal rule with six sub divisionsto evaluate

.We shall alao fmd the value of the integral by using Simpson's rule.

Area under r Curve

Using trapezoidal rule, we obtain

Using Sbqson' rule, we have

Exmaple 9: A river is 80 m wide. The depth d in metres at a distance x meters fiom one bank is given by the following table:

Let us find, approximately, the area of cross-section. Applying Simpson's rule with n = 8, we obtain the area of cross-section

As we have seen in this example, Simpson's rule is very useful in approximatingthe area of irregular figures like the corss-sections of lakes and rivers. See if you can do these exercises now.

E E 13) a) Use Sunpson's rule to evaluate the following, taking the given value of n

b) From the formula

Applications of C a l c u ~ u s

E E14) Acurveisdrawnthroughthepoints(l,2),(1.5,2.4),(2,2.27),(2.5,2.8), (3,3),(3.5,2.6)and (4,2.1).Estimate the area between the x-axis and the ordinates x = I, x = 4.

E E 15)A river has width 30 metres. If the depth y metres at distance x metres from one bank be given by the table.

Find the approximate area of cross-section.

E E 16)The velocity of a train, which starts from rest, iSgiven by the following table, the time being reckoned in minutes from the start and the speed in kmslhr.

Min knsthr

2 10

, 4 1

6 25

8 29

10 32

12 20

Estimate approximatelythh total distance run in 20 minutes.

18 2

20 0

Area Under r' Curve

Now let us quickly recallwhat we have done in this unit.

In this unit we have covered the following points: 1)

The knowledge of integration is helpful in finding areas enclosed by plane c w e s when their equations are known in

The area bounded by a closed c w e given by parametric equation is

When the integral cannot be exactly evaluated, we can use the method of numerical integration. The two methods given here are: a) Trapezoidal rule: b

1 I f ( x ) d x = ( i y o + y , + y 2 +...+y,,-, +-y.) 2

where n is the number of sub-divisions of [a, b], and Ax the length of each sub-interval. b) Simpson's rule: b

h If(x)dx=-3 [ ( Y ~ + Y ~ " J + ~ ( Y ~ + Y , 2m-2)+4(yl + . . . + Y +Y,+...+Y~,~)~

when [a, b] is divided into 2m sub-intervals of length h.

2.5

SOLUTIONS AND ANSWERS

Appllcntions of Calculus

~ 2 )I ex dx = ex]: = e2 - e.

E 5) Points of intersection of 9 =4ax and y = mx are (0,O) and (4a/mz,4alm).

E6) For a loop, we should find two dwtinct, consecutive values of 0, for which we get the same value of r. If

r, = a sin 39,, and r2 = a sin 3 4 , r, = r2*sin38,-sin3e2=O.

or 2 cos 391 2+ 8,) sin

3(9,

- 9,) = O

Thus, 8, = 0 and 8, = d 3 will give the same value of r.

Area Under .'Curve

E8) Points of intersection are given by 6-23 = 4 or w s 28 = la,i-e., cos28= 314

Because of symmetry w.r.t. the initial line, the required area A = 2 (the area under the lemniscate above the initial line ftom 8 = 0 to 0 =d 6 and fiom 8 = 5 d 6 to 0 = x minus the areaunder the wide from 8 = 0 to 8 = d 6 and fiom 8 = S d 6 to 0 =x.

=8& +28%/3

E9) x = a(3 sin 8 - sin38),y = a c o 3 8

-6 .-34 .-21 .-n.2 (reduction formula)

2 5

= 12a

E 11) A loop of this curve lies between t = 0 and t = n

- 4a'

sin t cos2 t dt

Applications of' Calculus

E 12) a)

A r ( 2i X ~ + 2 . 5 + 2 f i + ~ + ~ + ~ + J 3 ) = 9.76.

E 13) a)

= -+-

1+4& 3

= 1.0665

= 0.7846

d 4 a 0 . 7 8 4 6 3 3.1384. ~~

Now E 14)

1.5

2.5

3.5

2.4

2.27

2.8

2.6

2.1

A --- 0.5(1+2.4+2.8+2.27+3+2.6+ 1.05) = 0.5 (5.12)=7.56 (trapezoidalmle).

121

-3 (7.4 + 16.8)= -3 (24.2)= (Simpson's rule). 3

E16) S =2(5+18+2'5+29+32+20+11+5+2) = WlA7\=7OA

UNIT 3

FURTHER APPLICATIONS OF INTEGRAL CALCULUS

Structure 3.1

Introduction Objectives

Length of a Plane Curve

3.2.1 Cartesian Form 3.2.2 Parametric Form 3.2.3 Polar Form

3.3 3.4 3.5 3.6

Volume of a Solid of Revolution Area of Surface ef Revolution

3.1

INTRODUCTION

Summary

Solutions and Answers

In the last unit we have seen how definite integrals can be used to calculate areas. In fact, this application of definite integrals is not surprising. Because, as we have seen earlier, the problem of finding areas was the motivation behind the definition of integrals. In this unit we shall see that the length of an arc of a curve, the volume of a cone and other solids of revolution, the area of a sphere and other surfaces of revolution, can all be expressed as definite integrals. This unit also brings us to the end of this course on calculus.

Objectives After reading this unit you should be able to : find the length of an arc of a given curve whose equation is expressed in either the Cartesian or parametric or polar forms, find the volumes of some solids of revolution, find the areas of some surfaces of revolution.

3.2

LENGTH OF A PLANE CLTRVE

In this section we shall see how definite integrals can be used to find the lengths of plane curves whose equations are given in the Cartesian, polar or parametric form. A curve whose length can be found is called a rectifiable curve and the process of finding the length of a curve is called rectification. You will see here that to find the length of an arc of a curve, we shall have to integrate an expression which involves not only the given function, but also its derivative. Therefore, to ensure the existence of the integral which determines the arc length, we make an assumption that the function defining the curve is derivable, and is deriative is also continuous on the interval of integration. Let's first consider a curve whose equation is given in the Cartesian form.

3.2.1

Cartesian Form

Let y = f(x) be defined on the interval [a, b]. We assume that f is derivable and its derivative f is continuous. Let us consider a partition Pnof [a, b], given by

The ordinates x = a and x = b determine the extent of the arc AB of the curve y = f(x) Fig. 1 (a)]. Let Mi = 1,2, .........., n - 1, be the points in which the lines x = xi meet the curve. Join the successive points A, M,, M,, M3, ........,M,,, B by straight line segments. Here we have approximated the given curve by a series of line segments.

Appllcrtlons o f Calculus

If we can fmd the length of each line segment, the total length of this series will give us an approximation to the length of the curve. But how do we fmd the length of any of these line segments?Take M2,M,, for example. Fig. I(b) shows an enlargement of the encircled portion in Fig. I(a). Looking at it we fmd that

where A x, = M2Qis the length (x, - $),and AY, =M3Qtflx3)- fl$) =y3-y2. In this way we can fmd the lengths of the chords AMI, MlM2,........,M,,B, and take their sum

Sngives an approximation to thelength of the arc AB. When the number of division points is increased indefinitely, and the length of each segment tends to zero, we obtain the length of the arc AB as

provided this limit exists.

Our assumptions that f is derivable on [a,b], and that f is continuous, permit us to apply the mean value themtm meom 3, Block 21.

Thus,there exists a point P,' (xi', y,') between the points Mi_,and Mi on the curve, when the tangent to the curve is parallel to the chord M,, Mi. That is,

Ayi = f (xi*) Ax,

Hence we can write (1) as

, 2 J--~~~

o+IO

i-l

This is nothing but the definite integral h "

J(I*ll'Odx. ']

Further Appllcationr of Integral Calculus

Therefore,

Remark 1: It is sometimes convenient to express x as a single valued function of y. In this case we interchange the roles of x and y, and get the length

where the limits of integration are with respect to y. Note that the length of an arc of a curve is invariant since it does not depend on the choice of coordinates, that is, on the frame of reference. Our assumption that f is continuous on [a, b] ensures that the integrals in (2) and (3) exist, and their value L: is the length of the curve y = f(x) between the ordinates x = a and x = b. The following example illustratesthe use of the formulas given by (2) and (3). Example 1 :Suppose we want to find the length of the arc of the curve y = k x intercepted by the ordinates x = 1 and x = 2. We have drawn the curve y = lnx in Fig. 2.

Flg. 2

Using (2), the required length L,2 is given by

dx t If we put 1 + x2= t2, we get -= - ,and therefore,

Applications of Calculus

We can also use (3) to solve this example. For this we write the equation y = lnx as x = ey. The limits x = 1 and x = 2, h n correspond to the limits y = 0 and y = 1112respectively. ,

Hence, using (3), we obtain

as we have seen earlier. This verifies our observation in Remark 1 that both (2) and (3) give us the same value of arc length.

Now, here are some exercises for you to solve.

E E I) Find the length of the line x = 3y between the points (3,l) and (6,2). Verify your answer by using the distance formula.

E E2) Find the length of the curve y = In sec x between x = 0 and x =d2.

E E3) Find the length of the arc of the catenary y =C cosh (x/c) measuredfkom the vertex (0, c) to any point (x, y) on the catenary.

E E4) Find the length of the semi-cubical parabola ay2= x3 fkom the vertex to the point

Farther A p p ! i e a W of Integral Caleu!us

Applications o f c a l c u ~ u s

- .

E ES) Show that the length of the arc of the parabola y2 = 4ax cut off by the line 3y = 8x isa(ln2 + 15116).

In the next sub-section we shall consider curves whose equations are expressed in the panmetric f o m

Parametric Form

3.2.2

Sometimesthe equation of a curve cannot be written either in the form y = f(x) or in the form x = g(y). A common example is a circle x2+ $ = a2.In such cases,we try to write the equation of the curve in the parametric form. For example, the above circle can be represented by the pair of equations x = a cos t, y = a sin t. Here, we shall derive a formula to find the length of a curve given by a pair of panmetric equations. Let x = Nt), y = ~ ( t )a, 5 t S f3 be the equation of a curve in parametric f o m As in the previous sub-section, we assume that the functions $ and q~ are both derivable and have continuous derivatives $' and y/ on the interval [a,PI. We have dx = dt

I$'

Hence,

d~ (t), and - = y' (t). dt dy

W"t) =0 , ,and

- [$' (t)12 + [w' (t)12

4' (t)

Now, using (3) we obtain the length

(we assume that $' (t) # 0).

The following example ihows that sometunes ~tis more convenient to express the equation of a given c w e in the parametric form in order to find its length.

Example 2: Let us find the whole length of the c w e

By substitution, you can easily check that x =a cos3t,y = b sin3t is the parametric form of the given c w e . The m e lies between the lines x = f a and y = f b since -1 5 cos t 5 1, and -1 5 sin t S 1. The curve is symmetrical about both the axes since its equation remains unchanged if we change the signs of x and y. The value t = 0 corresponds to the point (a, 0) and t = d 2 corresponds to the point (0, b). By applying the m e tracing methods discussed in Unit 9 we can draw this c w e (see Fig. 3).

Fig. 3

Since the c w e is symmetricalabout both axes, the total length of the c w e is four times its length in the first quadrant.

($1 ($1 )' +

= 9 sin%0x4 (a2C O S + ~ b2 sin%)

Hence, the length of the c w e is

L = 4

7 '7

3 sin t cos t Ja2 cos2t + b2 sin2t dt

- 12

sin t cos t Ja2 cos2 t + b2 sin2t dt

Putting u2 = a2cos%+ b2sin2t,we obtain 2u = (2b2- 2a2)sin t cos t

dt du

and the limits t = 0, t = d 2 correspond to u = a, u = b, respectively.

Further Applications of Integral Calculus

Appllcatlons af.Caleulus

Thus, we have

12 b3 - a3 b2-a2 3

= --=

4(a2 + b2 + ab) a+b

Now you can apply equation (4) to solve these exercises. \

E E 6) Fmd the length of the cycloid x=a(8-sine) ;y=a(l-cos8)

E E7) Show that the length of the arc of the w e

x=ets~t,y=e'costftomt=Opt=d2isf i (e*-1).

3.2.3

Pobr Form

In this sub-section we shall consider the case of a curve whose equation is given in the polar form, Let r = f(0) determine a curve as 8 varies from 9 = a to 8 = b, i.e.,-the function f is defined in the interval [a, @] (see Fig. 4). As before, we assume that the function f is derivable and its derivative f is continuous on [a,b]. This assumption ensures that the c w e represented by r = f(8) is rectifiable.

Furtbct Applieatfonr o f lntcgral Calcolur

Transforming the given equation into Cartesian coordinates by taking x = r cos 0, y = r sin 0, we obtain x = f(0) cos 0, y = f(0) sin 0. Now we proceed as in the case of parametric equations, and get,

Hence, the length of the arc of the curve r = q0) from 0 = a to 0 = $ is given by

changing the variable x to 8.

hf'(0) cos 0 - f (0) sin 01' + If' (8) sin 8 + f(0) ms 81' dB

We shall apply this formula to find the length of the curve in the following example. Example 3: To find the perimeter of the cardioid r = a (1 + cos 0) we note that the curve is symmetrical about the initial line (Fig. 5). Therefore its perimeter is double the length of the arc of the curve lying above the x-axis.

dr Now, - = - a sin 0. Hence, we have a0

Appllcatlanr af Calculus

We know

In this section we have derived and applied the formulas for finding the length of a curve when its eqwtion is given in either of the three forms: Cartesian, parametric or polar. Let us rmnrmarise our discussion in the following table.

Table 1: Length of an arc of a curve

Using this table you will be able to solve these exercises now.

E E 8) Find the length of the curve r = a cod (013).

E E9) Find the length of the circle of radius 2 which is given by the equations x=2cost+3,y=2sint+4,O~t~2x.

Further Appllc8tions of Integral Calculus

E E 10) Show that the arc of the upper half of the curve r = a (1 - cos 8) is bisected by 8 =2 d 3 .

3.3

VOLUME OF A SOLID OF REVOLUTION

Until now, in this course, we were concerned with only plane curves and regions. In this section we shall see how our knowledge of integration can be used to find the volume of certain solids. Look at the plane region in Fig. 6 (a). It is bounded by x = a, x = b, y = f(x) and the x-axis. If we rotate this plane region about the x-axis, we get a solid. See Fig. 6 (b). -- -

Such solids are called solids of nvoluhbn. Fig. 7 (a) and Fig. 7@) show two more examples of solids of revolution.

Fig. 7

The solid in Fig. 7(a) is obtained by revolving the region ABCO around the y-axis. The solid of revolution in Fig. 7(b) differs from the others in that its axis of rotation does not form a part of the boundary of the plane region PQRS which is rotated. We see many examples of solids of revolution in every day 1ife.The various kinds of pots made by a potter using his wheel are solids of revolution. See Fig. 8(a). Some objects manufactured . with the help of lathe machine are also solids of revolution. See Fig. 8 (b).

Now, let us try to find the volume of a solid of revolution. The method which we are going to use is d t e d the method of slicing. The reason for this will be clear in a few moments. LetTn= {a=x,,<x, <x,< ......<x,,+x,,= intervals.

Furtbw Appllcatloa~of Integral Calculus

b) bea~tionoftheinte~al[a,b]intonsub-

Fig. 9

Let A xi denote thc length of the ith sub-interval [xi_,, 31.Further, let P and Q be the points on the c w e , y = f(x) corresponding to the ordinates x = and x = xi, respectively. Then, as the c)we revolves about the x-axis, the sbaded strip PQNM (Fig. 9(a)) generates a disc of thickness A xi.In general, the ordinates PM and QN nay not be of equal length. Hence, the disc is ac ally the frustum of a cone with its volume A vi, lying between II PM2MN and n Q N 2 dm* is,between

If we assume that f is a continuous function on [a, b], we can apply the intermediate value theoran (Theorem 7, ~ l & k1, Plso see margin remark), and exprtss thisvolume as A vi =~ ~ { f ( tA, )xi, )~ where ti is a suitable point in the interval [xc,, xi]. Now summingup over all the discs, we obtain .

Y . =

[f(ti11' A xi,

A vi = i -1

%,Sti dXi a~ mapproximation

to the volume of the solid of revolution. As we have observed earlier while defming a d e f ~ t e integral, the approximation ge$s bet* as the partition P, gets finer and finer and Axi tends to zero. Thus. we get the volume of the solid of revolution as n

lirn Vn = lirn

v=,

0-m

n [f(ti )I' A xi i-1

We shall use this formula to fmd the volume of the solia described in the following example.

Example 4: Let us fmd the volume of the solid of revolution formed when the arc of the parabola y2 = 4ax between the ordinates x = 0, and x = a is revolved about its axis. The solid of revolution is the parabolic cap in Fig. 10. The volume V of the cap is given by

IrPM2. MN is the volume of the disc with radius PM and thickness MN. rrQNIMNis the volume of the disc with radius QN and thickness MN. Iff is continuous og [a, b], fla)=candflb)=d,andz l i e between c and d, then 3x, €14b[ S.t. Rx,,) = Z.

Applicatloos of Calculus

Our next example illustratesa slight modification of Fonnula (6) to find the volume of a solid obtained by revolving a plane region about the y-axis. x2 y2 Example 5: Suppose the ellipse 2 + = I. (a >b) is revolved about the minor axis. AB

(see Fw. 11). La us tind the volume of the solid generated.

In this case the axis of rotatian is the yaxis. The am revolved about the y-axis is shown by the shaded region in Fig. 11. You will agree that we need to consider only the area to the right of the y-axis.

RI. 11 To fiad the v o h of this solid we intacbangc x a d y in (6) and get

We can also modify Formula (6) to apply to curves whose equations are given in the parametric or polar forms. Let us tackle these one by one. PIvruneMeForm

If a c w e is given by x = Nt), y = ~ ( t ) a , S t 5 @, then the volume of the solid of revolution about the x-axis can be found by substituting x and y in Formula (6) by Nt) and ~ ( t ) , respectively. Thus,

We'll now derive the formula for curves given by polar equations.

PdarForm Suppose a curve is given by r = f(8), 8 , I 8S 8,. The volume of the solid generated by rotating the area bounded by x = a, x =b, the x-axis and r = Q0) about the x-axis is d 1(r sin el2 (r cos 9) do d0

= n

er = II

If(@ sin 01' If' (9) cis 0 - f(0) sin 01dB

Let's use this formula to find the volume of the solid generated by a cardioid about its initial line. Example 6: The cardioid shown in Fig. 12is given by r = a (1 + cos 0).

Fig. I 2

The points A and 0 cornspond to 0 = 0 and 0 = IC, respectively. Here,again, we need to consider ody the part of the cardioid above the initial line. Thus, 0

1n (r sin 8)' d0 (r cos 9) d0

1 + con

- na3

0)' sin3 0 (I + 2 sos 0) dB, since r = a (I + COS 0)

Further Applications of Integral Calculus

Appliertions of Calculus

= 256 na3

sin3 ( cos9 I$ d( - 64na3

64 xa3 15

= ----

'j?

sin3 I$ cos7 ( d$, where ( = 9 12

8xa3 on applying a reduction formula from Unit 12.

In all the examples that we Save seen till now, the axis of rotation formed a boundary of the region which was rotated. Now we take an example in which the axis touches the region at only one point.

Example 7: Let us find the volume of the solid generated by revolving the region bounded by the parabolas y = x2 and y2 = 8x about the x-axis. We have shown the area rotated and the solid in Fig. 13 (a) and.(b), respectively.

Fig. 13

Here, the required volume will be the differencebetween the volume of the solid generated by the parabola y2 = 8x and that of the solid generated by the parabola y = x2.

Note the limits of integration. Here, we list some exercises which you can solve by applying the formulas derived in this section.

E,E12) Find the volume of the right circular cone of height h and radius of the circular base r. (Hint: The cone will be generated by rotating the triangle bounded by the x-axis and the h e y = (rlh) x).

: t't'r $1,

5, -: .

:,*'I

Further Applicatiolls o f Integral Calculus

i ,' a

154

i'ij

r j,i

1..? ,.# i t:t ii'i :*!

:ST* .%.

1.i,!

it;; -,,

?&it E

> ~ ? .

Iij k.f f

ij/

i,:!l ,.'l

:Qi

It.!

::I.: s

ti >i :

i4;

i.i.1 [;I;

;:ej

X'S

'. I

NI,~

1:;

E E 13) Show that the volume of the solid generated by revolving the curve x* + yU3= a2I3about the x-axis is 32m3/105.

i,?,! f!:!2

. 4:;I

:I" I

!.ij 5

i.I.; \.! .?! ;;,i

i ,,I i: t ,1

ill{

?;:I

b' ? i i') :i i j

i:i!i

$1p;,,

! 21

;iij ::.I

iii r ii!

:-$

~. .:;I ,._I

i .'i ,>!,I

: $1

jj;l

[:-.1

[ iii

* i:i

;ii; .,.

I:/ i i:j i$4 ? .I1

;;!!

;:$!

I!::? ;!.I

i,$j rI $81 73 6

!:it!

E E 14) The arc of the cycloid x = a (t - sin t), y = a (1 - cos t) in [O,~R]is rotated about the y-axis. Find the volume generated. (Caution:The rotation is about the y-axis.)

Applicrtlons of C ~ l c u l u s

E E 15) Find the volume of the solid obtained by revolving the limacon r = a + b cos 8 about the initial line.

E E 16) The serpicirculariegion bounded by y - 2 = J9 - xZ and the line y = 2 is rotated about the x-axis. Find the volume of the solid generated.

3.4 AREA OF SURFACE OF REVOLUTION Instead of rotating a p l h e region, if we rotate a curve about an x-axis, we shall get a surface of revolution. In this section we shall fmd a formula for the area of such a surface. Let us start with the case when the equation of the curve is given in the Cartesian form.

Cartesian Form Suppose that the c w e y f(x) [Fig. 141 is rotated about the x-axis. To find the area of the area of the generated surfilce, we consider a partition P, of the interval [a, b], namely, P l l = { a = ~ < x , < x 2........ < <x,,<x,,=b)

I I

Fig. 14

Let the lines x = xi intersect the curve in points Mi, i = 1,2. ......n. If we revolve the chord Ma Mi about the x-axis, we shall get the surface of the frustum of a cone of thickness Ax: = x: xL,. Let As;be the area of the area of the surface of this frustum. Then the total surfwe area of all the frusta'is 8

This Snis an approximation to the area of the surface of revolution. The area of the surface of revolution generated by the curve y = fix), is the limit of Sn(if it exist), as n +wand each Ax; +0.

To find the area A of the curved surface of a typical frustum, we use the formula A = IC (r, + r,) I. where I is the slant height of the frustumand r, and r. are the radii of its bases (F;~.1;). In the frustum under consideration the radii of the bases are the ordinates fixo) and fixi), I

while the slant height Ma Miis give* by ,/(Axi12 +'(A Y , ) ~,where Ayi = fixi)- fixk,).

We assume that f is derivable on [a, b] and f is continuous. Then by the mean value theorem (Theorem 3, ~ 1 & 2), k we obtain A yi f (ti) xi, for some ti E [xi_,,xi]. Therefore, -

= 2n

where (fixi_,)+ fix,))/2is the mean radius of revolution

Jw A x,

\zJ Fig. 15

Exunple 8: Let us find the area of the surface of revolution obtained by revolving the parabola $ = 4ax h m x =a to x = 3%about the x-axis.

The area of the surface of revolution.

Instead of revolving the given curve abour the x-axis, if we revolve it about the y-axis, we get another surface of revolution. The area of the surface of revolution gmmted by the curve x = g(y), c 5 y 5 d, as it revolves about the y-axis is given by,

Now let us look at curve rtprtscnted by parametric tquations.

Parametric Form Suppose a curve is given by the parametric equations x = Mt), y = yr (t), t E [a, PI. Then we know that

Substitutingthis in formula (lo), we get the area of the surface of~evolutiongenerated by the curve as it revolves about the x-axis, to be

Now we shall state the formula for the surface generated by a curve represented by a polar equation.

PoIar Form If I = h(8) is the polar equation of the curve, then the area of the surface of revolution generated by the arc of the curve for 8, 5 0 4 B2,as it revolves about the initial line, is

Study the following examples carefully before trying the exercises given at the end of this section. Example 9: Suppose the astroid x = a sin3t, y = a co$t, is revolved about the x-axis. Let us find the area of the surface of revolution. You will agree that we need to consider only the portion of the curve above the x-axis. For this portion y > 0, and thus t varies from - d 2 to.d2. dy

= 3a sin2tcos t, - = - 3a cos2t sin t dt dt

($1' (z) 2

Therefore,

+ i

=9 '.

sin%cos2t

We therefore get, nl2

-7 s = 2 1 J'acos3t J9a- T sm T tcos t d t-

nl2

= 271

a cos3 t / 3a sin t cos t 1 dt

-nlZ

= 6na2

c0s4 t I sin t ( dt

-zl2

Example 10: Suppose we want to find the area of the surface generated by revolving the cardioid r = a (1 + cos 8) about its initial line. Notice that the cardioid is symmetrical about the initial line, and extends above this line from 8 = 0 to 8 = n.The surface generatedby revolving the whole w e about the initial line is the same as that generated by the upper half of the curve. Hence,

Further Applicatio~~s of Integral Calc-ulus

Applications ~ ( C a l c u l u i

dr Simer=a(l +cosO),and - =-asin€), wehave de

f + ($J2

0 =a2(l + c o s 0 ~ ~ + a ~ s i n ~ 0 = 4 a ~ ~ ~ 2

Therefore,

- 4na2 J 4 sin - ms4 Bd0 2

- 32xa2

sin ( ma4 ( d(, *hie ( = 8 1 2 0

E- E 17) Find the area of the surface generated by revolving the circle r = a aboutthe x-axis, and thus verify that the surface area of a sphere of radius a is 4 d .

E E 18) The arc of the curve y = sin x, h m x =0 to x =n is revolved about the xlaxis. Find the area of the surface of the solid of revolution generated.

E E 19) The ellipse x21a2+ y'/b2= 1 revolves about the x-axis. Find the area of the surface of the solid of revolution thus obtained.

E E20) Prove that the surface of the solid generated by the rwolution about the x-axis of the ~ q o f t h e c u r v e x = ~y, =

(t -g)

is3n.

E E2-1)Find the surface area of the solid generated by revolving the cycloid x = a(8 - sin 8), '

y = a(1- cos 8), about the line y = 0.

Furtber Applientions of Integral Cnleulus

Appllcmtions of Cmkulus

Now let us quickly recall what we have covekd ih this unit.

In this unit we have seen how to find

the lengths of curves

volumes of solids of revolution and

the areas of surfaces of revolution.

In each case we have derived formulas when the equation of the curve is given in either the Cartesian or parametric or polar form. w e give the results here in the form of the following tables.

Length of an arc of a curve

Volume of the solid of revolution

w=w

Volume

x = g(y) about y - axis

x = +(t), Y = ~ ( t ) about x - axis

r = h(0)

about the initial line

1 dx

Y = f(x) about x -axis

a d

J x2 dx

1 1

[ul(t)12 +* (t) dl

[h(B) sin 01' [hl(B) cos 6 - h(6) sin 01 dB

Area ofthe surface of revolution

E<pm-

Arpa

Y = f(x) abut x -axis

x = g(y) about y - axis

c /

= Mt). y = W(t) about x -axis

r = h(0) about the initial line .

3.6

J W (0 ,/[v (t)lZ+ [w' (t)lZ dl

SOLUTIONS AND ANSWERS

By distanceformula,

secx13+tanx/3 sec 0 + tan 0

Further Applicatloar ef Integral Calculur

Appllcatloas of Calculus

H-

(0,O) and

9a 3a (16 T ) are the points of intersection

Further Appllcrtlons of Integral Calculus

= 2a2(1 cos 0) = 4a2s& (812)

- 8a

E9)

sin + d + = 8 a

dy = 2 cos t x=-2 sin t dt dx

.: L = 2 0 dt = 4x Note that L = 2m since, here, r = 2.

:. /r2 +

(:I2

= 2a sin -

The length of the curve in the upper half

2a sin

The length from 8 = 0 to 6 = 2 d 3

The arc of the curve in the upper half is bisected by 8= 2d3.

dr Ell) r=a(@- l), d0 -2a8

Note &t the total volume generated is equal to twice the volume generated by the arc of the curve between x = 0 and x = a.

.: v

n2] a2 (I-sintl2 asin tdt 0

(t2 sin t - 21 sin2 t sin3 t) dt

na3 0

dx E15) r = a + b c o s e 3 - =

d(r cos 8)

(The other two integrals are equal to zero since cos (x - 8) = - cos 0.)

(using a reduction formula)

2na2 f, sine de 0

=' 4na2 cos 8

":I

Further Applications of Integral Calculus

Applications 01 Calculus

dy E18) y = sinxa-=cosx

:. S E 2x

*. J l+cos x d x smx

E20) l%e loop is between t =

- & and J5 .Becauseofsymmetry, is enoughto t=

considerthccwebetwant=Oandt= &.

El) S = 271

a (I - cos 0) ,/a2 (I - EOS

= 4na2

0 (I - cos 0) sin -dB 2 0

+ a2 sin2 0 dB

Further Applications o f Integral Calculus

UNIT 1 SETS Structure 1.1 Introduction Objectives

- II

1.2 1.3

Sets Subsets

1.4

Venn Diagrams Operations on Sets

1.5

1.5.1 Complementation 1.5.2 Intersection 1.5.3 Union

, I

1.6

1.6.1 Distributive Laws

. , I

Laws Relating Operations 1.6.2 De Morgan's Laws

1.7 1.8

1.1 INTRODUCTION

)

Consider the collection of words that are defined in a given dictionary. A word either belongs to this collection or not, depending on whether it is listed in the dictionary or not. This collection is an example of a set, as you will see in Section 1.2. When you start studying any part of mathematics, you will come into contact with one or more sets. Thu is why we want to spend some time in discussing some basic concepts and properties concerning sets.

Cartesian Product surmnar~

In this unit we will introduce you to various examples of sets. Then we will discuss some operations on sets. We will also introduce you to Venn diagrams, a pictorial way of describing sets. I

As mentioned earlier, a knowledge of the material covered in this unit is necessary for studying any mathematics course. So please study this unit carefully.

And now we will list the objectives of this unit. After going through tho unit, please read this list again and make sure that you have achieved the objectives.

Objectives I

After reading this unit you should be able to :

iI t

I L

! I 1

identify a set; represent sets by the listing method, property method and Venn diagrams; perfonn the operations of complementation, union and intersections on sets; prove and apply the distributive laws; prove and apply De Morgan's laws; obtain the Cartesian product of two or more sets.

Solutions of Polynomlrl Equrnonm

1.2 SET You may have often come across categories, classes or collections of objects. In mathematics, a welldefined collection of objects is called a set. The adjective 'well-defined' means that given an object it should be possible to decide whether it belongs to the collection or not. For example, the collection of all women pilots is a set. This is because any person is either a woman pilot or not, and accordingly s/he does or does not belong to the collection. On the other hand, the collection of all intelligent human beings is not a set. Why? Because, a particular human being may seem intelligent to one person and not to another. In other words, there is no clear criterionfor deciding on who is intelligent and who is not. So, the collection is not well-defined. Now we give some more examplesof sets which you may havecome across. We will also use them a great deal in this course .

3 The set of natural numbers, denoted by N. ii)

The set of intergas, denoted by Z.

iiii The set of rational numbers, denoted by Q.

iv) The set of real numbers, denoted by R In the next unit we will be studying another set, namely, the set of complex numbers, denoted by C. .Youmay like to try this exercise now. El) Which of the collections mentioned below are sets? a) The collection of all good people in India. A prlme number is a natural

b) The collection of all those people who have been to Mars.

number other than one, whose only factom am one and itself.

c) The collectionof prime numbers. d) The collection of even numbers. e) The collection of all rectangles that are not squares. An object that belongs to a set is called an element or member of the set. For example, 2 is an element of the set of natural numbers. We normally use capital lettrm A, 8, C, etc., to donote sets. The small letters a, b, c, x, y, etc, are usually used to denote elemeno of sees. We ttymboliicrlly write the statement 'a Is an elemant of the set A' as a e A.

The rymbol a rundr for 'belongs to'. It was suggerted by the Italian mathematician Peano (1858-1 932)

If a is not at1 element of A or equivalently, a does not belong to A, wc write is a e A. So, for example, if A is the set of prime numbers, then 5 E A and 12 e l A.

Try the following exercise now. )

Which of the following statements are true? a) 0 . 2 ~ N b) 2 e N

e R

Any circle is a member of the set in E 1 (e).

Now, yo~uknow that a number is either rational or irrational, but not i )oh. So what will be the set of all1 numbers that are both rational as well as irrational? It will nc at have any element.

The set which has no elements is called the empty set (or the void set, or the null set). It is denoted by the Greek letter @ (phi). A set, which has at least ode element is d l c d 8 iwn-empty sat. We usually describe a non-empty set in two ways-the listing methodand the pmprty metbod. In the first method we list all the members of the set within curly brackets. For instance, the set of all natural numbers that are facton of 10 bit drnoGd by { 1,2,5,10) Por PI tO br able to write them all down? In this But what about the set that ha4 too inmy tQ axhibit m e pattern which its elements case we list some of the elemento of t b wt, follow, followed by three dats. For exsmple, tbsetN ofdalral numbers can be described as

and the set of all even numbers lying between 10 and 100 is

This method of representing sets is oallsd tlw Ilrtiaa method (or tabular method, or roster method).

In the second method of describing a set we describe its elements by means of a property possessed by all of them. As an example, consider the set S of all natural numbers which are multiple of 5. This set S can be written in the form

The vertical bar after x denotes 'such that'. (Some authors use ':' h t e a d of '1' to denote 'such that1)So (1) states that S is the set of all x such that x is a natural number and x is a multiple of 5. We can also write this in a slightly shorter form as

S = {5n)n E N). This method of describing the set is called the property method or the set - builder method.

In some case we can use either method to describe the set under consideration. For instance, the set E, of all natural numbers less than 10, can be d e s c r i i as

E = {x I x is a natural number less than 10) (by tfte propsrtymethod).

You can see that both these suta are the rmn, bbgth of thcm have precisely the same elements. This exempk leads us to th& following dtffrritisn. Deflnitian :Two sets S and T are d k d cqrwl,(iakotdby S = T, iff every element of S is an element of T and every element of T is an elrmeat of S. Now, while describing a set by the listing method, you must keep the following important remarks in mind. Remark 1:The set {1,2,3,3) is the same as {1,2,3). That is, while listing the elements in a set we do not gain anything by repeating them. By convention we do not repeat them. Remark 2 :Consider the sets {1,2,3) and {2,1,3). Are they equal or not? You can see that evey element of the first set belongs to the second, and vice versa. Therefore, these sets are equal. This example shows that changing the order in which the elements are listed daes not alter the set. We would also like to emphasise an observation about the property method. Remark 3: There can be severalproperties that define the same set. For example,

= {x

I x is an even prime number).

You may like to do the followirtg exe-

maw.

Solutions 01 ~olynomi:~I Equations

E3)

Describe the following sets by the listing method. a)

{x I x is the smallest prime member )

{x I x is a divisor of 12}

the set of all letters in the English alphabet.

FYI) Describe the following sets by the property method.

E5)

Give an example of a non-empty set which can be represented only by the property method.

While solving E3 you would have come across sets consisting of exactly one element. Such a set is called a singleton. The singleton with element x is usually written as (x). Remark 4: The element x is not the same as the set {x). In fact x E (x}

A set which has a finite number of elements is called a finite set. By convention, the empty set is considered to be a finite set. A set which is not finite is an infinite set. The selutlon set o f an equation is the set of solutions of the equation.

Some examples of infinite sets are N, Q, R and the set of points on a given line. The followinq exercise will help you in getting used to the notion of finite and infinite sets. E6) Which of the following sets are finite, and which are infinite ? a) Z b)

the solution set of 2x + 5 = 7

the set of points on the circumference of a circle

Now, given any two real numbers a and b, you know that either a < b or a = b or a > b. Is there a similar relationship between sets? Let us see.

In this section we shall see what we mean by the terms 'is contained in' and 'contains'. Consider two sets A and B, where A = the set of all students of IGNOU, and

B = the set of all female students of IGNOU. Every female student of IGNOU is a student of IGNOU. So, each element of B is also an element of A. In such a situation we say that B is contained in A. Of course, IGNOU also has some male students! So, there is an element x in A such that x does not belong to B. Mathematically, we write this as

sets

In this situation we say that B is properly contained in A. In gcneral, we have the following definitions.

3 'denotes' there ex~sts'.For a

Definitions :A set A is a subset of a set B if every element ofA belongs to B,and we denote this fact by A sB.

~ ~ on this, ~ ~see the , appendix to this block. J

In this situation, we also say that A is contained In B, or that B contains A, denoted by B a A. If A r;B and 3 y E B such that y (E A, then we say that A is a proper subset of B ( or A is preperly contained in B). We denote this by A sB. If X and Y are two sets such that X has an element x which does not belong to Y then we say that X is not contained in Y. We denote this fact by X Y. Let us look at a few examples of what we have just defined. Consider the set A = { 1,2,3). Is A G A? Since every element of A is in A, we find that A c_ A. In fact, this is true for any set. In other words, any set is a subset of itself. But note that no set is a proper subset of itself.

Wefindthat3(- 1) E A suchthat (- 1 ) B.~

:. A

Similarly, B P A. Note that given any two sets A and B, one and only one of the following possibilities is true.

Using this fact we can prove by contradiction (see the appendix to this block) that the empty set $ is a subset of every set.

Z; A for any set A.

To prove this consider any set A. Supose I$ A. Then there must be some element in I$ which is not in A. But this is not possible since I$ has no elements. Thus, we reach a contradiction. Therefore, what we assumed must be wrong. That is, I$ A is false. Thus I$ G A, for any set A.

I .

Try the following exercises now. While doing them remember that to show that A E B, for any sets A and B, you must show that if a E A then a E B,

Also, to show that A g B, you must show that 3 x E A such that x s~ B. E7) Write down all the subset {1,2,3). How many of these contain a) no element, h) one element, c) two elements, d) three elements? E 8)

Show that if A r;B and B s C, then A Z; C. This shows that 'z;' is a transitive relation.

E9) Show that ' SZ ' is not a transitive relation. For this, you need to find an example of three setsA,BandCsuchthatAcB, B a C , b u t A ~ ; C . Now, let us go back for a moment to the point before Remark 1, where we defined equality of sets. Let us see what equality means in terms of subsets. consider the sets

A = the set of even natural numbers less than 10, and

They are equal since every member of A is a member of B, and vice versa. That is, A s B and B sA.

Now, for some interesting exercises!

' -' denotes 'implies' (see the block appendix also).

The set of all subsets of a set A is the Power set 'of A.

Solutlonc of Polynomlrl Equrtlonc

El0

Consider the sets A = { x 1 x + 1 = 3) B = {1,2 ) and C = the set of all even numbers that are prime. What is the relationship betwren i) A andB?

ii)

A and C?

Ell) If A = B and B -

C, what is the relationship between A and C? -

So far you have seen two methods of representink sets. There is yet another way of depicting sets and the relationships between them. This is what we discuss in the next section.

1.4 VENN DIAGRAMS It is often easier to understand a situation if we can represent it graphically. To ease our understanding of many situations involving sets and their relationships, we represent them by simple diagrams, called Venn diagrams. An English logician, John Venn (1834 - 1923),invented them. To be able to draw a Venn diagram, you would need to know what a universal set is.

A universal set is not unlque

In any situation involving two or more sets, we first look for a convenient large set which contains all the sets under discussion. We call this large set a universal set and denote it by U. Clearly, U is not unique. For example, if we are talking about the set D of women directors, and the set S of women scientists, then we can take our universal set U to be the set of all earning women. This is because U contains D as well as S. But, we can also take U to be the set of all women. This would also serve the same purpose. Again, if we wish to work with the sets of integers and rational numbers, we could take the set of real numbers as our universal set. We could also take Q as our universal set, since it contains both Z and Q. We usually take our universal set to be just large enough to contain all the sets under consideration. Now, let us see how to draw a Venn diagram. Suppose we are discussing various sets A,B,C,..... We choose our universal set U. so, A sU, B sU, C sU, and so on. We show this situation in a Venn diagram as follows : The interior of a rectangle represents U. The subsets A, B, C etc., are represented by the interiors of closed regions lying completely within the rectangle. These regions may be in the form of a circle, ellipse or any other shape. To clarify what we have just said, consider the following example. Example 1: Draw a Venn diagram to represe3t the sets

Solutlon: See Figure 1.

Fig. 1 :Ven diagram

We have denoted A by a circle, B by an ellipse which intersects A, and C as another closed region. The points 8,9 and 10 dont't lie in any of A, B or C. All these regions and points lie in the universal set U, which is represented by the outer rectangle. Note that 3 belongs to both A and B. Therefore, it lies in the circle as well as the ellipse. Also note that A and C do not have any elements in common. Therefore, the regions representing them do not cut each other. For the same reason tho regions representing B and C do not cut each other. Of course, we could have drawn B and C in the shape of circles also. Now, consider the following situation : A and B are two sets such that A 5 B, that is, A is a proper subset of B. What will a Venn diagram corresponding to this situation look like? Well, we can just take B to be our universal set. Then the Venn diagram in Figure 2 is one such diagram. If we take another set U that properly contains B, as our universal set, then we get the Venn diagram in figure 3.

Fig. 2

Try this exercise now.

E12) How would you represent the following situation by a Venn diagram? The set of all rectangles, the set of all squares and the set of all parallelograms.

Fig. 3

Now that you are familiar which Venn diagrams, let us discuss the various operations on sets. During the discussion we will be using Venn diagrams off and on.

You must be familiar with the basic operations on real numbers - addition, subtraction, multiplicationand division. In using these we combine two real numbers at a time, in different ways, to obtain another real number. In a similar way we can obtain new sets by applying certain operations to two given sets at a time. In this section we shall discuss the operations of complementation, intersection and union.

1.5.1

Complementation

Consider the sets N and {0,1) ,There are elements of N that do not belong to {O,1) like 2,3, etc. The set of these elements is the complement of {O,1) in N, according to the following

definition.

Deflnltion : Let A and B be two sets. The complement of A In B, denoted B \ A, is the set { X E BIxe A), Similuly,A\B-{%a A l x t B). If B h the universal set U, the B \A is U \A. This set is called the complement of the set A and is denoted by A' or A'. The unshaded area in Figure 2 denotes the set B \A (or A', since B = U in this case). This diagram shows us that x E A0 if and only if x e! A. Note that in the situation of Figure 2, A\B = $, since every point of A is a point of B. Try this exercise now.

E 13) a) Represent the following sets in a Venn diagram : The set P of all prime numbers, the set Z and the set Q E. b) Is the set Z \ P finite or infinite? E 14) Let A be any set. What will A \ A,

\, -4,A \

9 and

be?

Let us now comider another operation on sets, namely, the ~ntcrsectionof two or more sets.

(A \ B) C A

Solutions of Polynomial Equations

1.5.2 Intersection Let A and B be two subsets of a universal set U. the intersection of A and B will be the set of points that belong to both A and B. This is denoted by A n B. T h u s , A n B = { x l U J x e A a n d x ~B). To clarify this idea consider the following expple.

Example 2 :Let S be the set of prime numbers and T be the set {x l Z} I x divides 10). I

What is S nT?

Solution :We take Z to be our universal set. Then S nT = set of those integers that are prime numbers as well as divisors of 10 = { 2 , 5 } .Note, 1 is not a prime number. You should be able to do the following exercise now.

While solving E 15 you may have noticed certain facts about the operation of intersection. We explicitly list them in the following theorem.

Theorem 1 : For any two sets A and B,

(f) says that the operation of intersection of sets is commutative'.

AnBsA

AsB*AnB=A

AnA=A

IfCiaasetsuchthatC~AandC~BthenCsAnB.

Proof :We will prove (a) and (b): and leave you to check the rest (see E 16). S o l e t x ~A n B . T h e x e A a n d x e B.Thus,AnB~;AandAnBsB..Sowehaveproved(a) and (b). Now, using (a) and (b), try to do the following exercise.

E 16) Prove (c) - (h) or Theorem 1. E 17) Is Theorem 1 (h) true if we replace 'E'by ' g ' everywhere? Why ? Now, consider the set Q- of negative rational numbers and the set Q+of of positive rational numbers. The Q - n Q+=$. This shows that Q- and Q+are disjoint sets, a term we now define.

Definition :Let A and B be two sets such that A n B = $. Then we say that they are mutually disjoint (or disjoint). Now let us represent the intersection of sets by means of Venn diagrams. The shaded region in Figure 4 represefits the set A n C, which is non-empty, as you can see. Also note that the regions representing A and B do not overlap, that is A nB = 9.From this diagram we can also see that neither is A G C, nor is C G A. Further, both C \A and A \ C are non-empty sets. See how much information a Venn diagram can convey!

Fig. 4 C

Now, go back to Figure 2 for a moment, What situation does it represent? It shows two sets A and B, with A 5 B, that is, A is a proper subset of B. Then the shaded area shows A n B = A. Try the following exercise now.

E 18) Le U = Z, A = {1,3,5) B =the set of odd integers. Draw the Venn diagram to represent this situation, and shade the portion A n B So far we have considered the intersection of two sets. Now let us define the intersection of 3, 4 or more sets.

Definition :The intersection of n sets A,, A,, ..., A,, is defined to be the set { X ) X Aiforeveryi= E 1, ....n). n

This is denoted by A, n AAt n .... n A, or n A,. i-1 Let us look at an example involving the intersection of 3 sets.

Example 3 :Let A, B and C be the sets of multiples of 3, 6 and 10 in N, respectively. Obtain AnBnC.

Solution :A n B n C will consist of all those natural numbers that belong to A. R and (3. Thus, 3

A n B n C = { X EN13,6and 10dividex)

= {xE N ( 30 divides x) = { 3 0 n I n N). ~

Note that 30 is the lowest common multiple (lo,,) of 3,6 and 10 Try the following exercise now.

E19) LetA={ 1,2,3,4),B={3,4,5,6)andC={1,4,7,8). Determine A n B B C. Also verify that

What you have shou I L In E 19 is i-, ,t oiily tnie 101. those sets. It s true for any tluec sets A, B and C. This property of r , i i called associativity. Using this propcrty we can obtain the intersection of any n sets by intersecting any two adjaceilt sets at a time.

Solutlonr of PolynomlrI Equrtloris

For example, if A, B, C, D are 4 sets, then A n B n C n D=[(AnB)nC]nD =An[(BnC)nD] =(AnB)n(CnD). Let us now look at another operation on sets.

1.5.3 Union

'5' denotes 'less than or equal to' and '2'denotes 'greater than u equal to'

SupposewehavetwosetsA={x~R ) x S 1 0 )a n d B = { x ~R I x 2 lO).ThenanyelementofR belongs to either A or B, because any real number will be either less than or equal to 10or greater than or equal to 10; and 10 will belong to both A and B. In this case we will say that R is the union of A and B.

In general, we have the following definition.

Definition :Let A and B be two sets. The set of all those elements which belong either to A or to B or to both A and B is called the union of A and B. It is symbolically written as A u B T h u s A u B - { x l x e A o r x e B). Before going fbrther we make a remark. Remark 5 :Since A u B contains all the element8 of A as well as B, it follows that AsAuB,BsAuB.

Infact,AnBsAsAuBandAnB~B~AuB. UsingE8, thisslkows(trPtAnBsAu811. Now let us look at an exmqk. Example 4 :Find N u Z Solution :N ={1,2,3,"...)and2 = {

....-3,-2,-1,0,1,2,3 ,....).

We want to fmd

NUZ={xlx E N o r x e Z)

B u t N ~ Z . T h u s x ~ N ~ x c Z . T h i r ~ l y ~ l b u s t hZI r t{xIxEZ) Nu -2. '4 denotes 'equivalence' or 'implies' and is 'implied by' (we appendix to block).

~ x ~ l e 4 i r a ~ m k * r s u o o f ~ ~ B-B. ~ ~ ~ ~ ~ ~ ~ s i ) o ~ Your can use thig fact while rolving the following exercire. E21) For any three sets A, B and C, show that a) A u A = A , b)

A u B = B u A that is, the operation of union is commutative

Au$=A.

i f A ~ C a n d B s C b n Au B sC.

ED) LetUbethedlineR,A={x E R l 0 S x S l ) d B- {x E RI 1 g x 1 1 ; 3 ) . ~ A u B . E23) What can you say a b u t A and B if A u B ==$ ?

iGwt8idcaEilyrt 5. In this d i a p we see foursetsA,B,C~ndD,~~uni~~~'U.'Rasdrrdtd~repre~~lltoAuB.

It is easy t~ vilsustise unitma dm@I@ Y&

C u D is the area enclosed by both C and D, which is just D, since C c D. Can you find the area

in Figure 5 that represents A u B u D ? You may be able to, once we have defined the union of 3 or more sets.

Fig. 5 Definiton :The union of n sets A,, A,....., A,, is the set {x Ix E A,forsomeisuchthat 1s i I n ) ,

This is denoted by A, u A2 u

..... u A or 1=1 .u Ai ,

So now you can see that A u B u D is represented in Figure 5 by the shaded area along with the area enclosed by D. Now, let us considerA u B u C, where A = {1,2,3), B = {2,3,4,5), C = (1). Then A u B u C = {1,2,3,4,5). You will find thatthis is the same as (A u B) u C as well as A u (B u C). You may also like to verlfy that

These statements are particular cases of the general facts that we ask you to prove in the following exercise. E24) a) For the sets A, B and C in Example 3, show that A u B u C = ( A u B ) u C = A u (BuC) b)

For any two sets A and B, show that A u B = (A \ B) u (A n B) u (B \ A). (We depict this situation in the Venn diagram in Fig. 6.)

What you have shown in E 24(a) is tme for any three sets, that is, the operation of union of sets is associative. Consequently, we can obtain the union of any number of sets by taking the union of any two adjacent ones at a time. For example, if A, B, C, D are four sets, then A u B u C u D = [(AuB) u C ] u D = A u [BuC)uD] = ( A u B ) u (CUD). By now you must be familiar with the operations that we have discussed in this section. Now let us try and prove some laws that relate them.

1.6

LAWS RELATING OPERATIONS

In this section we shall discuss two sets of laws that relate unions with intersections, Cartesian products with unions and Cartesian products with intersections. We start with the distributive laws. 4-9

Soludonc of Polynomlrl Equation1

1.6.1 Distributive Laws You muit be familiar with the law of distributivity that connects the operations of multiplication and addition of real numbers. It is

Whereas we have only one law for numbers, we have two laws of distributivity for sets, which we will now state.

he or em 2: Let A, B and C be three sets. Then a) A n ( B u C ) = ( A n B ) u ( A n C) b) A i l ( B n C ) = ( A u B ) n ( A u C)

Proof: We will prove (a) and ask you to prove (b) (see E 25). a)

We will show that

A n ( B u C ) s ( A n B ) u (A n C)and ( A n B) u ( A n C) s A n ( B u C ) .

a x e A a n d x ~B u C -

~ ( XAEa n d x ~B) or (XEAand X E C) -XE ~

A n B o r x ~A n C X ( AE n B ) u ( A n C )

So, we have proved the first inclusion, i.e., (2). To prove (3), let X E ( A n B ) u (AnC).

a x e A n B orxe A n C ~ ( X A E andx~ B)or ( X E A a n d x ~C )

a x €Aand(x~Borx~C) a x € A a n d x ~B u C x€An(BuC) So we have proved (3). Note that our argument for proving (3) is just the reverse of our argument for proving (2). In fact, we could have combined the proofs of (2) and (3) as follows :

W X E Aand X E

Bu C

a x e Aand ( X E Bor X E C) W ( X EAand X E B)or ( X EA a n d x ~C) a x e AqBor o r A n C W X E ( A n B ) n (AnC)

This proves (a). Now try to solve the following exercise, using a two-way implication.

E25) Prove (b) of Theorem 2.

Let us verify Theorem 2 in the following situation.

Example 5 :Verify the distributive laws for the sets N, Q and R in place ofA, B and C. Solution :We first show that N n (Q u R) = (N n Q) u (N n R) NowQuR=R,sinceQ~R Therefore, N n (Q u R) = (N n R) = N, since N E;R AlsoNnQ=NandNnR=N

Now, to verify that N u ( Q n R) = ( N u Q) n (Nu R), note that both sides are equal to Q. Hence the law holds. Remark 6 :Theorem 2 (a) says that n distributesover u , and Theorem 2 (b) says that u distributes over n .

Let us now consider another set of laws.

1.6.2 De Morgan's Laws We will now discuss two laws that relate the operation of finding the complement of a set to that of the intersection or union of sets. These are known as De Morgan's laws, after the British mathematicianAugustus De Morgan (1806-1871)

Theorem 3 :For any two sets A and B in a universal set U,

Proof :As in the case of Theorem 2, we will prove (a), and ask you to prove (b).

a) Letx e ( A n B y = U \ ( A n B ) e xeAnB e x e Aorx z B(because,if x E Aandx E B,thenx E A n B )

So (A n B)C= ACu BC. Now try the following exercises.

E 26) Prove (b) of Theorem 3. E 27) Verify De Morgan's laws for A and B, where A = {I, 21, B = {2,3,4). (You can take U = {I, 2,3,4), i.e., U = A u B. Of course, the laws will continue to hold true with any other U.) -

So far we have discussed operations on sets and their inter-relationship. We will now talk of a product of sets, of which the coordinate system is a special case.

1.7

CARTESIAN PRODUCT

An interesting set that can be formed from two given sets is their Cartesian product, named after the French philosopher and mathematician Rene Descartes (1596- 1650).He also invented the Cartesian coordinate system. Let us see what this product is.

F i g . 7: Rene Descartes

Let A and B be two sets. Consider the pair (a,b), in which the first element is from A and the second element is from B. Then (a,b) is called an ordered pair. In an ordered pair, the ord& in which the two elements are written is important. Thus, (ib)and @, a) are different ordered pairs. We call two ordered pairs (a, b) and (c, d) equal (or the same) if a = c and b = d. I

Using ordered pairs, we give the following definition. Definition: The Cartesian product A x B, of the sets A and B, is the set of all possible ordered pairs (a, b) where a E A, b E B. Thisis,AxB =(a,b) l a c X a n d b ~B). For example, if A = {1,2,3),B = {4,6), then

Also note that B x A = {(4,1), (4, 2), (4,3), (6,1),(6, 2), (6, 3),). You can see that (l,4) E A x B but (l,4) e B x A Therefore, A x B # B x A

Try these exercises now.

E 29) If A x B = {(7,2),(7,3), (7,4), (2,2), (2,3), (2,4),) determine A and B.

Now that we have defined the Cartesian product of two sets, let us extend the definition to any number of sets. Q denotes 'for every'

Definition :Let A,, %,....A,, be sets. Then their Cartesian product is the set Al,xAz,X .....X A , = { ( x , , ~....... , x,,)(x~,EA,, V i k l , 2,...n). For example, if R is the set of real numbers, then R x R = {(a, ,a, ) I a, E R, a, E R),

It is customary to write R~for R x R and Rnfor R x ...........x R (n times). In your earlier mathematical studies you must have often used the fact that R can be geometrically represented by a line. Are you also familiar with a geometrical representation ofRxR? You know that every point in a plane has two coordinates,x and y, and every ordered pair (x,y) of real numbers defines the coordinates of a point in the plane. Thus, R2 is the Cartesian product of the x-axis and the y-axis and hence, R2 represents a plane. In the same way R3 represents three-demensional space, and Rn represents n-dimensional space, for any n 2 1.

Try the following exercise now. -

E30) Which of the following belong to the Cartesian product Q x Z x N ? Why? a) (3.01, b)

1 1 1

(?, 2 ?), c) (l,1,1),

(1 -5

, e) I-2,2, 31.

E31) Give an example of a pro3er non-empty subset of R x R. E32) Prove that for any 3 sets A, B and C, Q x dism%utesover u as wall as over n.

a) A x ( B u C ) = ( A x B ) u ( A x C ) b) A x ( B n C ) = (AxB) n (AxC) --

We will end our discussion on sets here. Let us summarise what we have covered in this unit.

1.8 SUMMARY In our discussion on sets we have brought out the following points.

1) A set is a well defined collection of objects. 2) The listing method and property method for representing sets. 3) Given two sets A and B, what we mean by AG B, A 2 B and A= B. 4)

The pictorial representation of sets and their relationships by Venn diagrams, and its utility.

The operations of complementation, intersection and union of sets, and their properties.

The distributive laws : For any three sets, A, B and C, A u ( B n C) = ( A u B ) n ( A n C) A n ( B u C) = ( A n B ) u ( A n C)

De Morgan's laws: For any two sets A and B, ( A n B)C= ACu BE (A u B)c = ACn BC.

The Cartesian product of sets.

Now, we suggest that you go back to the objectives given in sec. 1.1, and see if you have achieved them. One way of checking this is to solve all the exercises in the wit. If you would like to know what our solutions are, we have given them in the next section.

1.9 SOLUTIONSIANSWERS E 1) (b)l (~)l(d)l(e) E 2) (c), (e) E3) a) {2) b) {l,2,3,4,6, 12) c) (29-2) d) (8) e) {a,b,c,.......,x , y , z )

E4) a) {xZ) x E Z) b) {x I x is a prime number ) c) {x I x is an even integer) d) We can have several representations (see Remark 2). For example, I$ = { x E N ~xisbothoddaswellaseven),or t$

= {x E N I X <O).

E5) R E6) (b) and (c), are finite. (a) and (d) are infinite. E7) { l l , {3), {1,2}1{1,3},{2,3} {112,3). If a finite set A has n elements, then its power set has 2" elements. 4 9

E8) Let x E A. Then A c B 3x E B. And then B G C ~ X E C :.XE

A3x EC

AGC

Sets

Solutions of Polynomial Equations

E9) Consider A = {0), B = {I), C = {O). ThenALB, B G C , b u t A c C . E10) NotethatA= {2) =C,B= {1,2).

.-.A = C a n d A s B Ell) A 2 C E12)

Fig. 8

S, R,P are the sets of squares, rectangles and parallcloyrams, respectively. Here we have taken

u = P.

E 13) a)

Fig. 9

The shaded area is Q \ Z. b) Infinite E14) $,$,Aand

(AC)C = A, since x E A e x e AC w x E (AC)'. E l 9 Z, Z, Z, $ El 6) c) From (a) we h o w that A n B S; A. We need to prove that A c A n B. For this, let x A. Then, since A c B Thus, x E A n B .

.-.x E B

:. A S ; A n B .

:. A = A n B . d)

AnAcA,applying(a). A c A n A, as in proof of (c).

:. A = A n A.

Also 0 s A n I$ since I$ is a subset of every set.

AnBsBandAnBcA.

:. A n B c B n A .

Similarly, B n A E A n B. :.An B - B n A . g)

X E A \ B e x € Aandxe B

h) Le x E C. Then C E; A a x E A. Similarlyx E B. Therefore, x E A n B. Hence, C s; A n B. E17)No.Forexample,ifA= {1,2,3),B={1,2,4),C= {1,2),theCs;AandCs;B,butCisnot properly contained in A n B; it is exactly equal to A n B.

Fig. 10

E19) A n B n C - (4) = ( A n B ) n C = A n ( B n C ) .

E20) A n B n A = A n ( B n A ) = A n ( A n B ) = ( A n A ) n B. = A n B = ( 3 0 n I n ~N).

x B ~u A .

:. A u B = B u A . c)

$cA*Au$=A

~ C,sinceA~CandBcC.Thus, d) L e t x ~A u B . T h e x ~A o r x B.Ineithercasex~ x c A u B a x C. ~

:. A u B G C . E22){XER I ) 0 5 x 5 3 )

E23) Since A c A uB = $, we see that A c $ also $ c A always.

:. A = 0.Similarly,B = 0. EX) a) A = ( 3 x 1 ~E N),B- ( 6 x 1 N),C= ~ ~ {lox ( x E N ) . Note that B G A.

:. A U B = A ,

.'. (A u B) u C = A u C = {XE N) 1 3 divideds x or 10 divides X}

Solutions of Polynomial Equations

A l s o A u B u C = { x ~N(3,6orlOdividex} = { X EN )3orlOdividex) =(AuB)uC. Similarly,AuBuC=Au(BuC). b)

A\BEAEAuB,B\AEBEAu B , A n B E A u B . :. ( A \ B ) u ( A n B ) u ( B M ) c _ A u B . Conversely, let x E A u B. Then x E A or x E B. Now, there are only three possibilities for x :

X E A b u t x e B , t h a t i s , x ~A\B,or

ii)

XE Aandx~ B,thatisx€ A n B , o r

iii) X E B b u t x ~ A , t h a t i s xB\A. ~ Thus,AuB~(A\B)u(AnB)u(BM). So we have proved the result

J25) X E A u ( B n C ) WXE A o r x ~ BnC e x € A o r ( x € B a n d x ~C) ~ ( X AEo r x ~ B ) a n d ( x ~AorxE C) ex^ A u B a n d x ~ AuC W X E (AuB)n(AuC). E26) X E ( A U B ) ~ W X ~ A U B WxeAandxe B W X E ACandxeBC. W X E A C nB. E27) A u B = U . :. ( A u B ) ~ = $ Also AC=(3, 4),BC=(1) A C n BC=$ (A u B)C= ACu BC FurtherAnB={2). .-.( A n B)C={1,3,4) .-.ACv BC= (A n B)C

.-. :.

m) A x B = ((2921, (2,319 (5,2), (5,319) B x A = ((2,213 (3,2), (2,519 (3,519) A x A = {(2,2),(2951, (521, (5,5),) EB) A={7,2)B={2,3,4) E30) Only (c), (a) is not since it is only an ordered pair, and not a triple. (b) is not since 1

-EN u

(d) is not, since e N (e) is not, since it is not an ordered triple, it a set 2 of three elements. E31) Any subset is A x B, where A S; R or B S; R. For a proper subset of RZ,either A or B should be a proper subset of R E32) a) (x, Y), E A x (B u C) Wx~AandB y u~ C w ( x , ~ ) EA ~ B o r ( x , y A ) ~x C w ( x , ~ ) E( A x B ) u ( A x C ) . b)

(X,Y)EA x ( B n C ) wx ~ A a n d B y n~ C

UNIT 2 COMPLEX NUMBERS Structure 2.1

Introduction Objectives

2.2

What A Complex Number Is

Geometrical Representation

Algebraic Operations 2.4.1 Addition And Subtraction 2.4.2 Multiplication And Divison

De Moiver's lleorem 2.5.1 Trigonometric Identities 2.5.2 Roots of A Complex Number

26 27

sUrmnary

Solutions 1 Answers

2.1 INTRODUCTION In your studies so far you must have dealt with numbers, integers, rational numbers and real numbers. You also know that a shortcoming in N led mathematicians of several centuries ago to define negative numbers. Hence, the set Z was born. For similar reasoiis Z was exteqded to Q and Q to R at various stages in history. Then came a point when mathematicians looked for solutions of equations like xZ+ 1 = 0. Since x2+ 1 = 0 has no solution in R, for a long time it was accepted that this equation has no solution. The Indian mathematicians Mahavira ( in 850 A.D.) and Bhaskara ( in 1150 A. D.) clearly stated that the quare root of a negative quantity does not exist. Then, in the 16th century the Italian mathematician Cardano tried to solve the qudatic equation x2- lox +40 = 0. He found that x, = 5 + and x, = 5 + f i satisfied the equation. But then, what is ,& ? He, and other mathematicians, tried to give this expression some meaning. Even while making mathematical models of real life solutions, the mathematicians of the 17th and 18th centuries were coming across more and more examples of equations which had no real roots. To overcome this shortcoming the concept of a complex number slowly came into being. It was the famous mathematician Gauss (1777-1855) who used and popularised the name 'complex number' for numbers of the type 5 +

In the early 1800s,a geometric representation of complex numbers was developed. This representationfinally made complex numbers acceptable to all mathematicians. Since then complex numbers have seeped into all branches of mathematics. In fact, they have even been necessary for developing several areas in modem physics and engineering. In this unit we aim to familiarise you with complex numbers and the different ways of representing them. We shall also discuss the basic algebraic operations on complex numbers. Finally, we shall acquaint you with a very useful result, namely, De Moivre's theorem. It has several applications. We shall discuss only two of them in some detail. We would like to reiterate that whatever mathematics course you study, you will need the knowledge of the subject matter covered in this unit So please go through it carefully and ensure that you have achieved the following objectives.

Soluhna of Polynomial Equatjons

Objectives After studying this unit you should be able to : define a complex number ; describe the geometrical, polar and exponential representations of a complex number ; apply the various algebraic operations on complex numbers ; prove and use de Moivre s theorem.

WHAT A COMPLEX NUMBERS IS

2.2

When you consider the linear equation 2x + 3 = 0, you know that it has a solution, namely -3 x= . But, can you always find a real solution of the equation ax + b = 0, where a, b E R 2 -b and a # O? Is the required solution x = -? It is, since ( ? ) + b = O a

Now, what happens if we try to look for real solutions of any quadratic equation over R ? Consider one such equation namely x2+ 1 = 0, that is x2 = -1. This equation has no solution in R since the square of any real number must be non-negative. From about 250 A. D. onwards, nathematicians have been coming across quadratic equations, arising from real life situations, which did not have any real solutions. It was in the 16th century that the Italian mathematicians Cardano and Bombelli started a serious discussion on extending the number system to include square roots of nagative numbers. In the next two hundred years more and more instances were discovered in which the use of square roots of negative numbers helped in finding the solutions of real problems. In 1777the Swiss mathematician Euler introduced the "imaginary unit", which he denotesby the Greek letter iota, that is i. He defined i = f i .Soon after, the great mathematician Carl Friedrich Gauss introduced the term complex numbers for numbers such as

Nowadays these numbers are accepted and used in every field of mathematics. Let us define a complex number now. Definition :A complex number is a number of the form x + iy, where x and y arereal numbers andi2=-1.

We say that x is the real part and y is the imaginary part of the complex number x + iy.

Wewritex=Re(x+iy)andy=Im(x+iy). Caution :i) i is not a real number.

ii) Im (x + iy) is the real number y, and not iy. We denote the set of all complex numben by C. S o , C = { x + i y ( x , y ~R). By convention, we will usually denote an element of C by z. So, whenever we will talk of a complex number z, we will mean z = x + iy for some x, y E R In facf z = Re z + i Im z. There is another convention that we follow while writing complex numbers, which we give in the following remark. Remark 1: When you go through Sec. 2.4.2, you will see that iy = yi v y E R That is why we can write the complex number x + iy as x + y i also.

By convcntion, wewrite any complex number x + iy for which y E Q, as x +yi. For example, 3 5 ' 3' 5 we prefer towrite 2 + i, 2 + - i and 2 + - i for 2 + il, 2 + i - and 2 + i - , respectively. 2 9 2 9

But if Z E C is the form z = a + i Ji; ,b e It, then we prefer to write z in this form and not as z = a + &i. Now that you know what a complex number is, would you agree that the following numbers belong to C ?

5+&5,3i,JZ,J-T Each of them is 'a complex number because

5+J-15=5+ifi 3i = 0 +i3

f i = f i+iO From these examples you may have realised that some complex numbers can have their real part or their imaginary part equal to zero. We have names for such numbers. Definition :Consider a complex number z = x + iy.

If y = 0 we say z is purely real. If x = 0, we say z is purely imaginary. We usually write the purely real number x + oi as x only, and write the purely imaginary number 0 + iy as iy only. Try these exercises now. El)

Complete the following table : z

Re z

Irnz

1 + ,I25 2 1

-1 + 5

Now, given any complex number, we can define a related complex number in a very natural way, as follows. Definition :Let z = x + iy E C. We define the complex conjugative (or simply the conjugate) of z to be the complex number

T h u s , R e ? = Re z a n d ImZ = - h n z Forexample, if z = 15 + i then Z = 15 - i. Try this simple exercise now. E3)

Obtain the conjugates of 2+3i,2- 3i, 2,-5

In section 2.4.2 you will see one important use of the complex conjugate. So far we have shown you an algebraic method of representing complex numbers. Now let us consider a geometrical way of doing so.

Complex Number

Mutioms of Poiylro~nlrlEgurtloas

2.3 GEOMETRICAL REPRESENTATION You know that we can geometricallyrepresent real numbers on the number line. In fact there is a one one correspondonce between real numbers and points on the number line. You have also seen that a complex number is determined by two real numbers, namely, its real and imaginaxy par@. This observation led the mathematicians WesseZ and Oauss to think of representing complex numbers as points in a plane. This geometric representation was given in the early 1800. It is called an Argand diagram, after the Swiss mathematician J. R Argand, who propagated this idea.

Let us see what an Argand diagram is. Take a rectangular set of axes OX end OY in the XOY plane. Any point in the plane is determined by its Cartesian coordinates. Now we consider any complex number x + iy. We represent coordinates (x, y). This representation is an Argand it by the point in the plane with Cartesia~~ diagram. For example in Figure 1, P represents the complex number 2 + 3i, whose real part H i 2 and imaginary part is 3. And what number does P' represent? P' corresponds to 2 - 3i.

Fig. 1 :An Argand diagram

You may have realised that in an Argand diagram the purely real numbers lie along the x-axis . and the purely imaginary numbers lie along the y-axis. So, you have seen that, given x + iy E C we associate with it the unique point (x,y) E R2 .The , can associate with it the unique complex converse is also true. That is given (x, y) E R ~we number x + iy. This means that the following definition of a complex number is equivalent to our previous definition. Definition : A complex number is an ordered pair of real numbers. In the language of sets, we can say that C = R x R. With the help of this definition can you say when two complex numbers are equal ? Definition :We say that two complex numbers (x,, y,) and (x,, y,) are equal iff x, = x, and y, = y,. In other words, xl + iyl = x2+ iy, iff x, = x,and y, = y2.

II 1

Thus,two elements of C are equal iff their real parts are equal and their imaginary parts arc equal. So, for example,

-1 + 2

J Z - --1 + i - ,J5 but 2

Try these exercises now.

E4) a)

Plot the following elements of C in an Argand diagram :

3, -1.t i, -1 + i, i

Ii /

-1

1 . r) . (2 , 0

E5) Write down the elements of C represented by the points (T

)

(0,2)in the plane. 1

E6)Forwhatvduesofkandmislr+)i=-+im? 2 While solving E4 you may have observed that in an Argand diagram the point that represents Z is the reflection in the x-axis of the point that represent z, for any z E C. Here are two more exercises about complex conjugates.

E7) For which z E C will z = Z ?

E8) For any z E C, show that

= z, that is, the conjugate of the conjugate of z is z.

Now consider any non-zero complex number z = x + iy. We represent it by P in the Argand diagram in Figure 2. We call the distance OP the modulus of z, and denote it by ( z 1.

Ft.2: Modulus .adargument

Using the pythagoras theorem, we see that

If z is real, what is lz I ? It is just the absolute value of z. Here's another important remark on the modulus.

Complex Number

Sdutlolu

Mmdd wu8tbm

z-Oifflzl-0

Remark 2 : Z E C, but 1 z 1 6 R Now, if you see Figure 2 again, you will see that L XOP = 8. We call 8 an argument of z = x + iy. For z = 0, ( z I = 0 and its argument is not defined. Now if z E C, z t 0, will it have a unique argument ? If 8 is an argument so are 2%+ 8,4x + 8, etc. If we insist the 8 lie in the range -n < 8 5 x, then we get a unique argument. We call this value of 8 the argument of z, and denote it by Arg z. Y X If, in Figure 2, we write ( z I = r. Then you can see that sin 8 = - and cos8 = - . r r

:. x=rcos8,y=rsin8.

......(1)

So, we can also write z as

z=r(cos8+isin8),wherer=~z~and8=Argz.

This is called the polar form of z. Note that, give z = r + iy we can use (1) to obtain ATg z =

(9-

HOW~VCI, as

than

one angle between -n and x have the same tan value, we must draw in argand diagram to find the right value of Arg z. Let us look at an example.

Example 1: (a) Obtain the modulus and argument of 1 + i. (b)

ObtainqifIz)=2andArgz==-. 3

Solution :(a) Let z = 1+ i. Then Re z = 1, Im z = 1.Thus, 1 + i corresponds to (1, I), which lies in the first quadrant. We find that

Thus, Arg z = - . 4

z = I z I (cos (Arg z) + i sin (Arg z))

= 1+ i

Try the following exercises now. - ---- -

E9) Write down the polar f o m of the complex numbers listed in E4 (a). ElO) Show that (z E C 1 z 1 = 1) is the equation of the circle x2+ 9 = lin R2,and

vice versa. There is yet another way of rqresenting a complex number. In fact this method is closely related to the polar representation. It uses the expression ez,where z E C. Let us define this expression.

Definition :For any z = x + iy E C, we defiie eL= ex(cos y + i sin y). In particular, if z= iy, a purely imaginary number, then we get Euler's formula : eiy = cos y + i sin y y e R This formula is d u ~.J the famous Swiss mathematician Leonhard Euler. You will be using it quite often w h i l ~dealing with complex numbers. Now consider an.

E C. We write it in its polar form,

z == r (cos8 + i sin B ). IJow, using Eule: * s fo~mulawe find that

z = re'@. This is the exponential form of the complex number z. 345 + 3i For example, the exponential from of z= 2 2

Try this exercise now. Fig. 3: Eulcr (1707-1713)

E 11) Write the following complex numbers in polar form and exponential from:

By now you must be thoroughly familiar with the various ways of representing a complex number. Let us now discuss some operations on complex numbers.

2.4 ALGEBRAIC OPERATIONS

In this section we will discuss the addition, subtraction, multiplication and division of complex numbers. Let us first coilsider '+ ' and '-' in C.

2.4.1 Addition and Subtration We will now define addition in C using the definition of addition in R. Definition: The sum of two complex numbers z,= x, + iy, and z, = x2+ iy, is the complex number z, + z, = (x, + x, ) + i (y, + y, ), that is

Let us look at an example.

Example 2 :Find the sum of i)3+iand-2+4i, ii) -5 and 5 - i. Solution :i) (3 + i) + (-2 + 4i) = (3 + (-2) ) + i (1 + 4)

Have you observed that any complex number is the sum of a purely real and a purely imaginary number ? This is because x + iy = (x + Oi) + (0 + iy). In the following exercises we ask you to verify some very important properties of addition in C.

b) Addition in C is commutative and associative.

Showthatz+Z=2Rezforanyz~C.

- - -

E13) Show that zl + z2 = z, + z2 V z l , z2 EC. E14) a)

Showthatz,+z2=z2+z,foranyz,,z2~C.

b) Show that (z, + z2)+ z3= z, + (z, + 2,) for any z,, z,, z, E C. E 15) Find the element a + ib E C such that

If you have solved these exercises, you must have realised that the addition in C satisfies most of the properties that addition in R satisfies. Also, because of what you proved in El5, we say that 0 + iO (= 0) is the additive identity in C. Now, can you define subtraction in C ? We give you a very natural definition.

Definition :The difference z, z, of two complex numbers z, = xl + iy and

2,-x,+iy,isz,+(-z2), where - Z, = (- x,) + i (- y,). Thus, z, - z2= z, + (-z,)

=(x,--i~,)+[(-~)+i(-~~)l = (x,

- x2) i (Y,- Y ~ ) ' +

So, what do you think z z is, for any z e C ? h t ' a see.TBka z -x + iy. Then z-zm(x-x)+i(y-y)=O,

For my z e C, (-2) is the

theadditiveidmtityiuC.

Try the following exercise now.

additive inverse of z.

E 16) Find (- 6 + 3i) - (- 3 - 2i).

t I

E 18) Find the relationship between b) Arg z and Arg ( z ) , for any z E C. (see Figure 4.)

fig. 4 : z r n d

-z

(-z)

We will now make a brief remark on the graphical representation of the sum of complex numbers. Remark 3 :The addition of two complex numbers has an important geometrical representation. Consider an Argand diagram (Figure 5) in which we represent two complex numbers (x,, y,) and (x2, yZ)by the points P and Q.

If we complete the parahelogram whose adjacent sides are OP and OQ, the fourth vertex R represents the sum (x,,y ,) + (x,, y2).

Complex Number

Fig. 5: Addition in C.

In vector algebra you will come across a similar parallelogram law of addition. So far you have seen how naturally we have defined addition (and subtraction) in C by ushg addition (and subtraction) in R. Let us see if we can do the same for multiplication.

2.4.2 Multiplication And Division We will now use multiplication in R to define multiplication in C. But the route is slightly circuitous. Consider the following product of two linear polynomials a + bx and c + dx, where a, b, c, d E R. (a + bx) (c + dx) = ac + (ad + bc) x + bd x2. Now, if we put x = i in this, we get (a + ib) (c + id) = (ac - bd) + i (ad + bc), since i2= -1. This is the way we shall define a product in C. Definition :The product z, z2 of two complex numbers z, = x, + iy, and z, = % + iy2is the complex number

Or, in the language of ordered pairs. ~ ~ , ~ Y , ~ . ~ ~ , ~ Y , ~ = ~ ~ , ~ , - Y , Y , ~ ~ , Y ~ + ~ ~ Y , ~ ~

For example, . (1,2)(-3,2)=[1.(-3)-2.2,1.2+(-3).2].

Let us check and see what i2 is according to this definition. i2 = i,j = (0 + i) (0 + i) = ( 0 -1) + i (0-0) = -1, which is as it should be! Multiplication has several properties, which you will discover if you try the following exercise.

E 19) Obtain (x, y) (1,O). (x, y) (0, I), (x, Y)(O,O), (x, 0) (Y,0) and fY.,;r)(J

,X!/X

X . I \ ~

E20) Prove that

4 'iz,=z,z, VZ,,z,E C.

Note that x2 +

# 0, since (x, y ) # (0, O).)

If you've solved these exercises, you must ,haverealised that z . 1 = z v z E C, This means that 1 is the multlpllcatlve ldentlty of C. Exercise El9 also says that 2.0-0 y Z E C, i(x+iy) = -y+ix y x , y R, ~ and that in case zl and 3 are purely real numbers. our definition of multiplication coincides with the usual one for R Also, fiomE20 (a) you can see that multiplication is commutative, and fiom E20 (b) you can see that multiplication is associative. And, what does E20 (c) say? It says that for any non-zero element z of C, 3 z' E C, such that 1 zz' = 1. In this case we say that z' is the multiplicative inverse of z. So z' =

-.z

Using E20 let us see how to obtain the standard form of the quotient of a complex number by non-zero complex number. We will use a process similar to'the one you must have used for a+bh rationalising the denominator in expressions like -Consideran example. c +d h 2+3i Example 3 :Obtain l_i in the form a + ib, a, b E R Solution :E20 (d) gives us a clue to a method for making the denominator a rcal number. Let 2+3i

us multiply and divide

by F i . What do we get ?

2+3i - --1 + -1. 5. So, -iq-- 2 2 If you've understood the way we have solved the example, you will have no problem in doing the following exercises.

a+ib

E 22) For a, b, c, d E R and c2+ d2# 0, write -as an element of C. . c+id 1

E23) a) Show that

b) Show that ; =1 ,z , v , ~ c \ { o j . If you have done E22,then you know how to write the quotient of one complex number by a non-zero complex number in standard form.

Complex Numb3

Suppose we lcnow z,, z2 E C in their polp forms, say z, = r , (cos 8, +isin0,)andz,=r2 (COS8,+isin0,). Then

z, z, = r, r, (cos 0, + i sin 0, ) (cos 0, + i sin 8, )

= rl rZ {(COS cos 0,

- sin 0, si%) + i (sin 8, cgs 8, + cos 0, sin 0, ))

= r, r, (cos (8, + 8, ) + i sin (8, + 8, ))

..,.(2)

S O , I Z ~=rlr2=lz1 ~~I lIz2land Arg (z, z2) = (8, + 8, ) + 2k x, where we choose k E Z eo that -n<(8, + e l ) +2kx 5 n . In Figure 6 we give a graphic illustration of what we have just said.

cos (A + B) = cos A cos Bsin A sin B sin (A + B) = sin A cos B + cos A sin B

Flgurc 6: Productlon polar form.

Let us consider an example. Example 4 :Obtain the product of z, = 2(cos 1 + i sin 1) and z2= cos 3 + i sin 3.

Therefore, z, z2 = 2 (cos (1 + 3) + i sin ( 1 + 3))

Note the Arg (z, z2)= 4, since 4 > x. We need to choose an integer k such that

- x < 4 + 2k x Ix, k = -1 serves the purpose. thus,

k g (2,Z2)= 4 - 2 ~ . Hence (z, z,) = 2(cos (4 - 2x ) + i sin (4 - 2x)) is the polar form of z, z,. We have a very nice method of finding the multiplicative inverse of a non zero complex number in an Argand diagram. Let us see what it is. Let z E C \ (0) be represented by a point P (see Figure 7). Let Q represent the real number ( z 12. Let R be the reflection of P in the x-axis, so that R represents 2 . Now, through (1,O) draw a line parallel to QR. Let it intersect the line OR in S. Then S 1 represents - . Z

Solutions of Polynomial Equations

Figure 7: Finding the multiplication inverse.

Try the following exercises now.

E 24) Find the polar forms of z, and z2where z, = - 6 and z2= 1 + i. Hence obtain I z, z2 I and Arg (z, 2,). 21

E25) Knowing the polar forms of z , , z2E C, z2 # 0,obtain the polar form of - *

E26) Obtain 2in the polar form, where zl and z2are as in E 24.Represent z, , z 2 , z 2 , z2 and 3 in an Argand diagram.

2 -2

We will use multiplication and division in the polar form a great deal in the next section. Before going to it, let us give you a rule that relates '+' and 'x' in C. Do you know of such a law in R? You must have used the distributive law often enough. It says that a (b + c) = ab + ac v a, b, c E R. The same law holds of C. Why don't you try and prove it?

E27) a) Checkthat (1 + i) { ( h - 3i) + (5 +i))

Multiplicatin distributes over addition.

b) Prove that z, (z, + z,) = z, z, + z, z, v z, z, z, E C. -

Now let us discuss a very useful theorem.

2.5

DE MOIVRE'S THEOREM

In the previous section we proved that if z, = r, (cos 8, + i sin 8,) and z2 = r, (cos 6, + i sin 8, ),

then z,zz r, r, {cos (8, + 0,) + i sin (6, +- 8, )).

In particular, if z, z,, then r , =; r,, 6, 8,: and we find that z: ==

I-: 8 , sin 2 (COS 2

8,).

- .-

In fact ~hisis a particular case of a very nice fourmula, namely, that if z = r (cose + i sine ),

1 1

Complex Number

then zn- i t ' (cosn0 + i sinne) for any integer n. To prove this result we need De molvre'r

theorem, named after the French mathematician Abrahamne Moivre (1 667 - 1754 ). It may amuse you to know that De Moivre never explicity stated this result. Rut he seems to have known it and used it in his writings of 1730. It was Euler who explicitly stated and proved this result in 1748.

1 I t

Theorem 1 (De Mulvre s theorem): ( cose + i sine )n= cos ne + i sin no, for any n E Z and any angle 8.

: Proof : Let us flrst prove it for n 3 0.We will prove this by using the following important

principle.

Princlple of Induction :Let P (n) be a set statements of one statement for each a positive integer n, such that You can study induction in greater detail in the cours "Absiract Algebra".

P(l)istrue,and

ii)

if P (m) is true for some m E N, then P (m + 1) is true.

-- .

Then, P (n) is true y n E N. How will we use this principle ? For any n E N,we will take P (n) to be the statement We will first prove that it holds for n = 1 that is, P (1) "(cose + i sine)" = cos ne + i sin ne is true. Then, we will assume that it holds for n = m for some m E N, and prove that it is true for n = m + 1. This will show that if P (m) is true, then so is P (m +I).

Now. for n = 1,

(cose + i sine )' = cose + i sine = cos 1 . 9 + i sin 10.

So the result is true for n =I.

(cose + i sine )m = cos me -t. i sin me. Now, (cosO + i sine)"'+'

: %

Assume that it is true for n = m, that is,

= (cos me + i sin m 8) (cose + i sine)

by (3)

= cos (d +e ) + i sin (mfl+ 0), by the formula (2) far products.

=cos(m+1)8+isin(m+~)9 Hence, the result is true for n = m + 1. Thus, by the principle of induction, the result is true y n E N. Now let us see what happens if n = 0. We define z0 = 1, for any z E C \ (0). (As in the case of R, O0 is not defined.) Therefore, (cose + i sin 8 )O = 1.

Also,cos0.8+isin0.8.=cosO+isinO=1. Thus, the result is also true for n = 0. .

Now, what happens if n < O? May be, you can prove this case. You can do the following exercises, which will lead you to the result.

E 28) Prove that cos

1 +

sine = cosO i sine, for any angle 8.

Use this fact and De Moivre's theorem for positive integers to prove that (cos8 + i sin8 )" = cos n + i sin 8 . So, De Moivre s theorem is true Q n E Z. Now, if z = r ( c o d + i sin8 ) E C, then

nE Z

ztl= r" ( C O S+ ~i sine)" = r" (cos n 8 + i sin n 8), using De Moivre's theorem.

What we have shown is that

[r (cos8 + sin8)ln = r" (cos n8 + i sin no) for r E R, 8 E R, n E Z. This result has several applications in mathematics and physics. We shall discuss two of them here.

2.5.1 TrigonometricIdentities One of the most useful applications of Theorem 1 is in proving identities that involve trigonometric functions like sin 8 , cos 8 , etc. Let us look at an example.

Example 5 :Find a formula for cos 4 8 in terms of cos 8 and sin 8. Solution :by De Moivre's theorem

.........(4)

(~os8+isin8)~=cos48+isin84 We can also expand the left hand side of (4) by using the binomial expansion. Then (cos 8 + i sin

= ( C O )4 S ~+ 4CI(COS8)l (i sin8 ) + 4C2( ~ 0 ~ (i8 ) ~

I ..

+ 4C3cos 8 (i sin 8 )I + ( i sin 8)4 = cos48 + 4 i sin 8 cod8

- 6 sin28cos28- 4 i sin38cos 8 + sin48

......(5)

Thus, comparing the real parts in (4) and (5), we get

cos 48 = cos48- 6 sin% cos28+ sin48 . You can try the following exercise on similar lines. --

E 30) Find formulae for cos 38 in terms of cos 8 and sin 38 in terms of sine. Now, for any m E N let us look at zm,where z E C such that ( z I = 1. Then, by De Moivre's theorem

z"' = cos me + i sin nl8 'and z-'" = cos m8 - i sin me, since cos (-8 ) = cos 8 and sin (-8) = -sin 8 for any angle 8. Thus ztl1+ Z-'I1 = 2 cos 8 m, and zm-z-'"=2isinm8. We can use these relations to express cosn18and sinn18in terms of cos me and sin me for m = h 1, +2,. .....Let us consider an example.

Example 6 : Expand 24"-2 ( ~

08 +ssin4"8 ~ )~in terms of the cosines or sines of multiples of 8.

Solution: Putting m = 1 in the equations (6), we get

I 1

2cos8

Complex ~ u m b c r

1 '1 :I.and 2sin8 = z - z Z

by the binomial expansion.

= 2{2 cos4n8 + 2 (4nc2) cos (4n - 4)8 +

.......) + 2(4nc2,), using (5).

The procedure we have shown in Example 6 is very useful for solving differential equations involving trigonometic functions. It is also usefbl for fmding the Laplace transform of such functions. Why don't you try this exercise now ?

E31) Apply De Moivre's formula to prove that i) cos 28 = cos28- sin28 ii) sin 28 = 2 sin8 cos8

E32) Expand cos68 sin68 in terms of the cosines of multiples of 8.

Let us now look at another area in which we can apply De moivre's theorem with great success.

2.5.2 Roots of A Complex Number In Section 2.2 we told you thaf the whole subject of complex numbers first arose in an attempt'. to find the square roots of - 1. By now you know that we can always find two distinct complex square roots of any non - zero real number. That is, given a E R \ {0),3 distinct z,z, E C such that zt = a, z; =a. In fact, the set of complex numbers nas a much strongerproperty, which is a major reason for its importance in mathematics. This property is : given any n E N and z E C, z f 0, we can find distinct z,,.......,znE C such that %n = z Each of these zk's is cilled an nth root of z. To extract all the nth roots of a complex number, we need De Moivre's theorem as well as the following result that we ask you to prove.

Solutlonn of Polynomial Equstionn

E33) Let x be a positive real number and n E N. Show that there is one and only one positive real number b such that bn = x. ( Hint : Let r,s > 0 be such that rn = x = sn.Suppose r # s. Then r" -s" = 0 and r - s # 0. Then you should reach a contradiction.) We denote the unique positive nth root obtained in E 33 by x"? . Now let us consider an example of extraction of roots of a complex number. Example 7 : Obtain all the fifth roots of i in C. Sotution :Let z = r ( cos0 + i sin0 )be any 5th root of i. Then z5 = i. The polar form of i is

n x i = cos - + i sin : . Therefore 2

rS( c o d + i

= cos - + i sin -

by De Moivre's thorem. Comparing the moduli (plural 'modulus') and arguments of the complex numbers on both sides of (7),we get n: 9-1 and50- -+Zknwherek=O,f 1, f 2 ,....... 2

r is the unique positive real fifth root of I (see E33 ). Since 1 E R, r e l , that is ( z I = 1. The possible values of 8 are

Thus, the possibe 5th roots of i are

From this it seems that i has infinitely many 5th roots, one for each k E Z. But this is not true. There are only 5 distinct ones among these. They will be the values of z for K=-2, -1,0,1,2. Let us see why.

4~)+isin(~-%) When k = - 2, = cos - - 7X 3X - i sin - = z - ~ say. , 10 10

= cos-

n When k = -1, z = cos--isin 10 X

When k = 0, z = cos-+isin 10 X

cos (2% + 6 1 - LO. 0 and sin (2n + 01 , (+

3x 10

- = z-, say. X

zosay.

When k = 1, z = cos- + isin - = z, say. 2 2 9n When k = 2, z = c o s E +isin - = z,, say. 10 10 13n: y'J+isin(2n-$)=z-2. When k = 3 , z = c o s - + i s i n - = c o s 2~-10 10 ,

(

.-. . ...--

Complex Number

Similarly, when k = 5, you will get z ,and so on. Thus, k = 5,6,7 ,..... don't give us new values of z.

[TI :-iT) In

- 7 '

NOW, ifwe put k = -3, we get z = cor

Simlarly, k = -4, -5.... will not give us new values of z. Therefore, the only 5th roots of i are

(;

cos - + 2k-

+ isitl - + 2k - for k = (;. '

;(

+ 2.

Remark 4: We also get the 5th roots of i by takifig h - I!. 1 , 2 . 3 , 4 in

(10

,as you ha\-c :;r:

-1.

Qnly note that for k = 1and k = 4,

the angles 0 will not lie in the range -rr < 0 S x. Thzt's why we had taken k = 0. + 1, 2. Now, look at all the fifth roots of i.. How are their rnodtl!i rels~ed:?They have the same rnodulus, namely, I i ~""=l). Thus they all lie on the circle \viih centre (0,O) and radius i. These points will be equally spaced on the circle, since the arguments of consecutive points differ by

2?r a constant. We plot them in the Argrnd diagram in Figure 8. 5

1 \111gthe same procedure as above we can obtain the dis~inctnth roots-of 4ny non-zero

complex number, for any 11 E N. Thus given, any non-zero c o r n p l e x ! ~ l ~ , awe : write~itinits polar fonn w = a ( c o s a + i s i n a ) , w h e r e a = /wIandQ=Argw,

By E33, there is a unique r E R, r > 0, such that r" =a, ahat is, r a"". Then the distinct nth roots of w are

zk = a''" [cos

a +2 k ~ + i sin - forli-dl; .........., n-I: n n

Geometrically, they lie on a circle of radius a"" and are equally spaced on it

Solutions of Polynonilul Equations

Note that a non :zero complex number has exactly n distinct nth roots for any n E N. If z is one root, then the others are al.........a,:, are the nth roots of unity. z a, ......,2% where

Now you can do some exercises. E 34) Find the complex cube roots of unity, that is, those z E C such that z3 = 1. Plot them in an Argand diagram.

E35 ) Solve the equation z4 - 4z2+ 4 - 2i = 0.

(Hint :The equation can be rewritten as (z2- 2)2= (1 + i)2.) --

The cube roots of unity that you obtained in E34 are vkry important. We usually denote the cube root

-I + i&i by theGreek letter w (omega). 2

the other non - real cube root of unity. Thus,

the three cube roots of unity are 1, w, w2, where

-1 + i h

Also note that 1+ w+w2=0. We will often use wand (8) in Unit 3. We will equally often use the following results, that we ask you to prove. root r, and the cube roots of a are r, rw, rw' E 36) a) Let a E R . Show that a has a real cube , .* b) Show that if a E R, a < 0 and n is an even positive'integer, then a will not have a real nth root. c)

Let z E C \ R. Thgn show that z has three cube roots, and if any one of them is y, the other two are y w, y w'.

With this we come to the end of our discussion on complex numbers. Thus does't mean that you won't be dealing with them any more . In fact, you will often use whatever we have covered in this unit, while studying this course as well as other mathematics courses.

in this Let us take a brief look at the points covered :.' . ' , unit. ..;

2.6

SUMMARY

In this unit on complex numbers you have studied the following points. 1) The deflmition of a complex number: A con~plexnumber is a number of the form x + iy where x, y E R and i = f i Equivalently, ~tis apair (x,y) E R x R. 2) x is the real part and y is the imaginary part of x + iy.

Complex Numhcr

x,+iy,=x2+iy2iffx,=x2andy,=y2.

The conjugate of z = x + iy is 2 = x -'iy.

The geometric representation of complex numbers in Argand diagrams.

f i y 2 and --

6) Thepolarformofz=n + i y i s z = r ( c o s R + i s i n R ) , w h e r e r = z =

0 = Arg z = tan-'

, where we choose the 0 that corrc.ponds to the position of z in

an Argand diagram. 7) Euler's formula: eie= cos 8 + i sin 8 t/ 8 E R.

8 ) The exponential form of z = x + iy is z = reie,where r = I z ( and 8 = Arg z *

9) Addition, subtraction, multiplication and division in C : t/ a, b, c, d, E R (a+ib)*(c+id)=(a*c)+i(b*d),

(a +ib). @ + id)-= (ac - bd) + i (ad + bc), -

a+ib c + id

(a+ib) (c-id) , for c + i d c2+d2

+ 0.

10) For zl, z2E C, ) z ,z2 ) = I z , ( ( z 2(,Arg(z,z,)=Argz, +Argz2+2kx

where k, m E Z are chosen so that - x < Arg ( z l z 2 ) < x and - X c Arg

(2) 2 x

11) De Moivre's theorem : (case + i sine )" = cos ne + i sin ne t/ n E Z and any angle 8. 12) The use of De Moivre's theorem in proving trigonometric identities and for obtaining nth roots of complex numbers, where n E N. 13) The cube roots of unity are 1, o,w2,where o =

-1 + i J 5 . 2

Now that you have gone through this unit, please go back to the objectives listed in Section 2.1. Do you think you have achieved them? One way of finding out is to solve all the exercises that we have given you in this unit. If you would like to verify your solution or answers, you can see what we have written in the following section.

2.7

SOLUTION 1 ANSWERS z

Re z

Imz

+J-23

J23

-i+h

-I+& 5

Solution6 of Polynomial Equatlonr

E 2 ) Yes,because every real number x P the complex w b e r x + Oi. E3) 2-3i,2+3i2,-3i. E.o)a)

Fig. 9 --.

P,Q, Rand S represent 3, - 1 + i, - 1 + i and i, respectively.

F la. 10 Id,,

L, and L, reprorent the setr S,, S,nrt: S,, rerpcrctivo!y,

Es) k - ~1 , m - 3 ,

E 7 ) Loto x+iy.Then 'i-x-ly. ,; o m Z e x + i y - x - i y * y m - y d y - 0 , ,*. y z e R z - 2

E8) Loto-x+iy.Tksn E-x-iy, 7

.'. 'i = x-iy = x + iy = z '

3-3(core+irine) Nowl-l+iJ-

a r/Z, =

and

Arg (-1 +i)=tan-'(-1) = - x / 4 o r 3 ~ / 4 , 3% Since -1 + i corresponds to (-I,] ), which lies in the 2nd quadrant, Arg (-1 + i ) = 4

... (-1 +i j = f i (cos

(2)+ (2)) i sin

-.71

i = cos- + isin 2 2

J5 ei ( exponential form ) 'I4

1 + i = f i (cos- = i sin --) (polar form) 4 4

= f i e i '14 ( exponential form) -1 = cos rc + i sin x (polar form) = ei w2 ( exponential from ) X

-X

i = cos - + isin (polar fom) 2 2 = eilX(exponential from ) . b) Let z = x + iy. Then z+ Z = ( x + i y ) + ( x - i y ) = 2 ~ = 2 R e z . E 13) Let z, = x, + iy, and z, = x2+ iy2. Then

-a'.

Z I

+ z2 =(x, + X ~ ) - ( Y I + Y * )

= %, + ZZ

E 14)a)Letz,=(x,,y, ) ~ d z , = ( x ~ ~ y ~ )

Then~l+za~(x,~~l)+(xa~~~) (~,+~,IY,+Y,)

= ( x a + x I I y Z + y , ) , s i n c e a + b = b +yaa , b e Ra '(x~~Y~)+(x~,YI) =Z2+ZI

E15)

b ) Letz,m(xla~,),z2m(xa~a)az,=(x~a~,)* Then, u8e the fact that ( a + b ) + c = r + (b + c ) yr, b, c E R to prove the resultl Letzmx+iy, Thenz+(a+ib)=z +(x+iy)+(a+ib)=x+iy *(x+a)+i(y+b)=x+iy x+a=xandy+b-y

):(

Arg (-2) = tan-'

= tan-'

):(

= Arg z - n ,because (-2) is the reflection of

z in the origin. El91

(&Y)(l,O)=(&Y) (x,y)(O.l)=(-~,x) ( x,y) ( 0 ) =(0,O)

E20)

(~o)(Y,o)=(xY,o) (x,y)(l, l)=(x-y,x+y). a) Let z, = (x,, y, ) andz, = ( x, ,y, ). Then 2, ? = ( x " Y ' ) ( x ~ , Y ~ )

'(XI 3-Y, Y ~ ' xY, Z + ~ 2 ~ 1 ) = ( ~ ~ ~ ~ - y ~ y ~ , ~ ~ y ~ + ~ ~ yv~a ), b, ~, ci ~R. n c e a b = ba =( 3 , Y2) (x',Y')

=z2z1. b ) If z ' = ( x , , ~ ' ) ~ = ( x ~ , ~ ~ ) z ~ = ( x ~ ~ ~ ) t h e n 2' z2=( XI 3 - Y ,

Y2'Xl Y 2 + X 2 Y l )and

? z 3 = ( 3 x 3 - ~ 2 ~ 3 ' ~ ~~3 3+ ~ 2 ) Therefore, ( z, ?) z3

E21)

= (x+iy) (x-iy) = x 2 + Y 2=

(,/=I2

= 1212.

-2 + i - -2 + i since i2=- 1. i jS+ i(2i) -2 + i j S 3

a +,ib is meaningful. . , c + id -a+ib - - (a+ib) (c-id) - (ac+bd) + i(bc-ad) -- - + bd) + i [ b c - a d ) c+id (c+id) (c-id) c2 + d 2 c- + d 2 c2 + d 2

E22)

c2+d2 # 0 means that c # 0 or d # 0. Thus, c + id # 0. Hence

Cc~mplcrNumber

E23)

a) L e t z = x + i y = O . T h e n , f i . o 1 ~ ~ E 2 0 ( d ) w e k n o w t l ~ a t z Z = z ~ ~ .

(iF I: I 1

Therefore,

1 b) F o r z # O , z . - = Z

E21)

I,:.

7 is the rnultipl;cative inverse of Z . I z l-

1x1. - = I l l = I.:.

n z l = 6 ( c o s n + i ~ i n n ) , ~ , = ~4 [ C 0 ~ - + i ~ i ~ -

:. 1 z, zz I = 6 & and

(

Arg (z, z2 ) = .rr + -

+ 2kn , where k E Z such that -n < Arg (2, z,) < x.

z, r, (COSO, + isin0,) then L= ? r2(~~~02jisin02) P A -

r2 r!

(cose, + i sine,) (cos 8, - i sin 8,)' multiplying and hviding by (cos

= - cos (8, + 8,) + i sin (8, r2

rl r2

= - (cose,

,261

- 8, ))

- 8, + 2kx) + i sin (8, -

5 = -- (cOs(" -

%ti

- i sin 8?).

+ 2kn)), where k E R such that

): + i sin (n - :)

Fie. - 11

1 21 The points P, Q, R, S and T in Figure 1 1 represent z l , 22, Z2, - and -, respectively. z2 22 Here OT = -I'? and LXOT = LXOP - LXOQ

E27) a ) L H S = ( l + i ) [ ( f i +5)-2i]=(7+ f i ) + i ( 3 +

RHS=[( f i + 3 ) + i ( & - 3 ) ] + ( 4 + 6 i ) = ( & + 7 ) + i ( & + 3 ) Thus, LHS -= M S .

LHS stands for let? hand side and RHS stands for right hand sids

Solutions of Polynomial Equntlons

Let 2 , x, T iy, , z, = X, r iy,, 2, = X, + iy3 . Thenz, (zl+z3)-=(xl+lYl )C(x2+x3)+i (y,+ Y3)1 = [ x ~ ( x ~ + x ~ () ~- Y 2 +~ ~ 3 ) I + i( [Y~ ~1 + Y ~ () xT ~Y+~x ~ ) I

=(xIx2-~I~2)+(xIx3-~,~3)+iCx,~Z+x~~,j+(x,~3+x3~1)1 =[(xi x 2 - ~ I ~ Z ) + i ( x I ~ 2 + x 2 ~ I ) l + C ( x , x ) - ~ I ~ 3 ) + i ( ~ 1 ~ 3 + x 3 ~ 1 ~ 1 = z l z2+z1Z3. You can also solve this by writing z,, z, and z3 in polar form, If you do, you must remember to be careful about zi = 0 for any i.

1 (COS- i sin@)n = = (case + i sin 0)"' (cosO = ( cose

+ i sin 6

- i sine)"

= [ cos (-0 ) + i sin (-8 )Im

=cos(-mQ)+isin (-me),sincem>O. = cos ne + i sin n8.

E30) (cose + i sine )) = cos 38 + i sin 38. Also, ( cose + i sine )3 = cos3 8 + 3 cos2@(i sine ) +3 cosC) ( i sin 8)2+ ( i sine )3 = ( cos38- 3 cose sin28) + i ( 3sinC) cos28- sin3@ ).

Thus, comparing real parts of the two equalities, we get cos 38 = cos39 - 3 cose sin28= cos38 - 3 cos0 ( 1 - cc-rs'Q ) = 4 cos38 - 3 cos 8.

Similarly, comparing the imaginary parts we get sin 38 = 3 sine ( 1 - sin28) - sin38 = 3 sine - 4 sin38 .

E 3 1) ( cose + i sine )2= cos 20 + i sin 20, and ( cose + i sine )2 = cos28+ 2i C O S sin8 ~ - sin28.

:. cos 28 = cos28- sin28,and sin 28 = 2 sine cose.

E 32) Let z = cos8 + i sine. Then, using (6)

= 4 cos 68 + 60 cos 28

1 Cos68 - sin68= - (cos 68 + 15 cos 28 ) 16 E33) Let r, s E R,r, s > 0 and rn = x = sn.We will prove the result by contradiction(sec appendix to this block ). Suppose r f s. Then rll - Sll ( r - s ) ( rl'-l + r"-2 s + .....+ mll-2 + sn-l ) = 0

Since r > 0, s > 0, rll-' + r"-' s + ......+ rs11-2 + sl'-' > 0. Also, r - s # 0.

But then how can the product of two non - zero numbers be zero ? So we reach a contradiction. Therefore, our assumption must be false. Thus, r = s.

E 34) Let z = (cos8 + Bsin8) be a cube root of 1 = cos 0 + i sin 0,

0+2kx

2kx

he^ r = 1113 = 1 0 = ----- - -for k = 0,1,- I. 3 3

2kn

2 k K ) for t = Q I . - I :. The roots are 1. (COs j + i sin 3

JJ and -I - JJ

-1 Thus, the roots are 1, - + i-

1 -,

Flgure 12: Cube root# o f unlty.

E35) We want to obtain those z E C for which

( z 2 - 2 ) = f ( 1 +i),thatis z 2 - 2 = 1 +iandz2-2=-(l+i),thatis, z 2 = 3 + i a n d z 2 = 1 -i. Thus, we want to find the square roots of 3 + i and 1 - i. 1

Now, 3 + i =

cos (tan-'

I I I

Thus.the square roots of 3 + i are (COS

8 2

-)2/ and 10"' {mr (:+n)

. . 8

- + ,sin

where 8 = tan-' Also 1 - i = f i (c:s

f n + x,\1j ,

isin -

;

(9 (9) +

sin

SO that the square mots of 1

- i are

These 4 square roots are the 4 roots of the given equation.

E 36) a) If a 2 0, then by E33, a has a real cube root, all3 .Now, a = a (cos 0 + i sin 0 ). Thus, the cube roots of a are

3 I

that is, all3, all3u ' / ~ a'/' , u2. If a < 0,then - a > 0. Thus, - a has a real cube root, say b. Then r = - b is a real cube root of a. And I r I = I a Ill3 that is, r = - I a ('I3 ( since r is negative).

Now a = I a I (cosx;+ i sin x;). Therefore, the cube roots of a are la ill3 (cos ( l k

3 'In + i sin (2k+i)n), k = 0, 1,2.

(since- 1 =cosx+isinx;)

Thus, the cube roots of a are r, rw rw2. b) Letn-2-me N. Then,foranyb E R, bn b2m = (bl)m 2 0. 3

Thus ,bn # a for any b E R.Hence, a can't have a real nth root. c) h t z = r ( cos0 + i sin0 ) ,in polar form.

Then its cube mots are r1I3 (Cos

0+2kn

!), then the other roots uo

n u s , if y = r 113 (cos - + i sin 3 3

UNIT 3 CUBIC AND BIQUADRATIC EQUATIONS Structure 3.1

Introduction Objectives

Let Us Recall 3.2.1 Linear Equations 3.2.2 Quadratic Equations

Cubic Equations 3.2.1 Cardano's Solution 3.3.2 Roots And Their Relation With Coefficients

3.4

Biquadratic Equations 3.4.1 Femri's Solutions 3.4.1 Roots And Their Relation With Coefficients

3.1 INTRODUCTION In this unit we will look at an aspect of algebra that has exercised the minds of several mathematicians through the ages. We are talking about the solution of polynomial equations over R. The ancient Hindu, Arabic and Babylonian mathematicians had discovered methods of solving linear and quadratic equations. The ancient Babylonians and Greeks had also discovered methods of solving some cubic equations. But, as we have said in unit 2, they had not thought of complex n;lrnbers. So, for them, a lot of quadratic and cubic equations had no solutions. In the 16th century various Italian mathematicians were looking into the geometrical problem of trisecting an angle by straight edge and compass. In the process they discovered a method for solving the general cubic equation. This method was divulged by Guolamo Cardano, and hence, is named after him. This is the same Cardano who was the first to introduce complex -numbers into algebra. Cardano also publicised a method developed by his contemporary, Ferrari, for solving quartic equations. Later, in the 17th century, the French mathematician Descartes developed another method or solving 4th degree equations.

In this unit we will acquaint you with the solutions due to Cardano, Ferrari and Descartes. But first we will quickly cover methods for solving linear and quadratic equations. In the process . we will also touch upon some general theory of equations. There are several reasons, apart from a mathematician's natural curiosity, for looking at cubic and biquadratic equations. The material covered in this unit is also usefid for mathematicians, physicists, chemists and social scientists. After going through h e unit, please check to see if you have achieved the following objectives.

Objectives After studying this unit, you should be able to solve a linear equation; solve a quadratic equation; apply Cardanio's method for solving a cubic equation;

1 '

apply Ferrali's or Descartes method for solving a quartic equation;

use the relation between roots and the coefficients of a polyr~omialequation for obtaining solutions.

3.2 LET US RECALL 7

You may be familiar with expressions ofthe from 2x + 5, -5x2 + --, f i x 3 + x 2 + I, etc. 2

All these expressions are polynomials in one variable with coefficients in P.!n general, we

have the following definitions. Definitions : An expression of the form a, x0 + a; x' +a,x2 +.........+ a, x" ,where

n e N and ai E C vi =I ,.....n, is called a polynomial over C in the variable x. a, a,, .....a,,are called the coefficients of the polynomial. If a, + 0 we say that the degree of the polynomial is n and the leading term is a, xn.While discussingpolynomials we will observe the following conventions. Conventions : We will i) Write x0 as 1, so that we will write a. for a. xO, ii) write x' as x, iii) Write xnlinstead of 1.xn(i.e. when a,,,= I), iv) omit terms of the type O.xnl.

Thus,the polynomial 2 + 3x2- 9 is 2x0+ 0. x' + 3x2+ (-1) x3. We usually denote polynomials in x by f (x) ,g (x), etc. If the variable x is understood, then we often write f instead off (x). We denote the degree of a polynomial f (x) by deg f (x) or deg f. Note that the degree off (x) is the highest power of x occurring in f (x). For example, 5 ix3 is a polynomial of degree 3, i ) 3x + 6x2+ 2 ii 1 x5 is a polynomial of degreee 5, and

i.i) 2 + i ia a polynomial of degree 0,since 2 + i = (2 +i) xo. Remark 1: Iff (x) and g ( x) are two polynomials, the11

deg (f (4.g (XI)5 deg f (x) + deg g (x) We say that f(x) is a polynomial over R if its coefficientsare real numbers, and f (x) is over Q if its coefficients are rational numbers. For example, 2x + 3 and x2 + 3 are polynom~alsover Q as well as R ( of degrees 1 and 2, respectively ). On the other hand & is a polynonlial (of d 0) over R but not over Q. In this course we shall almost always be dealing witti polynomials over R

Note that any non- zero element of R is a polynomial of degree 0 over R We define the degree of 0 to be Now, if we put a polynomial of degree n equal to zero, we get a polynoraial equation of degree n ,or an nth degree equation.

For exeample, ( i ) 2x + 3 0 is a polynomial eqtlnllon ofdcg! ce 1 and ;

(ii) 3x2 + J? x - 1 = 0 is a polynon~~.~i cquation of degree 2. Iff (x) = a, + a, x + .......+ a,, xl' is a polynomial and a E C, we can substitute a for x to get f (a), (r the value of the polynomial at x = a . Thus, f (a) = a, + a, a t a, a2 t ..........+ a,, an. For example, iff (x) = 2x +3, then f (1) = 2.1 + 3 = 5, f (i) = 2i t 3, and

f(2)

-3 2 Definition :Let f (x) be a non - zero polynomial, a E C is called a root ( or a zero) off (x) if

since

= 0 we say that - is a root o f f (x).

f(a)= 0.

In this case we also say that a is a solutuion (or a root) of the equation f (x)==O. A polynomial equation can have several solutions. For example, the equation x2- 1 = 0 The set of solutions of an equation is called its solutlon set. Thus, the solution set of x2 + 1 = 0 I \ 11,-i}. Another definition that you will need quite often is the following. Detlnition: Two polynomials a. + a, x + ......+ a,, xl' and b, + b, x ..... + bn,xnlare called equal if n = m a n d a , = b,, yi=O,1, ......,n. Thus, two polynomials are equal if they have the same degree and at their corresponding coefficients are equal, Thus. 2x3+ 3 = ax3+ bx2 + cx + d iff a = 2, b = 0, c = 0, d = 3. Let w now take a brief look at polynomials over R whose degrees are 1 or 2 and at their solutions sets. We start with degree 1 equations.

3.2.1 Linear Equations Consider any poly~lomialax + b with a, b E Rand a # 0. We call such a polynomial a linear polynomial. If we put it equal to zero, we get a linear equation. Thus,

is the most general form of a linear equation; -b You h o w that this equation has a solution in R, namely, x = -; and that this is the only a solution. Sometimes you may come across equations that don't appear to be linear, but, after sirnplification they become linear. Let us look at some examples. 3p-1 2p Example 1 :Solve -- - = P. (Here we must assume p # 1.) 3 p-1 Solution: At first glance, this equation in p does not appear to be linear. But, through crossmultiplication, we get the following equivalent equation:

On simplifying this we get

Two equations are equivalent if their solution sets are equal.

Solutions of Pulynomlal Equations

3p2 - 4p +1 -

= 3p2

- 3p,

thatis,7p-1 =O. The solution set of this equation is started with.

.Thus,, this is the solution set of the equation we

Example 2 : Suppose I buy two plots of land for total Rs.1,20,000, and then sell them. Also suppose that I have made a profit of 15% on the first plot and a loss of 14% on the second plot. If my total profit is Rs. 5500, how much did I pay for each piece of land 7

Solutlon :Suppose the first piece of land cost Rs.x Then the second piece cost

15 100

10 100

Rs.(1,20,000 x). Thus, my profit is Rs. - x and my loss is Rs. - ( 1,20,000 - x).

25x- 1,750,000=0 e x = 70,000. Thus, the first piece cost Rs. 70,000 and the second polt cost Rs. 50,000. You may like to try these exercises now. El)

Solve each of the following equations for the variable indicated. Assume that all denominators are non - zero. X

a) J (-k + a) = x for x where J, k and a are constants. 1 1 1 b) - = -+ - for R, keeping r, and r2 constant. R r1 12 c)

5 C = - (F - 32) for F, keeping C constant. 9

E2) An isosceles triangle has a perimeter of 30 cm. Its equal sides are twice as long as the third side. Find the lenghs of the three sides.

E3) A student cycles from her home to the study centre in 20 minutes. The return journey is uphill and takes her half an hour. If Ler rate is 8 km per hour slower on the return trip, how far does she live from the study centre?

E4) Simple interest is directly proportional to the principal amount as well as the time for which the amount is invested. If Rs. 1000, left at interest for 2 years, earns Rs 110, find the amount of interest earned by Rs. 5000 for 3 years. (Hint :S = k Pt, where k is the constant of proportionality, S is the simple interest, P is the principal and t is the time.)

Now that we have looked at first degree equations, let us consider second degree equations, that is, equations of degree 2.

3.2.2 Quadratic Equations Consider the general polynomial in x over R ofdegree 2: ax2+bx+c,wherea,b,c€ R,a f O . We call this polynomial a quadratic polynomial. On equating polynomial to zero, we get a quadratic equation in standard from. Can you think of an example of a quadratic equation ? One is x2 = 5, Which is the same as The word 'quadratic' comes from The latin-word ~~~~~~~p

xZ- 5 = 0. Another is the equation Cardano tried to solve, namely, x2- lox + 40 = 0 (see sec. 2.1). We are sure you can think of several others.

Various methods for solving such equations have been known since Babylonian times (2000 B.C.). Brahrnagupta, in 628 A.D. approximtely, also gave a rule for solving quadraratic equations. The method that can be used for any quadratic equation is "completing the square". Using it we get the quadratic formula. Let us state this formula,

Cubk m d Biquadratic Equation

Quadratic Formula :The rolutions of tho quadratic equation ax2+bx +c = 0, where a, b, c E R and a # 0, are X =

-b +

J-4rc 2a

The expression b2 4ac is called the dlscrlmlnantof ax2

+ bx + c = 0,

Note that this formula tells us that a quadratic equation has only two roots. These roots may be equal or they may be distinct; they may be real or complex. Convention :We call a root that lies in C\ R a complex root, that is ,a root of the form a + ib, a, b E R, b # 0,is a complex root. Let us consider some examples. Example 3 : Solve

x2-4x+1=0

ii)

4x2+25=20x

iii x2-l0x +40=0 Soiutlon :i) This equation is in standard form. So we can apply the quadratic formula immediately. Here a = 1,'b = - 4, c = 1, Substituting these values in the quadratic formula, we get the two roots of the equation to be. X =

+4 - 4 1 ) ( 1 a- 4 + f i = 2 + 2 2(1)

f i and

Thus, the solutions are 2 + 6and 2 - f i ,two distinct elements of R. Note that in this case the discriminant was positive.

ii) In this case let us fust rewrite the equation in standard form as 4x2- 20x+25 -0. Now, putting a = 4, b = - 20, c = 25 in the quadratic formula, we find that X =

20+ ,/a-4(4)(25)

2 (4)

20+& --8

- -, and 2

Here we find that both the roots coincide and are real. Note that in this case the discriminant is 0.

iii) Using the quadratic formula, we find that the solutions are

By a complex root we mean a . root in C \ R

Thus, in this case we get two distinct complex roots 5 + i Note that in this case the discriminant is negative.

a and 5 - i Jz.

Solutions of Polynomial Equations

In the example above do you see a relationship between the types of roots of a quadratic equation and the value of its discriminant? There is such a relationship, which we now state. The equation ax2+ bx + c = 0, a # 0, a, b, c, E R has two roots. They are

i) real and distinct if b2 4ac > 0;

ii) real and equal if b2 4ac = 0;

iii) complex and distinct if b2 4ac < 0. Now, is there a difference in the character of the roots of ax2+ bx + c = 0 and

dax2+ dbx + dc =0, where d is a non- zero real number ? For example, if b2- 4ac > 0, what is the sign of (db)2- 4 (&a) (dc) ? It will also be positive. In fact, the character of the roots of equivalent quadratic equations i s the same. Two equations are equivalent ifone is obtained from the other by mllltiplying each term throughout by a non-zero constant. Now let us consider some important remarks which will be useful to you while solving quadratic equations. Remark 2 : a and P are roots of a quadratic equation ax2+ bx + c = 0 if any only if ax2+bx+c=a(x-a)(x-p). Thus, a E C is a root of ax2+ bx + c = 0 if and only if (x -a ) I (ax2+bx + c). Remark 3 : From the quadratic formula you can see that if b2 - 4 ac < 0, then the quadratic equation ax2+ bx + c = 0 has 2 complex roots which are each other's conjugates. Remark 4 : Sometimes a quadratic equation can be solved without resorting to the quadratic formula. For example the equation x2= 9 clearly has 3 and -3 as its roots. Similarly, the equation (x Remark 2).

= 0 clearly has two coincident roots, both equal to 1 (see

Using what we have said so far, try and solve the following exercises

E5) A quadratic equation over R can have complex roots while a linear equation over K can only have a real root. True of falre ? Why ?

E 6) Solve the following equations: a) x 2 + 5 = 0 b) (x+9)(x- l)=O

E 7) For what values of k will the equation kx2+ (2k + 6 ) x + 16 = 0 have coincident roots ? E 8) Show that the quadratic equation ax2+ bx + c = 0 has equal roots if (2ax +b)I(ax +bx+c). E9) Find the values of b and c for which the polynomial x2 + bx + c has 1 + i and 1 - i as its roots.

' ~ 1 0 ) 1 f a a n d ~ a r e r o o t s o f a x ~ + b x + c = 0 , t h e n s h o w t h a t a-+ ~ =and a p = - . a a

El 1) is the converse of E l 0. We will use it in the next section.

Let us now consider some epuations which are not quadratic, but whose solutions can be obtained from related quadratic equations. Look at the following example.

Cubii md Biquadratic Equation

Exanple 4: Solve i) 2x4+xZ+1 =O,and ii) x = $5

- 2x.

Solution :i) 2x4+ x2+ 1 = 0 can be written as 2y2+ y + 1 = 0, where y = x2.Then, solving this for y we get y =

-1

* i J i that is, x = -I f i J i ,two polynomials over C. 4

Thus, the four solutions of the original equation are

ii) x = ,/= is not a polynomial equations. We square both sides to obtain the polynomial

equation x2 = 15 - 2x. Now, any root of x = ,/= is also a root the equation x2 = 15 - 2x. But the converse need not be true, since x2 = 15 - 2x can also mean x =

-.In.

So we will obtain the roots of x2= 15 - 2x and see which of these satisfy x = = /,.

Now, the roots of the quadratic equation x2 = 15 - 2x are x = 5 and x = 3. We must put these values in the original equation to see if they satisfy it. Now, for x = - 5, ~-,/iC%=(-5)-

JETX=<-~>-S=-~O#O.

So x = - 5 is not a solution of the given equation. But it is a solution of x2= 15 2x. We call it an extraneous solution. i.e., 3 = 3, What happens when we put x = 3 in the given equation ? We get 3 = ,/= which is true. Thus, x = 3 is the solution of the given equation. Now you may like to solve the following exercises. Remember that you must check if the solutions you have obtained satisfy the given'equations. This will help you

to get rid of extraneous solutions, if any, and

to ensure that your calculations are alright.

E 12) Reduce the following to quadratic and hence, solve them. a) 4p4- 16p2+5=0 b) (5x2- 6)'14 = x

E 13) Ameena walks 1 km per hour faster than Alka. Both walked from their village to the nearest library, a distance of 24 km.Alka took 2 hours more than ameena. What was Alka's average speed ? In this section our aim was to help you recall the methods of solving linear and quadratic equations. Let us now see how to solve equations of degree 3.

3.3 CUBIC EQUATIONS In this section we are going to discuss some mathematics to which the great 1lth century Persian poet Omar Khayyam gave a great deal of thought. He, and Greek mathematicians before him, obtained solutions for third degree equations by considering geometric nlethods

Solutions of Palynomlal Equarlons

that involved the intersections of conics. But we will only discuss algebraic metl~odsof obtaining solutions of such equations, that is, solutions obtained by using the basic algebraic operations and by radicals. Let us fust see what an equation of degree 3, or a cubic equation, is. Definition : An equation of the form.

is the most general form of a cubic equation ( or a thud degree equation) over R

For example 2x3 = 0, f i x3+ 5x2= 0, -2x = 5x3 !and x3+ 5x2+ 2x = - 7 are all cubic equations, since each of them can be written in the form ax3+ bx2+ cx + d = 0,with a # 0.On the

other hand x4+ 1 = 0,x3+ 2x3= x3 x and x3+ & = 0 are not cubic equations. There are several situations in which one needs to solve cubic equations. For example, many problems in the social, physical and biological sciences reduce to obtaining the eigenvalues of a 3 x 3 matrix ( which you can study about in the Linear Algebra course). And for this you need to know how to obtain the solutions of a cubic equation. For obtaining solutions of a cubic equation, or any polynomial equation, we need some results about the roots of polynomial equations. We will briefly discuss them one by one. We give the first one without proof. Therorem 1 : The polynomial equation of degree n, ,

a, + a, x + ........+ a, xn= 0,where a,, a ,,........a, E R and a,, # 0, has n roots, which are real or non - real complex numbers. If x,, ......x, are the n roots of the equation in Theorem 1, then

( Note that the roots need not be distinct. For example, 1 + 2x + x2 = ( x + 1 )2)

We will not prove this fact here; but we will now state a very important result which is used in c the proof.

Theorem 2 ( Dlvlrlon algorithm ) : Given polynomials f (x) and g (x) and g (x ) ( # 0 ) over R, 3 Polynomials q (x) and r (x) over R such that f (x) g (x) q (x) + r (x) and deg r (x) < deg g (x).

We will also use this theorem to prove the following result which tells us something about complex roots, that is, roots that are non- real complex numbers. Theorem 3 : If a polynomial equation over R has complex roots, they occur in pairs. In fact, if a + ib E C is a root, then a - ib C is also a root. Proof : Let f (x) =a, + a,x +.......+ a,, x" be a polynomial over R of degree n. Suppose a + ib E C is a root off (x) = 0,that is, ( x - (a +ib ) ) ( f (x). We want to show that (x - ( a -ib )) I f (x) too.

Also, by the division algorithin, 3 polynomials g (x) and r (x) over R such that f(x) I g (x) deg f(x) 5 deg g (x) (see Remark 1 ).

f(x)={(x-a)2+b2}g(x)+r(x),wheredeg(r(x))<2. Since x - (a + ib) divide f (x) as well as (x - a)2+ b2, it divided f(x)- ((x - a)2+ b2 ) g(x), that is, r (x). But as deg (r(x)) < 2, therefore, r(x) is linear over R or a constant in R. So (x - (a + ib) ) can only divide r (x) if r (x) = 0 ('; polynomial R has real coefficients)

We find that

Cubic and Biquadratic Equatio~~

f ( x ) = { ( ~ - a ) ~ + bg (~x)) . Since x - ( a - ib ) divides the right hand side of this equation, it must divide f (x ). Thus, a - ib is a root off ( x ) = 0 also. Note that Theorem 3 does not say that f ( x ) = 0 must have a complex root. It only says that if it has a complex root, then the conjugate of the root is also a root. Why don't you t q following exercises now ? In these we are just recalling some facts that you are already aware of. E 14 ) How many complex roots can a linear equation over R have ? E 15) Under what circumstances does the quadratic equation over R, x2 + px + q = 0, have complex roots ? If it has complex roots, how many are they and how are they related ?

Now let us look at Theorems 1 and 3 in the context of cubic equations. Consider the general cubic equation over R. ax3+bx2+cx+d=0,af O . Any solution of this is also a solution of

and vice versa. Thus, we can always assume that the most general equation of degree 3 over R is

Theorem 1 says that this equation has 3 roots. Theorem 3 says that either all 3 roots are real or one is real and two are complex. Let us find these roots algebraically.

3.3.1 Cardano's Solution The algebraic method of solving cubic equations is supposed to be due to the Italian, del Ferro ( 1465 1526 ). But it is called Cardano's method because it became known to people afier the Italian, O h l a m o Cardano, published it in 1545 in ' A n Magna'.

Let us see what the method is. We will first look at aparticular case. Example 5 :Solve 2x3+ 3x2+ 4x + 1 = 0 Solution :We fust remove the second degree term by completing the cube in the following MY-

Flg. 1: Cardnno

1 Put x + - = Y. Then the equation becomes 2 Assume that the solution is y = m + n, where m, n E C. Then.

SoluUon~of Polynornlal Equations

5 1 ( m + n ) 3 + - ( m + n) - - = 0 4

..........(I) Let us add a further condition on m and n namely, ...a.

.(2)

1 Then ( 1 )givesusm3+n3= 4' and(2)givesusm3d=

125 -1728 '

Thus, using El 1, we see that m3 and n3 are roots of

Hence, by the quadratic formula we find that

From Unit 2 ( E 36 ) you know that a and P have real cube roots, say u and v, respectively, Thus, m can take the values u, wu, d u and n can take the values v, wv, w2v. Now, wand 02are non - real complex numbers such that o (d) = 1. 5

Also, from ( 2) we know that mn = - -,a real number. I2

Thus,ifm~u,nmustbev;ifm=ou,nmustbedv;ifm=~,nmwtbe v.Hence,the possible values of y are u+v,wu+dvanddu+ov. To get the three roots of the original equation, we simply put these values of y in the relation

This example has probably given you some idea about Cardano's method for solving a general cubic equation. Let us outline this method for solving the general equation

Then ( 3 ) becomes

P P Now put y = x + - . Then x = y - - ; and the equation becomes 3 3

. . . . . . . . .

(.- $1 (.-:I

Cubic and Biquadratic Equation

+ (r-$

= 0, that is,

--E+, where A = q - P' a n d B = 2p3 27 3 Step 2 : Now let us solve ( 4 ).

Let y = a + B be a solution. Putting this value of y in ( 4 ) we get ( a + p ) ] +A(a+p)+B=O w a 3 + 3 a p ( a + p ) + P 3 + A ( a + p ) + B-0

............(5)

e a 3 + p 3 + ( 3 a p + ~(a+B) ) + B-0 Now, we choose a and B so that, 3 a a + A = 0. Then we have the two equations

A ( a M 3 = (--)3, 3

A' that is, a 3p3 = - 27 '

and from ( 5 ) a 3 + p 3 =-B.

............(7)

Thus ,using E 11, we find that a3and p3are roots of the quadratic equation

Hence, using the quadratic formula, we find that

- + - = U, say, and v, say,

Now, from Unit 2 ( E36 ) we know that any complex number has three cube roots. We also know that if y is a cube root, then the three roots are y, w y and d y. Therefore, if a and b denote a cube root each of u and v, respectively, then a can be a, a w or aw2,and can be b, bw or b d . Does this mean that y = a + p can take on 3 values? A Note that a and p also satisfy the relations a P = - - E R. 3 Thus, since w E C, w ' ~C, 03 = 1 E R, the only possibilities for y are a+b,a w+bw2,aw2+bw. Step 3 :The 3 solutions of ( 3 ) are given by substituting each of these values of y in the

P equation x = y - -. 3 So, what we have just shown is that P P P t h e r o o t s o f ~ ~ + ~ ~ ~ + ~ ~ + r- -=, a~ a w r le+aP+d ~- - , a d + p w - 3 3 3' where o =

----

-1 + i f i , a is a cube root of 2

[ - t i = } , A = q , B .= p2 3- -

,/=I,

2p3 27

Pq 3

-- +

a isacuberootof +

The formula we have obtained is rather a complicated business. A calculator, would certainly ease matters, as you may find while trying the following exercise.

...

Soludlom of Polynomial Equations

El 6 ) Solve the following cubic equations : a)2x3+3x2+3x+ 1=O b)x3+21x+342=0 c) x 3 + 6 x 2 + 6 x + 8 = 0 d) x3+29x-97-0 e) x3-30x- 133

B' A' In each of the equations in E 16, you must have found that - + 2 0. 4 27

B~ A3 But what happens if - + - < 0 4 27 This case is known as the irreducible case. In this case ( 9 ) tells us that a3and f13are complex numbers of the form a + ib and a ib where b # 0. From Unit 2 you know that if the polar from of a + ib is r ( cos0 + i sin 0) ,then its cube roots are

rl13 (ms

0 + 2 k n +isin +

2k n), k = 0, 1,2. 3

Similarly, the cube roots of a - i b are r113 (cos

0 + 2k x - isin 3

2k 3

'), k 0 ~ 1 ~ 2 . =

Hence ,the 3 values of y in ( 4 ) are zr113ms

0 + 2k n , where k = 0, 1, 2 . 3

All these are real numbers. Thus in this case all the roots of ( 3 ) are real, and are given by

This trigonometric fromof the solutipn is due to Francois Viete (1 550 - 1603 ). Now try an exercise. El7 ) Solve the equation x3-3x + 1 = 0 So far, we have seen that a cubic equation has three roots. We also know that either all the roots are real, or one is real and two are complex conjugates. Can we tell the roots or the character of the roots by just inspecting the coefficients ? We shall answer this question now.

3.3.2 Roots And Their Relation With Coefficients.

In this sub section we shall first look at the cubic analogue of E 10 and E 11. Over there we saw how closely the roots of a quadratic equation are linked with its coficients. The same thing is true' for a cubic equation. Why don't you try and prove the relationship that we give in the following exercise ?

E 18 ) Show that a , P and yare the roots of the cubic equation ax3 + bx2 + cx + d = 0, a # 0, if and only if

Cubic and Biquadratic Equation

(Hint : Note that the given cubic equation is equivalent to a(x-a)(x-p)(x-y)

= 0.)

The relationship in El8 allows us to solve problems llke the following.

Example 6 : If a,P, yare the roots of the equation x"7x2+x-5=0 find the equation whose roots are a + b, f4 + y, a + y. Solutlon : By E 18 we know that

Therefore,(a+b)+ (P +y) + ( a + y ) = 2 ( a + b + y ) -14. Also, a +

= 7 - y , P + y =7-a,

y + a =7-P,sothat

= 50, and

( ~ + P ) ( B + Y ) ( ~ + Y ) = ( ~ -7Y - B) )( ( 7 - a ) To evaluate the expression on the right hand side, we can use ( 10 ) or we can use the fact that

Therefore, ( a + P) (P + y) ( a + y) = 2.

................(13)

Now, E 18, ( 1I), (12) and ( 13 ) give us the required equation, which is,

x3- 14x2+50x-2-0. Why don't you try the following exercise now ?

E 19) Find the sum of the cubes of the roots of the equation x3- 6x2+ 1l x - 6 = 0 Hence find the sum of the fourth powers of the roots. Let us now study the character of the roots of a cubic equation. For this purpose we need to introduce the notion of the discriminant. In the case of a quadratic equation x2+ bx + c = 0, you know that the discriminant is b2 - 4c. Also, if a and P are the two roots of the equation, then a + b = - b, a p = c. Therefore. Thus, the discriminant = ( a - P )2,where a and 3j are the roots of the quadratic equation. Now consider the general quadratic equation, ax2+ bx + c = 0. Let its roots be a and f3. Then its discriminant is b2- 4ac = a2 ( a - P )2. We use this relationship to define the discriminant of any polynomial equation. Definition :The discriminant of the nth drgree equation

a 2 ( n - I) n

l l i <nj l n ( a i - a j ) ,

where a, ....., a,,are the roots of the polynomial equation.

hlu*ons of Polynomial Equations

In particular, if we consider the case n = 3 and a, = 1, we find that The discriminant of the cubic x3+ Px2+ qx + r = 0 is

p2 B = -- - ( 2 7 +~ 4~ ~ ~where ) . A = q - -, 3 27 3 L

Now consider Cardano's mlution of the cubic quation ( 3 ), namely, x3+px2+qx+r=0.

B~ A3 The expression under the root sign is - + 4 27

-D ,where D is the discriminant. -108

Now, ( 9 ) tells us $at the sign of the discriminant is closely related to the characters of the roots of the equation. Let us l o ~ at k the different possibilities for the roots a, P and y of (3 ). 1) 2)

The roots of (3) are all real and distinct. Then ( a - p )2 (0 - y)2( P - y)2 ,that is D, must be positive. Only one root of ( 3 ) are real. Let this root be a. Then P and y are complex conjugates. :. p - y is purely imaginary :. (P - y)2 < 0. Also, a - p and a - y are conjugates. Therefore, their product is positive. Hance, in this case D < 0.

Supposea=Pandy+a.Sincea-Q=O,D=O. Also, B # 0. Why ? Because if B = 0 , then A = 0 ( since D = 0 ).

P that is a ( a + 2 y ) = ( 2 a + Y )* B u t ~ = O a q =-, 3 3 ' [ Over here we have used the relationship between the roots,

since p = - ( a + p + y ) = - ( 2 a + y ) and q = ap + Py + a y = a ( a + 2 y ) ] . On simplifying we get a =y ,a contradiction. Thus, B # 0. So, if exactly two roots of ( 3 ) are equal, then D = 0 and B # 0, and hence, A # 0. 4)

If all the roots of ( 3 ) are equal, then D = 0, B = 0, and hence A = 0. Let us summarise the different possibilities for the character of the roots now. Consider the cubic equation x3 + px2+ qx + r = 0, p, q, r E R, 2p3 - -Pq+ andlet B = 27 3

r and A = q - P' . Then 3

B' A3 1) all its roots are real and distinct iff - + - < 0, 4 27 B~ A3 2) exactly one root is real iff - + - > 0, 4 27 B' 3) exactly two roots are equal iff -- + - = 0 and B # 0. 4 27 In this case all the roots are real. B~ A3 4) all three roots are equal iff - + - = 0 and B = 0. 4 27 You may now like to try the following problems to see if you've understood what we have just discussed.

Cubic and Blquadratlc Equmllon

E20) Under what conditions on the coefficients of ax3+3bx2 + 3 c x + d = O , a # 0, will the equation have complex roots ? E 2 1) Will all the roots of x3= 15x + 126be real ? Why? So far we have introduced you to a method of solving cubic equations and we have studied the solutions in some depth. We shall study them some more in Unit 6, as an application of the cauchy - schwarz inequality. Now let us go on to a discussion of polynomial equations of degree 4.

3.4 BIQUADRATIC EQUATIONS As in the case of cubic equations, biquadratic equations have been studied for a long time. The ancient Arabs were known to have studied them from a geometrical point of view. In this section we will discuss two algebraic methods of solving such equations. Let us fust see what a biquadratic equatiop is. Definition: An equation of the form a ~ ~ + b x ~ + ~ ~ ~ + d ~ + e = O , w h e r e aE, bRanda , c , d ,#0. e is the most general form of a biquadratic equation ( or a quartic equation, or a fourth degree equation) over R Can you think of examples of quartic equations over R ? What about x4+ 5 = f i x - x2 ? This certainly is a quamc equation, as it is equivalent to x4+ x2 - f i x + 5 = 0. What about & = x4 + 1 ? This isn't even a polynomial equation. So it can't be a quartic. Let us now consider various ways in which we can solve an equation of degree 4. In some cases, as you have seen in Example 4, such an equation can be solved by solving related quadratic equations. But most biquadratic equations can't be solved in this manner. Two algebraic methods for obtaining the roots of such equations were developed in the 16th and 17th centuries. Both these methods depend on the solving of a cubic equation. Let us see what they are.

3.4.1 Ferrari's Solution The first method for solving a biquadratic equation that we will discuss is due to the 16th century Italian mathematician Ferrari, who worked with Cardano. Let us see what the method is with the help of an example. Example 7 :Solve the equation

Solution :We will solve this in several steps. Step 1 :Add the quadratic polynomial ( ax + b )2 = a2x + 2abx + b2 to both sides. We get

Step 2 : Choose a and b in R so that the left hand side of ( 14 ) becomes a perfect square, say ( x2- x + k )2, where k is an unknown. Thus, we need to choose a and b so that

Equating the coefficients of x2, x and the constant term on both sides, we get a2-5=2k+1

.........._...(15)

Solutions of Polynomlrl Equations

(15)+a2=2k+6

This cubic equation is called the resolvent cubic of the given biquadratic equation. We have obtained it by eliminating a and b fromthe equations (15), (16) and( 17 ). We choose any one root of the cubic. One real solution of ( 18 ) is k = - 1. ( It is easy to see this by inspection. Otherwise you can apply Cardano's method.) Then, fiom(l5), (16) ahd(17) we get a2=4,b2=4,ab=-4. a = 2 and b = - 2 (or a = - 2 and b = 2 ) satisfy these equations. We need only one set of values of a and b. Either will do. Let us take a= 2 and b = - 2. Step 3: Put these values of k, a and b in (x2- x + k )2= ( ax + b )2. On taking square roots, we get two quadratic equation, namely, x2-X-l= f (2x-2),thatis, x2-3x+ 1=0andx2+x-3=0. Applying the quadratic formula to these equations we get

Does Example 7 give you some idea of the general method developed by Ferrari ? Let us see what it is. We want to solve the general 4th degree equation over R, namely, ~ ~ + ~ ~ ~ + q ~ ~ + r x + s =R~ , ~ , q , r , s , ~

..........(19)

The idea is to express this equation as a difference of squares of two polynomials. Then this difference can be split into a product of two quadratic factors, and we can solve the two quadratic equations that we obtain this way. Let us write down the steps involved. Sept 1: Add (ax + b )2 to each side of ( 19 ), where a and b will be chosen so as to make the left hand side a perfect square. So ( 19 ) becomes

~~+px~+(q+a~)x~+(r+2ab)x+s+b~=(ax+b)~ .................(20) Step 2 :We want to choose a and b so that the left hand is a perfect square, say 2 P (x + - x + k12,wherekisanunknown. 2 P . Note that the coefficient of x is necessarily - , since the coefficient of x3 in ( 20 ) is p. So we 2 see that , -,

Comparing coefficients of x2,x and the constant term, we have P -+2k=q+a2,pk=r+2ab,k2=s+b2. 4 Eliminating a and b from these equations, we get the resolvent cubic

8k3-4qk2+2(pr-4s)k+(4qs-p2s-r2)=0.

From Sec. 3.3 you know that this cubic equation has at least one real root, say a.

I i

Cubic and Biqurdrrtic Equation

Then, we can find a and b in terms of a Step 3 :Our assumption was that

Now, putting k = a and substituting the values of a and b, we get the quadratic equations

P x 2 +(--a)x+(a-b) 2

= 0, and

Then, using the quadratic formula we can obtain the 4 roots of these equations, which will be the roots of ( 20 ), and hence of ( 19 ). The following exercise gives you a chance to try out this method for yourself.

E 22 ) Solve the following equations : a ) x4-3x2-42x-40=O b ) 4x4-20x3+33x2-20x+4=0 c ) x4+12x=5 Let us now consider the other classical method for solving quartic equations.

3.4.2 Descartes' Solution The second method for obtaining an algebraic solution for a quartic was given by the mathematician and philosopher Rene Descartes in 1637. In this method we write the biquadratic polynomial as a product of two quadratic polynomials. Then we solve the resultant quadratic equations to get the 4 roots of the original quartic. Let us consider an example. In fact, let us solve the problem in example 7 by this method. Thus, we want to solve x'-~x'-~x'+ lox-3=0 ............(21)

Step 1 :Remove the cube term. For this we rewrite x4 2x3as

Thus, the given equation becomes

Now, put x - - = Y .We get 2

Step 2 : Write the left hand side of ( 22 ) as product of quadratic polynomials. For this, let us assume that

( Note that the coefficients of y in each of these factors are k and - k, respectively, since the product does not contain any term with y3. )

Equating coefficients, we get

Solutlona of Polynomial Equations

Eliminating rn and n from these equations we get

k6- 13k"+40k2- 16 = 0. If we put k2 = t, then &is becomes the resolvent cubic t3- 13t2+40t- 16-0.

This has one real root; in fact, it has a positive real root, because of the following result, that we give without proof. Every polynomial equation, whose leading coefficient is 1 and degree is an odd number, has at least one real root whose sign is opposite to that of its last term. So, using this result, we see that we can expect to get one positive value oft. By trial, we see that t = 4 is a root, that is, kz = 4, that is k = f 2. Any one of these values is sufficient for us, so let us take k = 2. Then, from the equations in ( 23 ) we get

Thus, ( 22 ) is equivalent to

Step 3 : Solve the quadratic equations 1

9 yZ+2y --=Oandy2-2y4

=O. 4

By the quadratic formula we get

-2f

,Is and y = 2fJ5 .

Step 4 : Put these values in x = y +

to get the hur roots of (2 1).

Thus, the roots of ( 21 ) are :

Let us write down the steps in this method of solution for the general quartic equation

~ ~ + a ~ ~ + b x ~ + ~ x + d = O , aR ,b,c,d,~

............(24)

Step 1 : Reduce the equation to the form x4+qx2+rx+s=0.

............(25)

Step 2: Assume that ~~+~x~+rx+s=(x~+kx+m)(x~-kx+n). Then, on equating coefficients, we get From these equations we get r m + n = k z + q , n - r n = -.

Substituting in mn = s, we get

(k3+qk-r)(k3+qk+r)=4sk2,thatis, k6+2qk4+(q2-4s)k2-$=O,thatis, e+2qtz+(q2-4s)t-?

=0,puttingk2=t.

This is a cubic with at least one positive real root. Then, with a known value oft, we can determine the values of k,m and n. So, ( 25 ) is equivalent to (x2+kx+m)(x2-kx+n)=O

Step 3 : Solve the quadratic equations x2+kx+m=Oandx2-kx+n=O. This will give us the 4 roots of ( 25 ), and hence, the 4 roots of (24 ). Now, why don't you try the following exercises to see if you have grasped Descartes' method ?

E 23 ) Solve the following equations by Descartes' method: a) x4-2x2+8x-3=0

b) x4+8x3+9x2-8x= 10 C) x4-3x2-6x-2=0 d) x4+4x3-7x2-22x+24 =0. E 24 ) Reduce the equation 2x8+ 5x6- 5x2= 2 to a biquadratic. Hence solve it.

While solving quartic equations you may have realised that the methods that we have discussed appear to be very easy to use; but, in practice, they can become quite cumbersome. This is because Cardano's method for solving a cubic often requires the use of a calculator. Well, so far we have discussed methods of obtaining algebraic solutions for'polynomial equations of degrees 1,2,3 and 4. You may think that we are going to do something similar for quintic equations, that is, equations of degree 5. But, in 1824 the Norwegian algebraist Abel (1802 - 1829 ) published a proof of the following result : There can be no general formula, expressed in explicit algebraic operations on the coefficients of polynomial equation, for the roots of the equation, if the degree of

This result says that polynomial equations of degree > 4 do not have a general algebraic solution. But there are methods that can give us the value of any real root to any required degree of accuracy. We will discuss these methods in our course on Numerical Analysis. There are, of course, special polynomial equations of degree 2 5 that can be solved ( as in E 24.) Let us now look a little closely at the roots of a biquadratic equation. We shall see how they are related to the coefficients of the equation, just as we did in the case of the cubic.

3.4.3 Roots and Their Relation with Coefficients In the two previous sub-sections we have shown you how to explicitly obtain the 4 roots of a biquadratic equation. Let us go back to Theorems 1 and 3 for a moment. Theorem 1 tells us that a quartic has 4 roots, which may be real or complex. By Theorem 3, the possibilities are

all the roots are real, or

ii)

two are real and two are complex conjugates of each other, or

iii)

the roots are two pairs of complex conjugates, that is, a + ib, a - ib, c + id, c - id for some a, b, c, d, E R

Now if r,, r2, r3, r4 are the roots of the quartic ax4+ bx3 + cx2+ dx + e = 0, then

Cubic and Biquadratic Equation

Sollrtions of Polynornlal Equatlons

a~~+b~~+~~~+d~+e=a(x-r,)(x-r~)(x-r~)(x-r~)

Comparing the coefficients, we see that b r, + r 2 + r 3 + r 4= - a

This means that sum of the roots = coeff. is short for coefficient.

coeff. of x3 coeff. of x4

sum of the product of roots taken two at a time =

coeff. of x coeff. of x

sum of the product of roots taken three at a time = -

coeff. of x

c ~ e f f of . X~ coeff. of x0 constant term product of the roots = , that is coeff of xJ coeff. of x4 These four equations constitute a particular case of the following result that relates the roots of a polynomial equation with its coefficients. Theorem 4 : Let a,...... ,a,,be the n roots of the equation a, x " + a , x n - l +.....+ a n = O , a i € R V i =O,l, ..... n , a o #O.Then

Xqa,=a L n

i,j=l icj

In E 10 and E 18 you have already seen that this result is true for n = 2 and 3. Theorem 4 is very useful in several ways. Let us consider an application in the case n = 4. Example 8 :If the sum of two roots of the equation

4x4-24x3+31x2+6x-8=0 is zero, find all the roots of the equation. Solution : Let the roots be a, b, c, d, where a + b = 0.

Cublc and Biquadratic Equation

.............(27)

Finally, abcd = - 2

:. ( 2 8 ) a c d = 8

.............(29)

Now using E 11, ( 26 ) and ( 29 ) tell us that c and d are roots of x2 - 6x + 8 = 0. thus, by the quadratic formula, c = 2, d = 4. Similarly, a and b are roots of x2

- -41 = 0 .'.a =-,21 b = - -21 '

Thus, the roots of the given quartic are

Try the following problems now. E 25 ) Solve the equation x4+ 15x3+70x2+120x+64=0, given that the roots are in G.P.,i.e., geometrical progression. (Hint :If four numbers a, b,c, d, are in G. P., then ad = bc.) E 26)

Show that if the sum of two roots of x4 - px3 + qx2- rx + s = 0 ( where p, q, r, s E R ) equals the sum of the other two, then p3 - 4 pq + 8r = 0.

We have touched upon relations between roots and coefficients for n = 2, 3,4. But you can apply Theorem 4 for any n E N. So, in future whenever you need to, you can refer to this theorem and use its result for equations of degree 2 5. Let us now wind up this unit with a summary of what we have done in it.

3.5

SUMMARY

In this unit we have introduced you to the theory of lower degree equations. Specifically, we have covered the following points: 1)

-b The linear equation ax + b = 0 has one root, namely, x = -. a

The quadratic equation ax2 + bx + c = 0 has 2 roots given by the quadratic fcrmula

3) Every polynomial equation of degree n over R has n roots in C. 4) If a + ib E C is a root of a real polynomial, then so is a - ib. 5) Cardano's method for soIving a cubic equation. 6)

A cubic equation can have :

i) three distinct real roots, or ii) one real root and two complex roots, which are conjugates, or iii) three real roots, of which exactly two are equal, or iv) three real roots, all of which are equal. 7)

Methods due to Ferrari and Descartes for solving a quartic equation. Both these methods require the solving of one cubic and two quadratic equations.

Solutions of Polynomial Equations

A quartic equation can have four real roots, or two real and two complex roots, or 4 complex roots.

If the n roots of the nth degree equation a, xn + a, xn-' + .....+ a, P, p2 ......pn, then

x + an= 0, are

That is, the sum of the prodllct of the roots taken k at a time is

As in our other units, we give our solutions and I or answers to the exercises in the unit in the following section. You can go through them if you like. After that please go back to

Section 3.1 and see if you have achieved the objectives.

3.6

SOLUTIONSIANSWERS

E 1) a) This has a solution provided J # k.

E2) Let the third side be x cm. Then the other two sides are each 2x cm long. Therefore,x+2~+2~=30 *x=6. Thus, the lengths of the sides are 6 cm, 12 cm and 12 cm.

E 3) Let her rate of travel to the study centre be x km per hour. Thus, the distance from her X home to the study centre is - km. While returning, her rate is ( x - 8 ) km I hr. 3 1 X :. - (x-8) = - o x = 24. 2 3 24 Thus, the distance is -km = 8 km. 3

11 WeknowthatllO=kx 1OOOx2*k= -. 200 11 :. s = - Pt. 200 So, the required interest is 11 200

- x 500 x 3, that is, Rs. 825.

E5) True . For example, x2 + 1 = 0 has complex roots. Any linear equation ax + b = 0 over R has -b only one root, namely, - E R a

E6) a)x2=-5 *x = i & and -i&.

b) This is (x (- 9)) ( x 1 ) = 0. Thus, by Remark 2, 9 and 1 are the roots. c) We rewrite the given equation is standard from as

J5+3 JS-3 :. x = and -. 2

EI) The roots are

The roots will coincide if the discriminant is zero, that is, (2k + 6)2- 64 k = 0. This will happen when k2 10k + 9 = 0, that is, k=lork=9.

* ax2+ bx + c = 0 has coincidentalroots. E 9) By Remnrk 2, we must have x2+bx+c=(x-(l+i))(x-( 1-i)) =x2-2x+2 Thus, comparing the coefficients of x' and xo, we get b=-2, c = 2

E 10) a and p are roots of ax2+ bx + c = 0

E 11) Substituting x = a in x2- px + q, we get a 2 - p a +q=a2-(a+$)a+a~=O,sincea+ f3 =paridup =q.

Similarly, f3 is a root of x2- px + q = 0

Cubic and Biquadratic Eqllarion

Solutlens nf Polynomial Equationr

E 12) a) qp4 - 1(jp2 + 5 = 0, Put p2 = X. Then the equation becomes 4x2- 1 6 ~ + 5 = 0 . Its roots are 2+-

Jii and 2--. Jii 2

These 4 values of p are the required rqots. I -

b)(5x2

- 6)4 =x.

Every root of this is a root of ox4--5x2+6=0 Put x2= y. Then y2-5y+6=0.

Nowx2= 3

5fl 2

5*,/-

Its roots are Y =

=-= 2

3, 2.

=JS ~ Or - JS,

Putting these 4 values of x in the given equation, we find that c)

and

JZ are its solutioqs.

Separating the radicals, we get

Squaring both sides, we get 2x+3=1 + ( x + 1 ) + 2 m

Again squaring both sides, we get x2-2x-300. Substituting these values of x in the given equation, we get ,/---=I,

and,/--m=l.

Thus, both x = 3 and x - 1 are roots of the given equation.

E 13) Let Alka's rate be x krn per hour. Then Arneena's is (x + 1) km per hour. 24 The time taken by Ameena of walk to the library = x + 1 hours. Thus, the time taken by Alka =

x+l *2 hours.

(24

*x(x+1)==12 *x=-4orx=3. Since (- 4) can't be the rate, it is an extraneous qulutioc. Thl~s.Lhc required spet.4 inust he 3km per hour.

Cubic and Biquadratic Equatlon

E15) If p2- 4q<0. There will be two such roots, and they will be conjugates. E16) a) 2x3+3x2+3x+1=0

Referring to Cardano's formula, we see that in this case

0-w2 w2-w :. the roots are --1 - -1 - - ,thatis, 1

Here we don't need to apply Step 1 of Cardano's Method, since there is no term containing x2 .Now, with reference to Cardano's formula,

and p = (-171

- 172)3 = -7.

Thus, the roots of the equation are 1 -7'0-702,02-7w,thatis-6,o-7w2,w2-70. c) x3 + 6x2+ 6x + 8 = 0. He.rep=6,q=6,r=8.

and p = ( - 6 - 2 f i ) '

= -2.243

( We have used a calculator to evaluate a and p to 3 decimal places.)

Then the required roots are a+p-2,ao+pw2-2,aw?+pw-2 d)x3+29x-97=0. Herep=O,q=29,r=-97.

Then the roots are

a -+ p ? a w + p o 2 : a o 2 + p o .

solution^ of Polynomlrl Equations

e)x3-30x+ 133=0.

:. the roots are - 7, - 2 o - 5d,-202 - 5 0 . E17) x3-3x+l=O. Herep=O,q=-3,r==l.

.. A=-3,B-1.

So we are in the irreducible case.

Thus, the solutions of the given equation are 2 cos k = O,1,2,that is

2n 8n 14n 2 cos -, 2cos -, Cos -. 9 9 9

E 18) a , $, yare the roots iff ax3+bx2+cx+d=a(x-a)(x-$)(x-y)

=a{x3-(a+p+y)x2+(ap+&+ay)x+apy)

On comparing coefficients, we get

E 19) Let the roots be a,B, y Then a+B+y-6, a P + ~ + u y - l l ,a h - 6 .

;. ~ ~ + $ ~ + ? = ( a + p + y ) ~ - 2 +$y+ay) (a$ = 36-22=14.

:. a 3 + p 3 + f = - 18 Now, each of a , p, y satisfy x3-6x2+ 11~-6=0.

Thus, they will satisfy x4- 6x3+ 1lx2 6x = O also.

.: a4-6a3 + l l a 2 - 6 a = O

p4-6p3+ 11p2-6p=O y-6++11f?-6y=O. Adding these equations, we get

( a 4 + p 4 + y ) - 6 (a3+B3+y3)+ 11( a 2 +p2+f?)-6(a+p+y)=O 3 a 4 + p 4 + y = 6 (- 18)-Il(14) +6(6)=-226

Therefore, the equation has complex roots if

B~ - + - > 0, that is, 4 27 (2b3 -3abc + a2d)' 4a

(ac- b2)l > 0,that is, a6

Thus, the equation has 1 real and 2 complex roots. E22) a) x4-3x2-42x-40=0. Adding (ax + b)2 to both sides, we get x4+ (a2-3) x2+ ( 2 a b - 4 2 ) ~ + b ~ - 4 0 = ( a x + b ) ~ Assume that the left hand side is ( x2 + k )2. ( Note that the coefficient of x in the given equation is 0.)

Thenx4+(a2-3)x2+(2ab-42)x+b2-40=x4+k2+2kx2 Comparing coefficients, we get a2-3=2k

p=2k+3

Eliminating a and b we get (21)2=(2k+3)(k2+40)=2k3+3k2+80k+120

:. 2k3+ 3k2+ 80k- 321 =O. 3 is a root of this equation. With this value of k we get a2=9,b2=49,ab=21. These equations are satisfied by a = 3, b = 7.

Thus, solving the given quartic reduces to solving the following quadratic equations:

Thus, the required roots are 4, - 1 and

- 3 f i f i 2

Cubic and Biquadratic Equation

Solutions of Polynomial Equations

b) The given equation is equivalent to

33 x4-5x 3 + -x4 4

-5x +1 = o

'The resolvent cubic is 8k3- 33k2+ 42 - 17 0. One real root is 1. With this value of k, we fmd that a=O,b=O.

Thus, the given equation becomes

Therefore, the given equation has the roots 1

2, - , 2,-, that is, two pairs of equal roots. 2 2 A real root is k = 2.

Then, solving the given equation reduces to solving ( x 2 + 2 ) = *(2x-3),thatis x2-2x+5=0andx2+2x - 1=0.

Thus,the required roots are 2iJ4-20 -2fJ4+4 and , that is, 2 2 1+2i,1-2i,-1+ J Z , - 1 - J Z . E23) a)

x4-2x2+8x-3=0. Since there is no x term, we don't need to apply Step 1. Now assume x4-2x2+8x-3=(x2+kx+m)(x2-kx+n).

Thenm+n-k2=-2, k(n-m)=8,mn=-3. Thus, eliminating m and n we get k6-4k"+ 1@-64=0. k2 = 4 is a root of this cubic in k2. Thus, k = 2 is a solution. For this value of k, we get n = 3, m = -1 Thus, the roots of the given equation are the roots of x2+ 2x - 1 = 0 and

This equation can be rewritten as (x+2I4- 15x2-40x-26=0. Putting x + 2 = y, we get y4- 15y2+20y-6=0. Then the cubic in k2 is k6- 30k" + 249k2- 400 = 0, that is.

6' - 30t2+ 249k2- 400 = 0, putting k2= t. One real positive root is t = 16. So we can take k =4. Then we need to solve the quadratic equations y2-4y+3 - 0 andy2+4y-2=O Thus,y=3,1,-2f$ Thus, the roots of the given equation are ( y - 2 ), that is, 1, - 1, - 4 f$

Cubic and Biquadratic E q ~ ~ a t i o ~ i

x4-3x2-6x-2=0. The cubic in k2 is k6- 6k4 + 17k2- 36 = 0. k2= 4 is a root. So we can take k = 2. Then we need to solve the equations. x2+2x+2=o,x2-2x-1=0, x=-1 *i,l*&.

1,2,-3,-4.

E 24) Putting x' = y in the equation, we get 2y4+5y3-5y-2=0. Then, by either Ferrari's or Descartes' method, we can find the four values of y, which are -1

1, - I, -2, -.

Putting these values in x2 = y, and solving, we get the 8 roots of the given equation. Thus ,the required roots are f h,f G,f

a,f E that is,

1, - ~ , i , - i , i A -i&,

-.,

1 -i -, -.

h JZ

E25) Let the roots be a, b, c, d. Then ad = bc. Now, we know that

..........(30)

i) a + b + c + d = - I 5 * ( a + d ) + ( b + c ) = - 1 5 ii) a b + a c + a d + b c + b d + c d = 7 0

* (a+d) (b+c)+ad+bc=70

..........(31) ..........(32)

iii) abc + abd + acd + bcd = - 120 iv j abcd = 64

Now,(32j * a d ( b + c ) + $ c ( a + d ) = - 1 2 0 * a d ( a + b + c t d ) *-15ad--120

=-I20

*ad=S.Thus,ad=8=bc.

Then(31) ( a + d ) ( b + c ) = 7 0 - 1 6 = 5 4 . This, with (30) tells us that a + d and b + c are roots of x2+ 15x + 54 = 0. Thus ,by the quadratic formula,

Then, ad = 8 and bc = 8 tell us that a and dare zeros of x2+ 6x + 8 = 0, and b and c are zeros of x2 + 9x + 8 = 0.

.i i

E26) Let the roots be a, b, c, d, where a+b=c+d. We know that

ab(c+d) +(a+b)cd=r abcd = s

Solutions of Polynomial Equations

MISCELLANEOUS EXERCISES This section is optional

We have listed some problems related to the material covered in this block. You may like to do them to get more practice in solving problems. We have also given our solutions to these questions, because you may like to counter check your answers.

Let 1, o and 53 be the cube roots of unity. Evaluate a) (1 -a + d ) (1 + o - a 2 ) b) (1-0) ( 1 - 0 ~ )(1-04 ) (1-as)

Give the equations, in standard form, whose roots are

For what value of m ( # -1 ) will the equation

x2-bx

8~=

m-1

have roots equal in

magnitude but opposite in sign ? Here a, b E R and a + b # Q.

Solve

x4 + 9x3 + 12x2 ~ O X 192 = 0 has a pair of equal roots. Obtain all its roots.

If a, b, c are the roots of x3 px2 + r 5 0, find the equation whose roots are

7) Solve x4- 4x2+ 8x + 35 0, given that one root ir 2 + 8) From the cubic whore rootr are a, b, c, where a+b+c=3,

a2+ b2 + c2= 5, and a3+b3+c3=11. Hence evaluate a4+ b4 + c4. 9)

Find the equation whose roots are 4 less in value then the roots of x4-5x3+7x2-17x+11=0. ( Hint: Write the equation as an equation in ( x - 4 ).)

10) From the polynomial equation over R of lowest degree which is satisfiedby 1 i and 3 + 2i. Is it unique ?

( Hint :Note that in this equation the coefficients of XI and x4 - r are the same "r = 0, 1,2,3,4,. So we can divide throughout by x2 and then write the equation as a

1 quadratic in x + - = Y say. Now solve for y, and then for x.) X

Solutions of Polynomial Equations

If you are interested in doing more exercises on the material covered in this block, please refer to the book 'Higher Algebra' by Hall and Knight. A copy is available in your study centre.

Solutions 1)

a) W e k n o w t h a t l + o + d - 0 1: :. 1 - o + d = - 2 o a n d 1 + a - d = - 2 0 2 . :. ( 1 - ~ + d )( I + @ + & ) =(-20) ( - 2 a 2 ) = 4 o 3 = 4 , since 0' = 1. b)

(1-0)(1 - d ) ( l -w4)(1 -0') = (1 - o ) ~(1 - d ) 2 , since m3 = 1. =(1-20+0+)(1 -2w2+w4) = (-30) (-303 =9

C) 0, since 1 - 03- 0. 2)

a ) The equation is

e ( X - 2 ) (2x+3) ( x - 9 ) = 0 e2x3- 19x2 +3x + 54 = 0, in standard form.

The equation is equivalent to ( 1 + m ) x 2 + { ( a - b ) - m ( a + b ) ) x+(m- 1 ) c = 0 . The roots are equal in magnitude and opposite in sign iff their sum is zero.

:. (a-b)-m(a+b)=O,thatis, m=4)

a-b a+b

Let y =

,/::

Than the given equation i.equivalent m

3 b2+6a2 2y+-= , that is, Y ab

b 3a Its roots are -, 2a b Thus, the roots of the given equation are

b2 9a3 (&$1 and (4t), that is. -and 4a b

B) Put y =

f i .Then our equation in y is

.'.y = 1 ;. Jz.-1 5)

x = 0 is the required root Let its roots be a, a, b, c. Then, the relations between the roots and the coefficients are 2a+b+c=-9

a2+2ab+2ac+bc=12

Miscellaneous Exrrcises

a2b+a2c+2abc=80

a2bc = 192 Using ( 1), (2) and ( 3 ), we get 4a3+ 27a2+ 24a = 80. a = - 4 is a solution. Then(l)+b+c=-l,and(4)

+bc=-12.

Thus, b and c are roots of x2+x-12=0.

:. b = 3 , c = -4.

Thus, the roots are - 4, - 4, - 4,3.

We know that a+b+c=p

abc = r

---

b+c c+a a + b NOW, + -+-= c a b

bc(p-a)+ac (p-b)+ab(p-c) , using (5) abc

a3 + b 3 +c3 abc

using (6).

Now, a, b, and c satisfy x3 px2 + r .- 0. Thus, on substituting each of a, b and c in this equation, and summing up, we get a3+b3+~3-p(aZ+b2+~2)t~3r=0. :, a 3 + b 3 + c 3 = p{ ( a + b + ~ ) ~ - 2 ( a b + a c + b c ) } - 3 r

=p3-3r

Also

(e) (y )(q) =

- -r using (6) and (7) 9

=-1

IJsing ( 8 ), ( 9 ) and ( 10 ), we see that the required equation is

2 + i f i is one root. So another must be 2 - i f i . Thus,(x2-4x+7)((x4-4x2+8x+35).

By long division, or by inspection, we see that x4-4x2+8x+35=(x2-4x+7)(x2+4x+5).

Thus the other two roots of the given equation are those of x2 + 4x + 5 = 0,that is -2 * i .

Now,a2+b2+~2=(a+b+c)2-2(ab+ac+bc)

;. abc - -23 .

Solutloas of Polynomial eguatlona

Thuc a, b, c are the roots of 2 . xs - 3x2+ 2x - - PO, that is, 3 3 x 3 - 9 9 i - 6 ~-2-0. Thus a, b and c satisfy this equation, as well as, 3x4 9x3+ 6x2- 2 ~ - 0 . 3(a4+b4+c4)-9(a3+b3+c3+ ) 6(a2+bZ+c2)-2(a+b+c)=0 +a4+b4+c4=25. Put y = x 4, that is, x = y + 4 in the given equation. Then the equation that we get in y will be the required equation. Thus, the required equation is (y+4)4-5(y+4)3+7(y+4)2- 17(y+4)+11=0

w)r+lly' +43y2+55y-9=0 ,lo) If 1 - i is a root, so must. 1 + i be. Similarly, 3 + 2 i and 3 - 2i are roots of the equation. Thus, the equation of lowest degree is the quartic [ x - ( 1 - i ) ] [x-(l+i)][x-(3+2i)l[x-(3-2i)]=O,thatis, x4- 8xs +27x2-38~+26.=0. This is unique, up to equivalence .That is, any other polynomial equation that satisfies our requirements must be equivalent to this equation. 11) x4+9x3+16x2+9x+1=0

o x + ( 2

+ 9 (x + - + 16 = 0,by dividing throughout by x2. x):

Putting x +

-1 = t, wo got X

Thur, wo got two equations in x,namoly

x2+2x+ 1=Oandx2+7x+ 1= O On solving these quadratic equations, we get the four solutions of the original equation, which are 1,1,

-7f 445 2

APPENDIX: SOME MATHEMATICAL SYMBOLS AND TECHNIQUES OF PROOF To be able to do any mathematical study, you need to know the language of mathematics. In this appendix we shall introduce you to some symbols and their meaning. We shall also briefly discuss some paths that you will often take to reach conclusions.

Symbols 1)

Implication (denoted by a):We say that a statement A implies a statement B if B follows from k

We write this as the compound statement, 'A a B' or of A, then B'. For example, consider A and B; where A :Triangles ABC and DEF are congruent. B: Triangles ABC and DEF have the same area. ThenA-B In this case 'A a B' is a true statement. Another way of saying A

* B is that 'A only if B', or that 'A is sumclent for B'.

The converse of the satement 'if A, then B' is the statement 'if B, then A', that is, B a A (which is the same as A e B ). For example, the converse of (1) is 'if two triangles have the same area, then they are congruent.' While studying geometry you must have proved that this statement is false. ( For cm. have the example, the right -angled triangles with sides 2,3, f i cm., and 1,6, same area; but they are incongruent. ) Thus, (1) is true, but its converse is not.

J37

Another way of saying A e B is that 'A is necessary for B'. 2)

Two way lmplicatlon (denoted by w): Sometimeswe find two statementsA andB for which A B and B a A. In this situation we save space and write A w B. This statement is the same as:

'A is equivalent to B' ;or 'A if and only if B', which we abbreviate to 'A iff B', or 'A is necessary and sufficient for B'. For example, let A:x+2=3and B:x=l. Then A a B and B a A. Therefore, A w B. Note that for the composite statement 'A iff B' to be true, both A a B and B a A should be true. Hence the statement. 'Two triangles are congruent iff they have the same area' is a false statement.

For all / for every (denoted by t, ) :Sometimes a statehent involving a variable x, say P ( x), is true for every value that x takes. We write this statement as :

' v xP(x)or'P(x) v x', meaning that P ( x ) is a true statement for every value of x. For example. let P (x )be the statement ' x > 0'.

There exist /there exists (denoted by 3) : If a statement depending on a variable x, say

P ( x ), is true for some value of x, then we write 3xsuchthatP(x). This says that there is some x for which P ( x ) is true.

Solutions of ?oIynomiaI Equations

For example, 3 x E R such that x - 3 = 2. Now, consider the two statements QXE

........(3)

R 3 y e Rsuchthat x=2y,and

Is there a difference in them ? What does ( 3 ) mean ? It means that for any real number x, we can find a real number y for which x = 2y.

In fact, y

- I. X

servos the purpose.

Now look at ( 4 ). It says that there is some real number y such that whatever real number x we take, x = 2y is true. This is clearly a false statement. This shows that we have to be very careful when dealing with mathematical symbols. So far we have looked at the meaning and use of some common logical symbols. Let us now consider some common tef hniques of proof.

Methods of proof In any mathematical theory, we assume certain facts called axioms. Using these axioms, we arrive at certain results ( theorems ) by a sequence of logical deductions. Each such sequence forms a proof of a theorem. We can give proofs in several ways. 1)

Direct proof : A direct proof, or step in a proof, takes the following form: A is true and the statement 'A

* B' is true, therefore B is true.

For example, ABC is equilateral and ( If a triangle is equilateral, then it is an isosceles triangle.), therefore, ABC is an isosceles triangle. One kind of result that you will often meet in this course and other methematics courses is a theorem that asserts the equivalence of a number of statements, say A, B, C. We can provethisbyprovingA*B,B=~sA,A*C,C*AandB*C,C* B.But,if A 3 B and B 3 C are both true, then so is A C. So, a shorter proof could consist of thestepsA*B,B*C,C*A.WewritethisinshortasA*B* C-A.

* *

Or, the proof could be A C B A. Thus the order doesn't matter, as long as the path brings us back to the starting point, and all the satements are covered.

Whenever you meet such a result in our course, we shall indicate the path we shall follow. 2j

Contrapositive Proof :This is an indirect method of proof. It uses the fact that ' A B' is equivalent to its contrapositive, namely, 'not B not A', that is, if B does not hold, then A does not hold.

(For example, x = - 2

* x2= 4 is equivalent to its contrapositive,

x2 # 4 *x #-2.) Sometimes, it is easier to prove the contrapositive of a given result. In such situations B', we we use this method of proof. So, how does this method work ? To prove ' A prove 'not B not A', that is, we assume that B does not hold, and then, through a sequence of logical steps, we conclude that A does not hold.

Let us look at an example. Suppose we want to prove that ' If two triangles are not similar, then they are not congruent'. We prove its contrapositive, namely, ' If two triangles are congruent, then they are similar', which is easy to prove. 3)

Proof by contradiction: This method is.also called reductio and absurdurn, a Latin phrase. In this method, to prove that a statement A is true, we start by assuming that A is false. Then by logical steps we amve at a known falseptatement. So we reach a contradiction. Thus, we are forced to conclude that A cannot be false. Hence, A is true. For example, to prove that f i tz Q, we start by assuming that f i E Q. Then

f o r s o m e p , q ~ Z , qi O , a n d ( p , q ) = l .

Appendix : Some Mathematical Symbols and Teehniquec of Proof

Letpa2m Then 2q2 *p2 = 4m2. q2 .=2m2 a 2 1 q232 1 q. Thus 2/p and 2/q and hence 2/(p, q), which is not possible because we assumed that ( p, q ) = 1.

Thus, we arrive at a contradiction. Hence, we conclude that f i g Q. Proof by counter-example: Consider a statement P ( x ) depending on a variable x. Suppose we want to disprove it, that is, prove that is false. One way is to produce an x for which p ( x ) is false. Such an x is called a counter-example to P ( x ). For example, let P ( x) be the statment 'Every natural number is a product of distinct primes' Then x = 4 is a counter- example ,since 4 E N and 4 = 2 x 2 is not a product of distinct primes. Jn fact. we have several counter- examples in this case. This is not always the best method for disproving a statement. For example, suppose you want to check the truth of the statement 'Given a, b, c E Z, 3 n E Z such that an2 + bn + c is not a prime number'. If you try to look for counter- examples, then you're in trouble because you will have to find infinitely many - one for each triple ( a, b, c) E R3. SO,why not try a direct proof, as the one below. Proof :For fixed a, b, c E Z ,take any n E Z and let an2+ bn + c = t. T h e n a ( n + t ~ ~ + b ( n + t ) + c = t ( 2 a n + b + a t1),whichisapropermultipleoft. + Thus our statement is true.

UNIT 1

SYSTEMS OF LINEAR EQUATIONS

Structure Introduction

1.1

Objectives

Linear Systems

1.3

Solving by Substitution SolvingbyElimination

1.4

1.1

INTRODUCTION

In the last unit we introduced you to polynomial equations in one variable. In this unit we will start by considering linear equations in one or more variables. After that we shall consider ways of obtaining common solutions for several such equations. We call a set of linear equations a system of linear equations. Such systems of equations can arise while studying many practical problems. These include studying oscillations, the flow of currents, migration patterns, chemical contents of various solutions, input-output models of industrial production, and so on. Therefole, it is important that you spend some time studying them. The fust definite t r a c ~of systems of linear equations is found in Chui-chang Suan-shu, that is, Nine Chapters on the Mathematical Art. This is an ancient Chinese mathematical text which was probably written in 1100B.C. Much later, in the third century B.C., the Greeks used some methods for solving certain systems of equations; Further notable developments in this area of mathematics took place in the 17th century. The Japanese mathematician Seki Kowa (around 1683) contributed greatly to the theory of systems of linear equations. About the same t h e the European mathematician Leibniz also discovered a method for solving systems of linear equations. In the next century the mathematic& Gauss and Cramer published methods that use the concepts of matrices and determinants for solving simultaneous equations. In this unit we will discuss two methods for solving systems of linear equations. We will explain the method due to Cramer in the next unit. Let us now list the objectives of this unit.

Objectives After studying this unit you should be able to : obtain the solution set of a linear equations in one or more variables; deftne a system of m linear equations in n unknowns; apply the methods of substitution and elimination for solving simultaneous linear equations; choose the appropriate method, of the two methods discussed, for solving a given linear Let us now start our discussion on linear equations.

Equation8 and Inequalities

1.2

LINEAR SYSTEMS

You know that the most general forin of a linear equation over

in one variable x is ax + b = 0,

a, b E R, a # 0. You also know that this has a unique solution, namely, x = - - . a Now, can you think of a linear equation in two variables? What about 2x + 5y + 5 = O? According to the following definition, it is linear in two variables. Definition: A linear equation in two variables x and y is an equation whi~hcan be written as

where a, b, c E R and a and b are not both zero. For example, 1 - x + - y = 0, x = 25 and 2s - 4t = 2 are linear equations in two variables. 2 What about xy = 1, the equation of a hyperbola? Is it a linear equation in 2 variables? It is not, since x and y are both variables; and hence, it is not of the form

Try the following exercises now.

E 1) Which of the following equations are linear in 2 variables? Can you explain why ?

c) J; + v = 2, where u and v are variables.

E 2)"Every linear equation in one variable is also a linear equation in two variables." Is this statement true? Why or why not?

Now, what would any solution of the linear equation 2x + 3y + 1 = 0 look like? It would consist of an ordered pair of real numbers say (a, b), such that 2a + 3b + 1 = 0.For example, (1, - 1) is a solution, since 2(1) + 3(Ll)+ 1 = 0. 1 You can check that - ad 0) are also solutions.

(i* i) (-

In fact, the given equation has infinitely m h y solutions given by (X* - ( 2 i + I)) ,as x varies in R. How do we get this general form of the solution? We can rewrite the equation as - (2x + 1) . Then, for any value that we give to x ,say x = a, we get a corresponding value Y' 3 - (2a + 1) for y. T ~ U S(al , - (2: + I)) is a solution a E R ~ o t that e the solution set is a 3 subset of It2.

Now, we could also have rewinen 2x + 3y + 1 = 0 as x = - '3Y + I) . hen the solution set 2 would have been hecause from -

{(-(3i + Y) Y R} .Are the two solution sets different? Not at all. If P

+ 1) (3y + 1) = r: we get - (2x T = v throu~hsuitable arithmetic owrations.

This shows us that we can either obtain the solution set in terms of x or in terms of y.

Systems o f Linear Equations

Now, consider the equation x - 2 = 0 as a linear equation in two variables. What is its solution set? Whatever value y takes, x will always have to be 2. Thus, the solution set is ((2, y) 1 y E R) . It is the set of all points on the line x - 2 = 0 (Fig. 1). In fact, any linear equation in 2 variables can be geometrically represented by a straight line in the ~y-~lane. Now let us define a linear equation i n n variables, where n E N. Definition: A linear equation over R in n variables x,, x,, ..., x,, has the general f o m a l x l +a2x2+...+a,,xn+b=0. where a,, a,, .........,a,,, b E R and not all of a , , a,, ........a, equal zero. Thus, 2x + 3y = 1 l z is a linear equation in 3 variables x, y and z. What does a solution of this look like? It will be ordered triple of real numbers that satisfies the equation. For example, (0,0,0) and (22,0,4) are solutions. But, (I, 1, 1) is not a solution. Let us see what a solution of general linear equation looks like. Definition: An n-tuple (b,, b,, ......, b,,) in R"is called a solution of the linear equation alxl + a2x2+ ...+ allxll= b, if h;b, + a 2 b 2 +...+ anb,,=b. In this case we also say that x, = b,, x, = b,, ....., xn = bn satisfy the linear equation. Note that the first element of the n-tuple is substituted for the first variable, the second for the second variable, and so on. Remember that a linear equation in two or more variables has infinitely many solutions. In general, m linear equations in n variables has infinitely many solutions, if m < n. Now why don't you see if you have absorbed what we have done so far.

E 3) Which of the following are solutions of 3x - 2y + 5z = 80?

E 4) Find the solution set of x = y. Also give its geometrical representation.

Studying only one linear equation at a time has been found inadequate for interpreting and solving real-world problems mathematically. The mathematical models of many problems consist of a set of several linear equations which need to be solved at the same time. For example, suppose the Indian Government has to suddenly send supplies of blood, medical kits, food and water to a quake-hit area. It knows the volume and weight of each unit of these items. It also knows that each aeroplane can take a maximum capacity of 600 cubic nietres and q maximum weight of 20,000 Kg. These facts, put together, lead to the two equations

where x,, x2, x,, x4denote the number of containers ofblood, medical kits, food and water respectively. We need to find common solutions to both these equations so as to get the amounts that can be sent. In other words, we need to solve these equations simultaneously. That is why we call such a set of equations simultaneous lines! equations. Definition: Any finite set of linear equations is called a system of linear equations, or a linear system, or simultaneous linear equations.

Fig. 1 : x = 2

Equations and Inequalities

You have just seen one example involving emergency airlifting. For another example, consider the three equations. I

They forma ~inear's~stem. This system is the mathematical formulation of the following problem: A company produces 3 products, each of which must be processed through 3 divisions, A, B, and C. The number of hours taken by each unit of the product in each division, and the total number of hours available for production each week is given in Table 1

Table 1 -

Product Division A

B C

2 3 4

3.5 2.5 3

3 2 2

Total number of hours per week 1200 1150 1400

What is the number of units of each product that should be produced so as to exhaust the weekly capacities of the 3 divisions? How is the system(1) obtained from this problem? Well, if x, y andz denote the number of units of each product, we get the system (1).

ln the following exercises you can see some more examples of linear systems arising from practical problems.

. E 5) A dietitian is planning a noon meal for school children. It consists of 3 food types. He wants to ensure that the minimum daily requirements (MDR) for 4 vitamins are satisfied.

In Table 2 we summarise the vitamin content per unit of each food type in milligrams, and we give the MDR.

Table 2 Vitamin contentiunit (in mg.)

Food 5 p e

1 b

3 5 2

1 7 3

0 2 0

1 6 2

MDR

2 3

What is the mathematical formulation of this problem?

E 6) Thirty litres of a 50% alcohol solution are to be made by mixing 70% solution and 20% solution. We want to know how many litres of each solution should be used. Translate the problem into a linear system. Let us now discuss what the set of solutions of a system of linear equations looks like. Consider the following liear system in one variable:

32, a # 0';

where a, b, c,

- 2 -

. ,

This will have a solution if and only if the two equation@ave a,cpmmonsolution, that is, b d b d - * iff- - = - - .And the%-i = - '(or - ) is the uniq* solutim. a c a . . c < ..a

I '

For example, the systerp x+l =o 3x+3 * o

i' "

- ,. ..-1

7 1, whik tbesystem

has the.l.lnique s o h t i ~'k n 0 3x 2x+5 =O

t b

has no solution. . .

Now consider the !ystem x+2y=5 . , x+y=3

5,.

:....(2)

.(*

From the second equation we get y = 3 - x. substihtinikii value in the first equation we get

, Theny=3-i=2

' ,#',

So, (2) has a solution, namely, x = 1 and y = 2, that is, the ordered pair (l,2). Now, recall that the solutions of a linear equation in two variables correspond to the points on the line representing h e equation. Thus, the solutions of (2) would correspond to the points of intersection of the two lines representing the two equations. From Fig. 2(a) you can see that they intersect in only one point, namely, (1,2). Thus,(2) has a unique solution.

(a)

(b)

-,.I!?&L~-

,%"

~ .(c)

--j ."- - -

- d k ~n s :

n~ y e t i o n . .

Now consider the system x+2y=5 -- -- .2* y4ygm

(c)

... ;! t .'

: -.!"31 YOUean&ck?hat-farany FR&=oidered pair (3 - fy,j.7 is a sblution of (3). Thbs, his *systemhas infinitely many solutions. e n " - .

fr;

[ fi ;p

Geometrically, since both the equations of (3) are multiples of eachother, they represent the same line in the plane (see Fig. 2Cb)). Thus, every point on the line is a common point. Hence >: , the system (3) has infmitely my'conimon points. 4 ,

Finally, consider the system 3

,~<,:bk+,ym&

x+y=4

::y!j

nj - .tf, a i r ! :; : d t < j 4 , , t ~ : , ; < , i i l t '

, .

.-,

,,'8

....(4)

' i

Equations a d InequaJities*

YOUcan see that this system of equations has no solution, since any sohtion would lead to the false statement 3 = 4. Geometrically, the two equations of (4) represent distinct parallel lines (see Fig. 2(c)). Thus, they have no point of intersection. So you have seen three situatidns, riahelp,

a linear system can have a unique solution, or

ii) a linear system can have infinitely many solutions, or

6) a linear system can have no solution. Innfact,these are the only situations possible fix my system of linear equations. We shall not

. prove this statement here. Now let us go back to a general linear system. We give the following definition. 4

Definition: If a systenl of linear equations has a soiution, we call it consiste~~t; otherwise we call itinconsistent. I

Thus, (2) and (3) are consistent systems, while (4) is not. Why don't you try the lollwin exercise now?

, P:

E 7) Give the geometrical view'dthe following system of equations. Hence find out which of them are consistent. , a). x + y = 3

I I

x=i, 1

11'

y"Q3

7 6

<. .

b) x + y = 2

2xi2y=10 x=y c) Z x + y = O I

3x-~=O x-y=O

d) x = 3 y=4

*Nowlet us discuss a method of solving a kystem of linear equations.

1.3 SOLVING BY SUBSTITUTION

I -

Let us consider the following system of linear equations in one variable: 3%+'5=d , 6x+10=0

.:............(5)

5 From the first equation, we get x - - . Substituting this value of x in the second equation, we 3 ., get , , I

- 5 + 5 = 0, a true statement.

Systems

Thus, the equations in (5) are consistent. and the unique solution is x = - - . The method we 3 have just used for solving (5) is called Ge substitution method. Let us see how this method can be used for solving linear systems in two variables. Consider the system 2x+y=7 5x+3y=18

.....(6)

We want to solve the equations in (6) simultaneously, that is, at the same time, by substitutiw. For this we first write one variable in terms of the other by using either of the equations. We will use the first oneto write y in terms of x, as y = 7 - 2x.

Then we substitute this value of y in the second equation, to get 5x + 3 (7 2x) A 18, thau is, 21 -x= 18. /

This gives x = 3. Substituting this value of x in y = 7 - 2x, we get y = 1. But, is (3, 1) a solution? We kust double check by substituting these values i ~ ( 6 )We . get 2 x 3 + 1 x 1 = 7, which is m e , and (5 x 3) + (3 x 1) = 18, which is true. Thus, the system (6) has the unique solution (3, 1). 18 -3y We can also solve (6) by using the second equation to write x = -' . Then 5 18 - 3y substituting in the fist equation, we get 2 + f = 7, giving y = I.

(-;3-)

18-3y 18-(3x1) And then x = -= 3. 5 5 To get some practice in solving by substitution, try the following problems. , I

E 8) Find solutions (if any) of the following sets of simultaneous equations by the substitution method.

a) x + y = - 2 y=3

The sul>stitutiohmethod that wehave employed for two"equations in two unknowns can also be extended for solving several equations in several unknowns. But it becomes more and more difficult.to apply as the number of equations and variables increase. In the next section we will

VW. ,.

.--.---.-----..--..-

SI -7

. I'

1.4 SOLVING BY ELIMINATION

r : Thk~fncrthddi~f sdlving s i m u ~ dtinem l ~ cg~ra&h$.isrdrPeto.the p a t Ghrman mathematician Carl Friedrich G w s 1,4777- 1855). Qecauseof his impense contribytion to the 'i ' c el, 'I developmsnt of mathematics, hd ~~%o~h~ki&~'~'prince';bf $atfiernatic$?ii inetlio'dbf : 1

'ix

me$od we use multiplication add addition to,eliminatethe variables, one by wg,fraptb&.' equations. At each stage we transform the system of equations into an equivalent one.

' i

- - Any of the foll6wing tramformWionsare allowed:

. \. \

' . '> :*\#''

i3-4:.:.> ;

* --

q,habogingthe order of the equations of t)le system;

t 1) , . i I' , , rr . . ,, Y F :L I ,. . -99-44 2) ' d h l ~ ~ @ ~ ~ & @ g f & f ~ ~ ~ ~k ~ sysmrn~by e q & i ~ la nawarc~ ~ & f maimxr&% L

. *

-;-

-I

ib'14.sJ

I , -

a--

Fig. 3 : Gauss in 1803

- r '

(

. - - ..

..,>.

3) replacing an equation'by thk sum of that equation and a non-zero multiple of another ' &,A in-tht"sp&&., ' ' . ' 2 ' ." ,'. ., . t , ' .'; < ,: )

Two systems of equatrons are equivalent if they have the same solution set.

Let's work out a simple example, using this method. Consider the system 2

...(7)

x+2y + x = 4 3x-y-4z=-9

!:.(@,

. *

Now let us eliminafey from (7)'hnd (8).For Sis we add(7)m Zrimes (8). We get " ' . .

Dividing throughout by 7, we gtt

Substitutingthis value of z in (111)we get

x = - 2 + 1=-1. Substitutingx =- 1, z = 1 in (9),we get y=2. We must verify if the ordered triple (- 1,2,1) satisfies all three equations.

On substituting this triple in each of the equations, we find that it is indeed the solution. *

whenever we use this method, or any method for solving a linear system we must keep the , following remarks*mi n d . c

Remark 1: wenever we solve an equation or a system of equations, we must'trhVhysZvefify our solution. I

.,.

Remark 2: m i l e solving a linbar . . system, if &eiei&a false.statement, it rnegns d that , . ,.the ' .. ... . !'. system:.h k n o solution. <

! ".

Now why don't you try to solve some linearsystems.

E 10) Solve

system that you got in E5, by elimination. ,l

,b;r'..:

. *

;

- E 1 1) D ~ r ~ n n i nby e , elirnhtiion, thesolution set of the system

In E 10 and E 11 you came naross system in which the humber of equations wps;ipdre &n the number of variables. In such a situation also the system can havea uniqie solbtion: infmitely many s~lutionsor no solurion. '11 ~

.")

not ~ ~ ~ . ~ q ~ , ~ + t i R p r ; ~ ~ j. ..f , r iywimW y i scti~ttor b , itwill q . kt

. so/?j?~np.~!;;.::i3;>Ii), .(.;:

F;yj? .$,. ?! :; ;<.:

; ( f A ;ii,

,&'.<)Ji$;;:.;1

.,. t i r . ,

: :.," .., ..,!;:

::: .,:>> ::.L,,:

.:i,,

'. ?+*,I. .>.

Let us consider the folI6Umg e~a&"l&""'~'"

-:.I:.,..*. , ? . . ;

'*'

. .. , .,,? , ..

3 * 3

, ,

5 , . : , ,*.. , : y .

::j;)~:,,::.

,:.:

,;:,

?<::?::.;\

~ . . ,f d'

w e can't eliilinate any myc( 1 1v, ~ p a;;:.::.::. y eks a? -l, Cq; e;. p k y i,e. q, (mypd.?$$/J+~.nd4hq,yrn. : systemwill only end in reuiirducing x.lnstead, we use (11)to.write y in terms of z. ,:!,d*~

; ,

;

. ,,

. : .

.A-,.

; ,

;

. . 2

For example, when z = 0 we ger a solution Gii, 4,&)ymd wben.~%;t we~get:acWtibnf, I ;

, 5' 5 9'

1) and so on. ~ h u sthe , given ~AMSW

5 -22 20- 32

w e say thsi the f

Now consider the systern . : i , " . - . , - . v

;,,-,Ls,

!,l2~ ,?<,[i ;.;i . ); . '

,-xI2jt+~.-l

. .

~31Witifii!fj%ari$ &liihai& '' '-' y ~ i i s ~ p r M t r a q r R B ~a a ~ b r

: , . , ..,<4.).!;j;ylr,!i, +:.: ? ? , .i;i:2[f,[ A

;

,. . ..... . !

r ~ t r i ,

; : ,!',

. j ; ; , , , j o. ; ; ;i

,,:-*i \.-

,, > 3 -<. :; . - ;> , s j t'! ,

.. !,!;it-;i i; q ( j ; :;..,, y;.;::!! r;j:L , fi

.;: <.,;

:.; ..,,,. , .: i.:. : i.:..:, >,

x+2y+z=-1

...(IS)< ,:

, -

...(16) :A

15)7 (16) s 0 = 2, a false stater,nent. ,

.~ ' .. . , ,

- ,

70,.

3 # 8 '

. ,' If. r T j < * , . * y : ~ *.* t=

EqlPations rod ~ ~ g u a ~ i t i e sNOW why don't you practicBthe elimination method some more?

E 12) Solve the system you got in E6, if it is consistent.

E 13) Solve the system (1) that we gave at the beginning of Sec. 4.2. E 14) Solve the system

E 15) Solve the systems a) x + y + z = O y+2z .=3 4

b) x + y + z = O y+2z =3 z =4

So f a wk have discussed two methods of solving lineai systems.'In the next unit we will consider yet another method, which is specifically meant f ~a rsystem of linear equations in which the nuinber of equations is the same as the number of unknowns. r

Let us now sumparise what wu have covered in this unit. I

1.5 SUMMARY

In this unit we have discussed systems of linear equations. In pa&u~ar yoli studied 1) what a linear system is andkow it can arise from practical problems.

2) that a linear system can have a unique solution, infmitely many solutions or no solution. 3) the substitution method for solving "small" linear systems simultaneously. 4) the Gaussian elimination method, which is the methog that is the most widelyped. have tried all the exercises in the unit. You may like to see what out . .We hope that solutions t 0 . b are.

1.6 SOLUTIONS/ANSWERS , 4.

E 1) (a) is not, since the quadratic term xy occurs in it. (b) is not, since the quadratic:&Iy2 occurrs in i t

1 a

(d)ds linear, sioce it k equiWalent to the linear equation 3x + 2y 4 =6. E 2) It is true because any linear equation in one variable is ax + b = 0, a c O . This is ecpivalent' to ax + 0 y + b = 0, a # 0, which is linear in two variables.

-4

)

E 3) (b) and (d) are, (a) e R ', and hence can't be a solution. (c) is not, shce 3(2) 2(3) + 5(11 . 80.

i I i

E 5) Let, x, y, z d&te the units of each food type.Then 3x+5y+2z =55 x+7y+3z =45 2y 110' x+6y+2z =45

Systems of-Llnear Equptlons

-1

E 6) Say, we take x litres of the 70% solution and )Ilitres of the 20% solution to make 30 litres of . the 50% solution. s

'O 20 Then -X+-y=-X 100 100

100

30 3 7x + 2y = 150.

Also x + y = 30.

FIE. 5 : An laconsistent system,

Thus, the problem reduces to solving the linear system ?x+2y=150 x+y=30

YA \

E 7)a) From Fig. 5 you can see that there is no point common to all three lines. Hence the system is inconsisient. b) We have given the geometrical representation of this system in Fig. 6. Again, you can \ see that the system is inconsistent. , k a

C) In Fig. 7 you can see tharthe three lines have a unique point of intersection, namely, ' (0,O). Thus, the system has the unique solution (0,O).

\ b

Fig. 6 : An inconsistent aystem

d) From Fig. 8 you can see that this system has the unique solution (3.4).

E 8) a) The wand equation says y = 3. Substituting this value in the first equation, wegetx+3=-2*x=-5.

:. (- 5,3) is the golution.

*, , b) a + 3 b = 13 *a= 13-3b. ;. 3a+7b=33*3(13-3b)+7b=33*2b=6?b-3. ;. a = 13-3(3)=4. :. (4,3) is the solution. r

c) .2s+t=20&t=20-2s .

2s-5t=30*2sT5(20-2s)=30*s-

I ,

(q - $1 ,

. .Fig. 7 :

is the solution.

Note that the second equation is equivalent to the fmt one. Thus, any solution . of the sysfem is a solution of x + y = 2.

Thus, for any value of x E R, (x, 2 - x) is a solution. For example; (0,2) is a solution. This system has id3nitely many solutions.

e) 3x=y+S*y=3x-S ;. 9 + y = 3 ~ * 9 + 3 ~ - 5 = 3 ~ ~ 4 = 0 , a f a l s e s t a t e m e n t .

Fig. 8 :

This the system is inconsistent. E9) a) 2 x + y + ~ = 9 -x-y+z =1 , ~x-Y+z=~ ,. To eliminate y we add (17) and(l8). We get

ii7) ...( 1s)

....(19)

TO eliminate z wc subtract

( ~ $ifd&-(ig) j Gt- ,c.gtt?#,

. :t ?

4x=8,thatis,x=2. Substituting this valve of x in (ZO), we get

2~=10-2=8*~=4. Then (17) gives us 4$2)+y+4&4..9*l;'y%1: ' Thus, (2.1.4) is the solution. (Verify this!) I

. r :

't x

-1

j;>:;

,',!,>.:;j,

>.:. 'i"

;,:f.,:,,L.,'?

2.0

.. ,.

::-ti

.,r,,,

. ,

. , >, l.

. : ! > h ; ! i ,-?.{?*!*

:. '? ,, . -$

, ,

.I , > ' L ).T , * Now we need to check if (f5,5,0) satisfies all the equations in the system. It doesn't satis@ (11). But our calculations have been right %

Conclusion: the system io'&cons;itent!

'%-

t :

The dietitian will have' to qlter his constraints!

..'.

1'"

*,. . I

E 12)The system is , ~

7x+2y7 =I50

3 ;

x + y =30. It has a uniquesolution, &ly,

(18,'12$

,,.,, . $,. ;:

, , $ , . ~ , ~ , ' j ~ ~ l j ~ ~ : c1 ; , , :

I ,

- !+

b) (1, - 5,4) is & h i q q solution. ,,<.

: i

.,'

;,:I .?....;;.::.

. =- : -

-. , -.-

,+:.

.J+

. %

.. $

- .

,'- .

,,,:: ;

<:;.;

1 , i >{! j

_ . .

;.?

.:,.:,, .* ;{!3$>:t

I -

d i . C :.,::,

." 0,, s.. -,-.. :*;

: .

,- ..

,. "..,. ~ " ' ,, >;; :.,?.. ,..,,, "$.', ; ::ij ; ; , / j , ; ,

.-L.

;

:-jI~,<,!;],~!:e{<!

UNIT 2

CRAMER'S RULE

Structure 2.1

Introduction Objectives

2.2

What is a Matrix ?

2.3

Determinants

2.4

Cnuner's Rule

2.5 2.6

Summary Solutions 1 Answers

INTRODUCTION

2.1 ,

In the previous unit we introduced you to linear systems and two methods for solving these. In this unit we discuss a method for solving a particular type of linear system. We shall first briefly introduce you to an effichnt notation for dealing with systems of linear equations, namely, a matrix. You can find a detailed study in our course on linear algebra. After that we shall explain a concept that is intimately linked with a certain type of matrix, and hence with the solution of certain systems of linear equations. The concept is that of a determinant, which seems to have been first used by ancient Chinese mathematicians for solving simultaneous linear equations. In 1683 the Japanese mathematician Seki Kowa started developing the theory of determinants for the same purpose. About the same time the German mathematician Leibniz also defined detcrminantsand developed their use for solving simultaneous linear equations. So, you can see that mathematicians through the ages and around the globe have felt that determinants are very important. Nowadays scientists and social scientists also increasingly feel the need to understand and use this concept. We shall end this unit with a discussion on a method which uses determinants for solving certain systems of linear equations. This method is due to the eighteenth century mathematician Cramer. It only applies to some of those linear systems in which the number of variables equals the number of equations. 'Let us,now list theobjecti..--7 of this unit.

Objectives

A .

After studying this unit you should be able to define a matrix, and a square matrix, in particular, evaluate any determinant of order 1,2 or 3; identify a non-singular matrix; identify the linear systems which can be solved by using Cramer's rule, and apply the rule to solve them. And now let us look at a convenient way of representing a linear system.

2.2 WHAT IS A MATRIX? Consider the set of linear equations

While writing (1) we have had to write each variable three times. Wouldn't it be more --L-C-,.r,..-..

:C*I.--

. -..+..L-.. *I..-.& ...,... 1.4 ..--l-l..

..-

6,.

--.-:A

&L:- --..r:r:--Q

A CI-- -11

:r :-

Equations and Inequalities

the coefficients that really matter in obtaining their solutions. Let us do away with writing x, y, z each time, and only write their coefficients in a table in the following manner :

How did we prepare this table? The first row consists of the coefficients of x, y and z, respectively, in the first equation; the second row consists of the coefficients in the second equation; and the third row consists of the coefficients in the thil-d equation. In fact, we can symbolically rewrite (1) as

Each group of numbers or variables enclosed in the square brackets is an example of a matrix. Writing systems of equations in matrix notation leads to a saving of effort in deallng with them, especially as the number of equations grows. Nowadays computers are being used increasingly for solving large systems of linear equations. Their efficiency increases considerably if matrix methods are used. You may be interested to know that matrices (plural of "matrix") were first used in 250 B.C. for solving systems of linear equations in the Chinese mathematics text "Nine Chapters on the Mathematical Art". But the developnlent of matrix theory is mainly due to the 19th century British mathematicians Arthur Cayley and J.J. Sylvester. We will discuss matrix theory in great detail in our course on linear algebra. In this section we shall only acquaint you with matrices. Let us start with the definition. Definition :A matrix is a rectangular arrangement of numbers in the foim of horizontal and vertical lines. The numbers occurring in a matrix are called its elements or entries. The set of entries in one horizontal line of a matrix is called a row of the matrix; and the set of entries in a vertical line is called a column of the matrix. For example, is a matrix with two rows and three columns, and [-I] is a matrix with one row and

L2 51 one column. i

Note that each row of a matrix has the same number of elements. Similarly, each column of a matrix has the same number of elements. This is why we say that the arrangement of numbers in a matrix is 'rectangular'. Now a few words about notation. As you have seen in the examples of matrices given so far, we use square brackets to enclose the entries of matrix. We usually denote matrices by capital letters. Far instance, the general matrix with m rows and n columns is

a 2 , a22............ A=

21,

= [a,,] , in brief.

Here a,, denotes the element lying in the 1st row and the 1st column, a,2 denotes the element

lying in the 1st row and the 2nd column, and, in general, aildenotes the element lying in the ith row and the jth column. We also say that ail is the (i, j)th entry of A.

[ -;

Thus, the ( l,3)th entry of 0

Cramer's Rule

isO,andthe(3,l)thentryis3.

3 0.3 - 1

We can also denote a matrix A consisting of m rows and n columns by A,, and we say that it has order m x n or is an m x n matrix.

,,or A'~.");

We will often refer to the ith row of a matrix, meaning the ith row from the top. Similarly, the ith column of a matrix refers to the ith column, counting from left to right. If the number of rows in a matrix equals the number of colu.ms, the matrix is called a square matrix. Isn't the name appropriate?

[::

0 0 0

Some examples of square matrices are [2], [3 0 5 ] and

:].Such

matrices are very important in matrix theory. You may like to try some exercises on matrices now.

E 1) Write the order of each of the following matrices.

Also write their (2,3)th and (1, 1) th entries and their 3rd columns, if they exist. E2)

Write down a 3 x 4 matrix in which the (i, j)th entry is 0 whenever i <j, and non-zero otherwise.

E 3) a) Rewrite the linear systems that you obtained in E5 and E6 of Unit 4, in matrix notation.

b) What would happer these matrices if the first and second equations in each of the linear systems we,: interchanged?

Now, go back to the system of equations (1). We rewrote them in matrix notation in (2). We can also write (2) in shorthand notation as AX = B, where A is the 3 x 3 coefficient matrix, X

If the number of equations in a linear system equals the number of variables, the coefficient matrix will be a square matrix. In this situation we can sometimes use the concept of a determinant to solve the system. Let us see what this concept is.

2.3 DETERMINANTS You have just seen that we can represent a set of n linear equations in n variables by a matrix equation AX = B', where A is an n x n matrix. Associated with this square matrix A, we can define a unique number - its determinant. In this section we will discuss determinants of matrices whose elements are real numbers. We will also discuss some of their properties. Let us start with a definition.

The coefficient rnatril o f a linear system is the matrix formed by taking the coeffic~entso f the equations in 'the system.

Equations and Inequalities

Definition: The determinant of a lx 1' matrix A = [b], denoted by I A 1 or det (A), is b. 1

For example, if A = [3], then I A I is 3. Similarly, if A = Now let us consider the determinant of a 2 x 2 matrix.

Definition: The determinant of the 2 x 2 matrix A =

[ ::: ] a12

is the number

a, az2- aI2a,,, and is denoted by I A 1. This is simply the cross multiplication

For example, if A =

,then(A(=OxS-2x(-l)=2

Another common way of writing 1 A ( is to write its elemenfs within parallel verticallines, instead of within square brackets. For example, we can write the determinant of

We will often see this way of writing a determinant. You may like to calculate some determinants now.

E4) Evaluate 1 1

Compare the determinants obtained in (a) and (e), and (a) and (0.

Now let us use determinants of 2 x 2 matrices to obtain the determinant of a 3 x 3 matrix.

Definition: The determinant of the 3 x 3 matrix

The determinants of A is denoted by I A I or det (A)

. where AIj is the matrix obtained from A after deleting the first row and the jth column, for

j=1,2,3. In obtaining 1 A 1, we expabded by (or along) the first row. We codd also have expanded by the second row, third row, or either of the columns. So, for example, expanding along the third column. we aet '

Crswr's Rule

where A denotes the matrix obtained from A after deleting the ith row and thefhird column. All 6 ways of obtaining 1 A ( lead to the same value. We will not prove this here. However, let us consider an example. 1 2 6 I

Let A = [I 7 3 2

]

.Then, expanding by the fvst row, we get

Now, why don't you try this exercise?

E 5) Obtain 1 A 1, for A in the example above, by expanding alongthe 3rd row and by expanding along the 2nd column. As you have seen, we define the determinant of a 3 x 3 matrix in terms of the determinants of 2 x 2 matrices. In the same way we can obtain the determinant ofany n x n square matrix (n 2 2) in tenns of the determinantsof (n - 1) x (n - 1) square matrices. Definition: The determinant of an n x n matrix A = [aij],where n > 1, is given by

where Alj= matrix obtained from A on deleting the 1st row and the jth column, tj j = i, ..., n. What we have stated for the 3 x 3 case is true for the n x n case (n 2 2). namely, that we can expand along any row or column to obtain the detenninant of A. Thus,

( A I = ( - I ) ~ ' '. a i , ) A i l ~ + ( - 1 ) ' + 2 . a i 2 ~ ~ i 2 1 + . . . + ( - l ) i + n . a i n i A tj i n Ii = 1...n, and

We call the determinant of an n x n matrix a determinant of order n (or size n). So.far we have spent some time on evaluating determinants of orders 1,2 and 3. In this course we shall not go to higher orders. (They are discussed in great detail in our course on linear algebra.) We will only introduce you to some elementary properties of determinants now. While solving E4 you may have realised some of them. These properties help us in evaluating a determinant in a shorter time. Let us see what they are. Theorem 1 :Let A be a square matrix. Then 1 A I satisfiesthe following properties. PI: If all the elements of any row or column of A are zero, then ( A I = 0 P2: If B is the matrix obtained from A by interchanging any two rows (0.r any two colunps), thenIBI=-IA(.

P3: If B is the matrix obtained from A by multiplying all the elements of a row (or all the elements of a column) of A by a number c, then ( B I = c I A 1. A multiple of a row (or df a

P4: If B is the matrix obtained from A by adding the multiple of a row to another row (or the multiple of a column to another polumn), then I B I = 1 A 1.

P5: If two rows lor two columns) of A are equal, then 1 A 1 = 0. i I

We shall not prove these properties here. If you are interested in the nroofs, you can study Block 3 of our next level course on linear algebra. In this course we shall only see how to apply

column) by a non-zero number K IS the row (or column) obtained by multiplying each of its entries by K

Equations end Inequalities

these properties. Let us first verify them in some cases.

If A = [3

[

1 -1

If A = 2

and B =

I -

3 1

,that is, B is obtained by interchanging the columns of A,

] and B is obtained by multiplying the second row of by 5, that is, A

Now, let us take an example which satisfies the hypothesis of P4. Let A = -3 [ I and B =

[ - - 1. 4

-,'I

Can YOU make out if B is related to A in any manner suggested in P4?

How have we got B?-We multiplied the elements in the second row of A by (- 1) and added them to the corresponding elements in the first row, that is, we subtracted the second row of A from the first row.

Now, I A I = (1) (- 1)- (2) (- 3) = 5 and I B [= 5 = I A I. So P4 seems to work in this case. Now, suppose we add 3 times the first column of A to the second column. We get the matrix

Then, ( C I = I A 1, verifying P4 again. Now let us see if P5 holds for a general 2 x 2 matrix satisfying the hypothesis. One such matrix is

Then I A I = ab - ab = 0, verifying P5. Now, let us try and obtain the determinant of the 3 x 3 matrix

9 -9 The third column is (-1) times the second column. So let us add the second column to the third corumn. By P4 we get

So, you have seen some ways in which P1 to P5 can be used for calculating determinants. Why don't you try this exercise now?

E6) Using P1 to P5, evaluate the determinants of ra b c l O O O ,where a, b, c, d, e, f E R a) A = [d e f J

Cramer's Rule

e) E=L:

2b 2c

ef ] , w h e r e a , b , c , d , e , f ~ R

As we said earlier, the importance of the properties P 1 to P5 lies in their use for decreasing the effort involved in computing a determinant. You must have realised this while doing E6. Let us look at another example of their use. For our convenience we will use Ri to denote the ith row and Ci to denote the ith column. . To evaluate 1 A 1 by expanding along any row or column will require us to

Let A =

evaluate 3 determinants of order 2. But, if we use P4, we can multiply R3by (- 2) and add it to ' R,,to get

ro 1

Now, R, has two zeros in it. So we can obtain I B 1, which has the same value as I A 1 by expanding along R,. This means that we only need to evaluate ( B I 21. Thus,(AI=/BI=(- l).l. lB,,(=-23. The following remark will be very useful to you for evaluating a determinant. Remark 1: Whenever you have to compute the determinant of a matrix, it is best to expand along the row or column with the maximum number of zeros. Therefore, one should use the property P4 so as to get as many zeros as possible in some row or column. You may like to use the remark above to obtain the following interesting results.

A and B are the general forms o f 3 x 3 triangular matrices over R.

where a, b, ...,f, p, q, ..., u E R. ,.,

Show that a) 1 A I = adf, that is, the product of the elements lying on the principal diagonal b) ( B ) = p s u . r a 0 01

C is the general 3 x 3 diagonal matrix over R.

Show that I C 1 = abc. E9)

Obtain

-9

6 12

(

E 10) Show that the analogues of E7 and E8 are true for general 2 x 2 triangular and diagonal matrices.

And 1 # O. So 1 I I # 0. Because of this property I belongs to the clasi of matrices that we will now defme. ~eflhitlon:A square matrix A is called non-singular if I A I # 0. Otherwise A is called singular.

You can check that some more examples of non-singular matrices are [5],

-O,],

[: i] and

:] [i I]

2 2 0

;and

is an example of a singular matrix.

Why don't you try someexercise now?

El 1) Which of the following matrices are non-singular? Give reasons for your choice. 1

J-2

E12) When are [a],

[: :] and

J-2 1

-1

a b c

singular? here a, b, c, d, e, f, g E R.

Non-singular matrices form an important part of matrix theory. In the next section we shall introduce you to a rule for solving any system of linear equations whose coefficient matrix is non-singular.

2.4 CRAMER'S RULE I' 7 German I

In 1750the mathematician Gabriel Cramer published a rule for solving a set of n linear equations in n unkaowns simultaneously. Though this rule is named after Cramer, it seems to have been discovered by the British mathematician Colin Maclaurin twenty years earlier. Let us see what this rule is. Consider the general system of 2 equations in 2 unknowns: ax+by+c=O dx+ey+f=O

Fig. 1 : Crrmer (1704-1752)

where ae - db # 0. Then, if you use the substitution method, what solution do you get? We get

cd af bf - ce x= a e - d b s y = ae-db Notice that this is the same as -C b

But how did we get x and y in the determinant form? What we did was to first write the system of equations as.

Cramer's Rule

Cramer's rule can only be applied if

i) the number of equations in the linear system equals the number of variables; and ii) h e determinant of the coefficientmamx is non-zero.

Equations and Inequalities

We first rewrite the system as 2~-3y+z =I x+y+z =2 3x-0.y-4z =17 This is of the form AX = B ,where

Now let us first see whether 1 A I = 0 or not. Let us expand along the fhird row. We get

So we can go ahead and apply Cramer's rule. For this we evaluate

1 1

-3

I )

D D D = 3 y = 1 z = -2 =-2. D D D On verifying, we find that (3, 1, -2) is indeed the solution of the given system.

Thenx=

You may like to try your hand at applying Cramer's rule now. E 13) Solve the following system by Cramer's iule, if applicable. Otherwise use the Gaussian elimination method (see Sec. 4.4). a) x + y + l = O 2x-y =7. b) x + y - z + 2 = 0 2x-y+~+5=0 x-2y+3z-4=0. c) 3 x + 5 y + 2 z = l 4x+y-7=0 9x+ 15y+6z=3.

E 14) Consider the n x n linear system a l l x l + a 1 2 x 2...+ + alnxIl = O

a,,lx, + an2x2+ ... + an,xll

If A is the coefficient matrix and I A I # 0, can we apply cramer's rule to solve the system? If so, obtain the solution set. E15) I have Rs. 24801- in five, ten and twenty rupee notes. The total number of notes is 290 and all the teH rupee notes add up to Rs. 601- more than the sum of the twenty rupee notes. How many of each type of note.do I have?

In this unit and previous one we have discussed three methods for solving linear systems. The examples and exercises we attempted involved a maximum of 3 equations and a maximum of four unknowns. But, in practical applications in the sciences and social sciences, one needs to solve very large systems. These may consist of 15,20 or more equations in as many or more variables. As you may have guessed, these systems require computers for solving them. Then the best method to apply is the Gaussian elimination method. We have discussed this method as well as Cramer's rule in greater detail in our course on linear algebra. Now why don't you try solving the following exercise by any of the three methods we have covered in this unit and the previous one?

E 16) Which of the following linear systems are consistent? Obtain the solution sets, wherever possible.

Let us now summarise the contents of this unit. -

2.5

SUMMARY

In this unit we 1)

defined an m x n matrix, and a square matrix, in particular.

2) 3)

introduced you to the concept of a determinant of a square matrix. discussed some p'roperties of determinants.

used the definition and properties of determinants to evaluate determinants of orders l , 2 and 3.

defined a non-singular matrix.

applied Cramer's rule for solving a linear system of equations whose coefficient matrix is non-singular.

With this unit we finish our discussion on simultaneous linear equations. In the next unit we shall look at some commonly used inequalities. But before going to Unit 6, please go back to the objectives given in Sec. 5.1 and check if you have achieved them. You may also like to go through the next section, in which we have given our solutions to the exercises in the unit. This may be useful to you for counterchecking your solutions.

E 1) Their orders are 3 x3, 1 x 1,3 x 3,3 x 1 and 1 x 3, respectively. The (2,3)th and (1, i)th entries of the fust matrix are 9 and 1. The (1, 1)th entry of the second one is 5. It has no (2,3)th entry. The (2,3)th and (1,l)th entries of the third one are go and 1. The (1.1) th entry of the fourth one is 4; it has no (2,3)th entry. The (1,l)th entry of the fifth one is 2; it has no (2,3)th entry.

Cramer's Rule

Equations and Inequalities

Only the first and third and fifth matrices have third columns, which are and [1] respectively.

E 2) The required matrix will be of the form

[: ::][:I[ii] 3 5 2

E3) a)

0 2 0

givtsusthesysteminE5.

[i :][;] [y,"] =

gives us the system in E6.

b) The first two rows of both the coefficient matrices, as well as of the matrices on the right hand sides, would be interchanged.

E5) By expanding along the 3rd row, we get

1 A 1 =(-I)~+'.7.1A311+ (-11~'~.3.1A321+ ( - I ) ~ +.2.~ lA331.

By expanding along the se~ondcolumn, we get (A1=(-I)~+'.2.1A211+ ( - I ) ~ +.4.1A22( ~ +(-I)'+~ .3.IA,,I =- 79. E6) a) ObyP1.

c) C is obtained from B above, by interchangingthe first and third rows. ICl=-IBI=O.

d) D = 0, as in (b). la a dl

E 7) a) We expand along a row or column which has the maximum number of zeros. ' So, expanding along C, ,we get.

b) We can expand along Rl to get

Note that we would get the same answer if we'd expanded along any other row or column. E 8) Expand along R, to get 1 C I = abc.

We want to make some entries zero. Looking at C,,you may have noticed that replacing

R,by R, + (-2) R , and R, by R, + 3R, will make the (2, 1)th and (3,l)th entries zero. Then, by P, we get

. where a, b, c e R. In

E 10) The geheral 2 x 2 real triangular matrix A =

both cases their determinant is ac, the product of the diagonal elements. The general 2 x 2 real diagonal matrix is C =

[iI],

where a, b E R.

( C ( = ab, the product of the diagonal elements.

[A :]

is singular, since its determinant is rem.

( [-3](= - 3 f 0; thus, [-33 is non-singular

1 -: A1

is not a square matrix, and henc'e can't be non-singular. Note that it is not

2 w

singular either, since a singular matrix has to be square too. The last matrix has zero determinant, and hence is singular. E 12) [a] is singular iff a = 0.

Ic d l

is singular iff ad -- bc -= 5.

The diagonal elements cF a:> non matrix are its ( i , !)!I> entries v i = I .... n

Equations and Inequalities

the given matrix is singular iff a = c or

E13)a)

[i:]

is singular.

x+y=-1

Thus, the solution is (2, -3).

b) The given system is equivalent to

NOWD-I:

-1

-2

= 3 # 0. So we can apply Cramer's rule. We calculate

Crsmer's Rule

7 22 Therefore,x=- j , y = ~ ~ z - 7 . c) The system is equivalent to

Note that the 3rd row of the coefficient matrix is 3 times the 1st row. Thus, its determinant is zero. So we can't apply Cramer's rule. Let.as solve the system by successive elimination. The system is

Note that ( 5 ) is equivalent to (3). So we can drop (5). Now let us eliminate y from (3) and (4). For this we calculate (3) - 5 x (4), which is

Now, we can't eliminate any M e r , between (3), (4) and (6). So let us use (4) and (6) to write y and z in terms of x.

Thus,we get a 1-parameter set of infinitely many solutions.

E 14) since ( A ('#0, we can apply Cramer's rule. Now, to apply this rule, we calculate Di

for i = 1, ..., n by substituting the ith column with the constant column

Iil.

Thus, each Di = 0, by P1 of Theorem 1.

The sol&on (0,0, ..., 0) is called the trivial solution.

E 15) Let x, y, z denote the number of five, ten and twenty notes, respectively. Then we know that

x+y+z=290 lOy-20z=60 We can solvethis by Cramer's rule to get x = 164, y=86,z=40.

Equations and InequnlitIes

E16) a) This is a 3 ,x 3 system. First let us see ifwe can apply Cramer's rule. Since the determinant of the coefficient matrix is zero. we can't apply Cramer's rule. So let us my to solve it by elimination. Adding the first two equations, we get 3x-By+ llz=9. Subtractingthis from the third equation in the-system,we get 0 = 2, a false state ment. Thus the system is inconsistent. b) The given system is 8

2x-y+3z-5w

...(7)

- 7 y + 3 ~ - 7 w =-I3

...(8)

3x+4y+2z

....(9)

We shall try and solve by elimination. To eliminate w from ('1) and (8), we calculate 7 x (7) - 5 x (8) We get 14x+28y +&= 16,thatis, ...(10)

7x+ 14y+3z=8 Eliminating z from (9) and (lo), we get

Now, we can't eliminate any further. So we shall try and obtain all the variables in terms of the minimum number of variables possible.

Thus, we get a 1-parameter set of solutions, namely,

TOcheck that these are the solutions, we substitute the 4-tuple

1(5(

- Y).Y, ~ ( 1 -4241, ~ :(Y - I)) in each of the equations of the system

and find that it satisfies them.

(1 ) I

-3x+ y-42-0

[ , 1)

Cramer's Rule

We check that (O,0,0)satisfies all the equations. Thus the system only has the trivial solution. b

d) Since the determinant of the coefficient matrix is non-zero, we can apply Cramer's rule as well as the elimination method. Let us apply Cramer's rule. For this we calculate

: ,

-32 28

-100 28

x = - y = -,z = 0,that is, the unique solution is

-25 (i-8 -7 0) . 9

WIT 3

INEQUALITIES -

Structure 3.1

Introduction Objectives

Inequalities Known to the Ancients 3.3.1 lnequality of the Means 3.3.2 Tnangle inequality

Less Ancient Inequalities 3.3.1 Cauchy-Schwarz lnequality 3.3.2 Weierstrass' Inequalities 3.3.3 Tchebychev's Inequalities

3.4

summary

3.1 INTRODUCTION So far we have discussed equations of various kinds. vow we shall consider some inequalities; not of the social kind, but between real nuhbers. A mathematical inequality is a mathematical expression of the condition that of two quantities one is greater than, greater than o r equal to, less than or less than or equal to the other. An inequality that holds for every real number is called an absolute inequality. In this unit we shall restrict ourselves to such inequalities. We will discuss six famous absolute inequalities. We have divided them into two sectionsthose that have been used for centuries and those that were discovered by some famous nineteenth century European mathematicians. These inequalities have several applications also. We will discuss a few of them. You may come across some applications in other courses too, at which time we hope that you will find that you didn't study this unit in vain! Let us list our unit objectives now.

Objectives After reading this unit you should be able to prove and apply the inequalities of the means; the triangle inequality; the Cauchy-Schwan.(Bunyakovskii)inequality: Weierstrass' inequalities: Tchebychev's inequalities. Let us discuss the ineqyalities one by one.

3.2 INEQUALITIES KNOWN TO THE ANCIENTS In this section we shall discuss two inequalities handed down to us by ancient mathematicians. But first we will give a list of some properties of inequalities you must be fmiliar with. They are the following: for a, b, c, d E R i)

aZb,cZO*acZbc

ii)

a2be-as-b

lnequnlities

iii)

a2b~-5-provideda;tO,bzO. a b

iv)

a2b,c2d*a+c>b+d

an>-bn,a2O*a2b,wherene N.

We will often use these properties implicitly while proving the inequalities mentioned in the unit objectives. Now let us discuss the inequality that relates three averages.

3.2.1 Inequality of the Means An important part of arithmetic that can be traced back to the Babylonians and Pythagoreans (approximately 6th century B.C.) is the theory of means or averages. The word "average" comes from the Latin word "havaria", which was the insurance paid to compensate for damage to goods in transit in the olden days. All of us are familiar with compensate for damage to goods in transit in the olden days. All of us are familiar with the term "average". In fact, all of us must have often calculated fie average of a fmite set of numbers by addhg them up and dividing the sum by the total number of these numbers. But this is only one of Inany types of averages. We will discuss three of these types here. Let us start with the "usual" average.

Definition: the arithmetic mean (AM) ofn real numbers x,, x2, ..., X, is XI

+ X2 + ...+X" n

, that is n

[A . i = ~xi]

1 -1 +o 1 For example, the AM of - - and 0 is 2=2 3 18 3 A

The AM is often used in statistics for studying data. Another type of average is the geometric mean. This is the best mean to use if we want to find the mean of any finite set of positive numbers that follow geometric progression. Thus, this mean is very useful for studying population'growth. Let us see how the geometric mean is defmed.

Definition: The geometric mean (GM) of n positive real numbers XI, %,

......,Xn 1s

For example, the GM of 3 and4 is .= = f i ,and the GM of 2 , 4 and 8 is (2x4x8)lB=4. . Yet another kind of average of numbers is their harmonic mean, which we now define.

Definition: The harmonic mean 0of n non-zero real numbers

Eauations and Incaualiticr

1 3 42 -For example, the HM of -2, - and 7 is 3 37 ' - - -1+ 7 + - 1 The HM is the most appropriate type of average to use when we want to find the average rate of a set of varying rates. Thus, it is the best average to use for obtaining the average velocity of a vehicle covering various distances at different speeds.. At this point we would like to make a remark. This proof is due to Cauchy] who you will meet again in See. 6.3.

Note: We can obtain the AM of any n real numbers. But, we only define the GM of n positive real numbers; and the HM of n non-zero real numbers. Now let us look at the three different means together. To do so, we clearly need to restrict ourselves to positive real numbers. What is the AM of 2,4 and 8? How is it related to their GM? And, how is their GM related to their HM? The following result answers these questions.

Theorem 1: Let {x,, xz, ...,x, ) be any finite set of positive real numbers, and let A, G and H denote their arithmetic, geometric and harmonic means, respectively. Then

We will only give a broad outline of the proof here. The inequality A 2 G is first proved by induction (see Unit 2) for all those integers n that are powers of two That is,

and equality holds iff x, = 3 = ...= x,m. Now, given any n E N, we can always choose r E N such that 2' > n. We apply (1) to the 2' numbers x,, x,, ...,x,, A, ...,A, where the number of A's is 2' - n We get

(with equality iff x, = 3 = ...=xn =A.) n

, since

Cxi nA. =

i-I

* A 5 G, since A and G are positive real numbers. =a.

Note thatA = G iff x, = 3 = ...

Thus,.the result is true v n E N. 1 Now let us consider the n positive numbers - 9 XI

1 X2

1 . Xn

Since their AM is greater than or equal to their GM, we get

Thus, A 2 G 2 H, with equality f i x , = x, = ...= x,.

In about 320 ~ bthe.geometer Pappus of Alexandria gave a geometric construction of the AM, GM and HM of two numbers. His construction is as follows: Draw a semicircle with a+b as diameter (see Fig. 1). Let its diameter be AC, with mid-point 0.

Flg. 1. Xbe AM, CM and HM of 8 and b arc DO, DB a11d DE, respectively.

Then OA is the radius of the circle. Mark off the point B on AC such that AB = a. Then BC = b. Draw BD I AC to meet the semicircle in D. Then draw BE I DO, as in Fig. 1. Then Pappus proved that DO is the AM of a and b. DB is the GM of a and b. DEistheHMofaandb.

Since DO 2DB 2 DE, this gives us a geometric proof for Theorem 1, when n = 2. Now let us apply Theorem 1 to prove some more inequalities. Example 1: Show that

[T~ I ]

1= 1

> nn (n!)r,

n! denotes fwtorial n and r > 0.

Solution: Let r be a fixed positive real number. Consider the n positive qmber lr, 2', ..., nr' By Thkorem 1

Since the numbers lr, 2: ...,n'are not equal, their AM is strictly greater thantheir GM. Thus

(1' +2'

fn. .+

.'1"

2 (n!)'

We can prove several inequalities, which are particularly useful in mathematics, by using Theorem 1. We ask you to provdsome of them in the following exercises. E 1) Show that jab + xy) (ax + by) 2 4abxy, where a, b, x, y are positive real numbers. Under what conditions on a, b, x and y would the equality hold?

Equationr and lnequalltier

E2) For any n E N and positive real numbers x and y, show that

E 3) IsTheorem 1 true if we remove the condition that the numbers are positive? Why? Now, you know that the inequalitiesin Theorem 1 become equalities when x, = x,= ...= x,,. When this happens, then xi = A = G = H v i = 1, ..., n. Thus, we see that ifx,, x,,...,x,are n positive real numbers such that x, + x,+ ... + x, is a constant, then their arithmetic mean attains its lowest value and their geometric mean attains its maximumvaluewhenx,=x,= ...- x =A = G Let us see how to use this fact for obtaining some maximum and minimum values. For convenience, we shall denote the set of positive real numbers by R". '

Example 2: Find the greatest value of xyz, where x, y, z E R+are subject to the condition yz+zx+xy= 12. ~ o l h i o nxyz : has greatest value when (xyz)' = (yz) (ZX) (xy) has greatest value. Since yz+zx+xy is a constant, we know that the maximum value of (yz) (zx) (xy) is attained when yz = zx = xy, that is, when x = y = z.

Hence, the maximum value of xyz is 23= 8.

Example 3: If the sum of the sides of a triangle is the constant 2s, prove that the area is greatest when the triangle is equilateral. Solution: Let a, b, c be the sides of the triangle, where a+b+c= 2s, and let A denote the area of the triangle. ThenA= J s ( s - a ) ( s - b ) ( s - c ) . So, A will be greatest when (s - a) (s - b) (s - c) is maximum. Now (s - a) + (s - b) + (s - c) = s, a constant. Thus, (s - a) (s - b) (s - c) is maximum when

s-a=s-b=s-c,thatis,a=b=c. Thus, the area is maximum when the triangle is an equilateral triangle.

Why don't you try these exercises now? E4) a) Prove that if the sum of two positive numbers is given, their product is greatest when they are equal. '

b) Is (a) m e if the words 'sum' and 'product' are interchanged? Why? .

E 5) Find the greatest value of (5+x13(5-x14, for -5 < x < 5.

(Hint: The greatest value of ( 5 + ~(5-x)' ) ~ occurs when the greatest value of

($I3 (y] 4

occurs.)

E 6) When docs a cuboid, with dimensions x, y and z such that x+y+z is fixed, have maximum volu@e? E7) Under what conditions on the dimensions, will a cuboid with fixed volume have minimal surface area? (Hint: Use the inequality G 2 H.) You can study other techniques for obtaining maximum values in our course on calculus. Let us now consider another inequality, which follows from Theorem 1

Theorem 2: If x,, ..., xn E R+such that not all of them are equal, and m E Q, m # 0, m # 1, then

The proof of this result uses Theotem 1. We shall not give it here. A result that follows from Theorem 2 (and Theorem 1) is that

If x, + xz + ... + x, = c, a constant, then n

for 0 < m < 1 the maximum value of

is n'-rn cm,and

i-1

f o r m < O o r m ~1 the minimum valueof

C xisnl-"' ~c

is1

These values are attained when x, = x2 = ...= x, Again, we shall not prove this result in this course. But let us consider an example of its use for fmding some maximum and minimum values, 1 1 1 Example 4: Find the least value of + - + - ,where x, y, z E and x + y + z = 27

Solution: - + - + -

= isoftheformxm+ym+zm,wherem=-1<O.Sincex+y+z=27,the

least value is obtained when x = y = z. Andthenx+y+z=27givesusx=y=z=9. 1 1 1 1 1 1 1 Thus,theminimumvalueof - + - + - is - + - + - = x y z 9 9 9 3 Note that we could have also obtained the answer by applying the inequality G 1 H (c.f Theorem l), exactly on the lines of the solution of E7.

Now for some exercises. ES) Show that the sum of the mth powers of the first n even numbers is greater than n(n+l)'", ifm> i.

E 10) Let a,, %,

a,, E R+and p, q E N such that p > q. Show that

a: +a! +...+a:

< n H (a1 + a , +...+a$)

-9 (Hint: Put m = -,xi = a! in Theorem 2.) P

Equatloor and Ioequalltler

So far we have discussed various inequalities related ta the arithmetic, geometric and harmonic means. Now let us consider an inequality that had its origin in ancient Greek geometry.

3.2.2 Triangle Inequality If you look up any translation of the ancient Greek mathematician Euclid's "Elements", you will find that Proposition 20 of Book 1 says: t

"In anytrimgle, two sides taken together in any manner are greater than the remaiping one.". This result is the basis of the triangle inequality, which is a statement about the absolute value of numbers. Recall that absolute value of x E R is defined by

Thus it satisfies the following properties

Ixl=I-xI v x c R

and

xSJXV ( XER

You can study the absolute value of real numbers in more detail in our course on cafculus. Now let us state the triangle inequality.

Theorem 3: Let x,, x,,...,x,,E R. Then Ix,+x,+ ...+ x,ISIx,I+Ix,I+ ...+Iql. Moreover, equality holds only when all the non-zero xi's have the same sign.

ProoZ: Let us prove the result for n = 2 first. Now (lx, +x21)' = ( ~ , + x , ) ~ , s i r r c ( x ( y~ =X xE ~R = x: + 2x, x, + x,2 ~ ~ x , ~ ~ + 2 ~ x ~ ~ ~ ~ ~ ~ + ~vxX,E~R~ , s i n c e x < ~ x (

=(lx11+1x~l)~. Now we take the square root on both sides, keeping in mind that I x t 2 0 V X E R We get

I x, + x, 1 5 1 x, I + ( x, I, which is what we wanted to prove. Notethatifx, c 0 , sayx, =-a,and~2>0sayx,=b,wherea,b>O,then Ix,+x,I=Ib-a),while)x, (+Ix,)=a+b.T@, whenx,andx,haveoppositesigns

So, Theorem 3 is true for n = 2. Now, let us prove the result for the general case, by induction. So let n > 2 and assume that Theorem 3 is true for any n-1 numbers. Now consider Ix,+$+ ...+\I

=I~~+x,+...+x,~)+x,I ~lx,+x,+...+~lI+I~nI S I X, ( + ( X, 1 + ...+ 1 xWl( + ( X, 1,since the result is true for n - 1 numbers.

Further, just as we have shown for the case n = 2, strict inequality holds if all the non-zero 5's don't have the same sign. Theorem 3 is not only m e for real numbers. In our course on linear algebra we have proved thatifz,,~,,...,z , ~ C,then(z,+z,+...+z,IIlz, (+Iz,I+ ...+I z,I, where I z 1 is the modulus of z. Let us verify Theorem 3 for the numbers - 2 , l , 5 , O .

Since1-2+1+5+OI=141=4,and

we find that strict inequality in Theorem 3 is true in this case. Note that -2 and 1 have opposite signs. I

Why don't you try some exercises now?

I E 11) The absolute value of the AM of n numbers is less than or equal to the AM of their absolute values. True or false. Why? E 12) Prove or disprove that

( To disprove a statement means to show that it is false. See the appendix of Block 1.)

E 13) Prove or disprove that I X - Y I ~ ~ I ~ I - I Y v xI J, Y e R.

.(Hint: W r i t e ( x I = ( ( x - y ) + y I,andalsousethefactthat(xI=I-XIv X E R.) Now let us discuss some "newer" inequalities.

,-=%

3.3 LESS ANCIENT INEQUALITIES In this section we shall discuss four important mequalities which are due to some mathematical giants of the nineteenth century. We start an mequality due to three mathematicians. I I I

3.3.1 Cauchy Schwarz Inequality Augustin - Louis Canchy, the famous French mathematician, was responsible for developments in infinite series, function theory, differential equations, determinants, probability and several other areas of mathematics. One of his contributions was result, which was later generalised by the German mathematician H.A. Schwarz (1848- 1921). We now state this result, which was also proved independently by the Russian mathematician Bunyakovskii.

: C8uchy (1789-1857)

Theorem4 (Cauchy-Schwarz Ipequality): Let a,, a,, ....,a,, b,, b,, ..., b,, E R. Then ( a , b , + a 2 b 2 + . . . + a n b n ) 2 5 (+a: a ~ + ....+az)(b:+b:

+ .... +bz),

with equality iff ai = cb, v i = 1, ...,n, where c is a fixed real number. Proof: To help you understand the proof we shall prove it for n = 3 fust. Then you can try and generalise it (see El 4). 1

NOW (a: +a: +a:) (b: + b? + hi)-(a,b, + a2b2+ a3b312

When a, = cb, y i = 1, .., n, where c is a constant, we say that the n-tuples (a, . ., an) and (b, ... b,,) a n proportional.

Equations and lacqualltfes

When will the equality sign hold? Equality holds iff a,b2- a2b, = 0, %b3- a3b2= 0 and a,b,- a,b, = 0, that is, a, = cb,, a2 = cb2,a, = cb,, for a fixed real number c. Thus, we have proved the result for n = 3. Now, to complete the proof of Theorem 4, why don't you try this exercise?

E 14)Prove Theorem 4 for any n E N. Let us consider an application of Theorem4 for locating the roots of a polynomial. Before going further, you may like to keep Unit 3 nearby for easy reference.

Theorem 5: If all the roots of the real polynomial equation

x" + a, x"' + a2xW2+ ... + an= 0 are real, then they lie between

and

n - l Ja: -

-y

(z)a2 n-1

(%Ia2

Ja: n+e n-1

Proof: From Theorem 1 of Unit 3, you know that the given equations has n roots. Let x be a root. If x,, ..., x", are the other roots, then by Theorem 4 of Unit 3,

Also, by Theorem 4 of Unit 3,

(a, + X)2 ~ ( n 1)(a: -2a2-x2)

by the quadratic formula. This holds for all x such that

Thus, any root of +e given polynomial equation must lie between the bounds of the given statement of the theorem. Before giving an example of the use of Theorem 5, we shall make some related observations.

Remark 1 :Consider the polynomial equation a, + a,x + a2x2+ ... + anxn= 0, where a, E Z.

Then any rational root of this equation is of the form

,where d is a factor of a,. an

Remark,Z: For cubic cases, we know from the discriminant (Sec. 3.3.2) when the roots are all real. An4 then Theorem 5 can be very useful, especially if we know that the roots are rational. Let us now consider an example of the use of Theorem 5. Example 5: Solve x3- 23i2+ 167x- 385 = 0. Solution: The discriminant of this equation (see Sec. 3.3.2) is posltlve. Hence the given equation has three distinct real roots. 1f the roots are rational they must be integral factors of - 385. Thus, they musthelong to the set

23 4 23 4 But, by Theorem 5 the roots must lie between - - - f i and - + - f i . Thus, lithey are 3 3 3 3 rational, they can only be 5,7, 11. On substituting these values in the equation, we h d that they are indeed roots of the given equation. Also, you know that the equation can oiily have 3 roots. ~ence,'thesevalues are the only roots. You may now like to try to apply Theorem 5 yourself.

Now let us consider another example of the use of Theorem 4. In this example, we shall apply the Cauchy-Schwarz inequality twice to get an inequality that we want. Example 6: Let x, y, z E R+such that x2+ y2+ z2= 27. Show that

312

Solution: Let us first apply Theorem 4 to the two triples of real numbers, (x ,y , z ) and (x1I2,y1I2,z1I2).We get

r y + z z s (x3+ y3+ z3)(xtY+z),that is (x + y 2 + z ~ ( x 3 + y 3 + z 3 ) ( x + y + z ) ( X PXl12

312

112

312

112 2

)

Now let us apply the Cauchy-Schwarz inequality to the triples (x,y,z)and(l, 1,l).Weget !x.l+y.l + ~ . l ) ~ S ( x ~ + ~ ~ 12+ + z 12),thatis ~)(l~+ ( x + y + ~ ) ~3 S( ~ ~ + ~ ~ + z ~ ) = 8 1 *x+y+zS9 Thus, by (2) (X2+y2+z2)S9(x3+Y3+z3) But x2+ y2-+z2= 27. Thus,

Why don't you try some exercises now?

Equations and Inequalities

E16) If a, b, x, y E R such that a2 + b2 = 1 andx2 +y2 = 1, then prove that ax + by S 1. El 7) Prove that if a,, ...,a, E R", then a) ( a , + a 2 + ...+ a,)

(&+A+ ...+A) ' ~ n ( a , + a , + ...+ a,)

E18) (Another form of the triangle inequality) If a, b, x, y E R, then show that (X - y)f 5

JZ?+ JW

(Hint: Write (a - b), + (x - y)2= (a2+ x2)+ (b2+ Y2)- 2(ab + xy), and then apply Theorem 4 to (a, x) and (b, y). E 1 9 ) I f i , y , z ~~"suchthatx~+~~+z~=81,then~rovethatx+~+z19. E2O)Prove or disprove the following generalisationof Theorem 4 : Letp~ N , p # l , a n d a ,,..., a,,,b,,b2,,...,~ , E R .

+...+ a!)(bp +b,P+...+b;)

Then(a,b,+a,b,+ ...+ anbn)'<(a: +a!

The Cauchy-Schwaxz inequality has several applications in physics and mathematics, especially in the context of inner product spaces. Fig. 3 : K.rt Tbcodor Weierstrasr

Now let us consider another useful set of inequalities.

3.3.2 Weierstrass' Inequalities It is generally through that a good mathematician must have started serious mathematical studies at an early age. But the German-mathematicianWeimtrass (18 15 - 1897) is an exception to this rule. This outstanding mathematician started serious mathematics at tl.ie age s rigorous, and is considered to be the of forty. He was responsible for making ~ a l y s imore "Wher of modem analysis". He is responsible for the following result. Theorem 6 (Weierstrass' Inequalities): Let a,, a,, ...,a, be positive real numbers less than 1 and S, = a, + a2 + ...+a,,. Then

(ii)

l + S , S ( l + a , ) ( l + a J ...(l + a n ) <

where it is assumed that S, < 1.

ProoZ:We prove (i) by induction on n, a principle that we introduced you to in Unit 2.

So, ( i j is true when n = 1. Let us assume that (i) is true for x i = m, where m E N.

We will see if it is also true for n = m+l .

-a;)

...

(l -a,)

- S , by ouras+,p~m.

:dp

Th&,i(,l-al)(l-%) .A (1 -a,,,)(l -amlj ~ (-sm)(i-.a,,) 1

,.,

= 1- S,

+ sma,,

:.<-$ ..,

- f<.:,~ :,; :y 2. -. &,.& . ... ,

, ..<.f-. >:,& ,.;.

~l-S,l,sinceSma,,>O.

~%qliIhbe.

1s.i rC+? ,;j~su:rni8

RRS.=I-(S~+~,,)+S~~,, ; !),

*siiiisu@s5r

. .

r.d,a5 . 4s .-- I rj !h:

- ;sj

I ..

,i~6irisdi i;L !,.d--

!j.,i 4 , !i >.;.

",. -,,:.,-, ..,:, ..

S.o,(l-a,) ...(I-a,,,+,)>l-Swl

... , .,,. . ::: ', ,.ii$bprl , ,..~

!'-; . .

: Funher, Since (1 - a,) (1 - p ) t , j , ~ r ? & l & ~,..kvflw@? . . & ~ WQ- %?&=&+ ; lnj ; . i. d ; , ~z :. ; , d t - ,. . *i . a i , & s j i (..*.. A . . ;-!. . ., 6 - : . i . , ~ + , ~. .. . i ' . '1 ' I

we fmd that

.. .<I .

,.: ... L .;

, .,

l.:

,.;.;a

;.,,>z

,:.

;

.d . i. .: ,4>;,B,:t

:..;::&.-

(3) and ( 4 , takentogether, tell us W t (i) is m e for n = m + l!!Yenc% . py -. ..indy~fio#~.(i)@Qq~ '-" , ... . . . \ I ~ N. E

Now, to comple@the proof you can try E21.

E 21) Rove (ii) of Theorem 6. ~ 2 2 ) or 0 c a,, i,,:..,a, < 1,prove that n

I ,

P n. -

i=l

i-I

, I

~i.2,.

!$ ,.

. - +:-.-.:.

;

[,ni,:?;,:/. f:'!:r, . ,

Is,ai

:j.,:; - o. ci ., - ,..:.> 5: >a$,,r, .: , ,.

,i- y j ; : ~ , . : ~-, ;;&, ,

fi%i ,:ii>.?>l~ ?j~ijj?;A

,.&:'i$;,,.?,f;, . .' . - ,.';,,rd7f,., . .

.-. .!

, .

, i

(Hint: 0 < 1-a, c 1.)

r$ ,::

, ... j 2'.,IX .1

.c.

- B)+,( ( .j ~ .. :1 ,. ;[,i.z;:;?cr:

b.1.

',Ti!

E23) Does Theorem 6 hold if ai < 0 or a. > 1 for any i = 1, ...,n? Give reasons for your answer. . r $ , .:<.a) ~ .i#FlVIi)911$3d$fo,:~)-tq$!ti %:i)iqru~%:t ir3 ..vo:x bnA ..

1 :

generalisations are very useful. .

,Fig. 4 : Tchibychev

i;t , r3iq i ~ k . ,,..~y:,,;>

,~;,... it$ . .>I... i-. .

#..

;>.

And fmlly, we shall discuss some inequalit&due. c t . ~ ~ ~t o~ $~ leadi%grR9sip ~ ~ . ~ , ~ ~. . , , . : w . t h e w . . , X I L . . ~ . ,

, ,

The mathematicuqPafnutii L. Tchebychev (prono+ced Che-bee-chefO ir most,@oyn for his tremendous work-manalytic number theory and the theory of o r t h o g o n t a $ d ~ & d @ ~ @ .. .. here we shall prove and apply some inequalities that are;ng@d afterhim. . <,>. <:;,...? .j .L,; .+. Theorem 7 (~iheb~chev's Inequalities): If a,, ....an,b,,,:. b j s R such that . <

8 . . !!

a, , n ( a ~ b 2~ ' +

ii)

, , " , : ; : : ,

n +

?nb&!+efl~$

-; '

, -

.. T.9. " :~ii~ls2

+b ,) %+..:$.Q + b2 +-.: fiii ,;

a, 2 a, 2 ... 1 an,b, Ib, 5.... 5 b,, then, . ': ..F . .* n(a,b, + a b + ...+ a;,bn)z5 (a, + a,.* .:I 2 2. . .

- : c -, .,

. ,.

...

;

, . ..

:.. 7 .

,-:f b 2 ; ,< ; .,: i - . (b :%.b,+--. !

I .

;; ,

.. s'

f :l j

. , r 2 .: . . - l i:

n '

Proof: Let us prove (i) for the case n = 3, so that you can understand the proof more easily.

.-.

- .*

J<,.

;a3bj;:

i>?<jtj ci,! :,><>:

-;>!,9i:g.*

~ ( :icj;;~'T 9 k

(1821-1894)

Now an add albl + +bz + %b, to both sides d simp@ to get

.Theproof of (i) f a my a E N is on exactly the sank lines. Let bs prove it by induction on n. The result is tne for n = 3 (uul, in fact, for p = 1 and 2). Assume thatitistrpCf&n- 1. Then +b2+...+b,,)

(n- l)(albl +...+a,+lbwl)2(al+++ ... Also a,b, + Q, 2 a,ba + s b , , since

(a, -

(b,

- bn)2 0-

Adding up the left brnd 8id- d the Gght hand sides of these n inequalities, we get

~d now,tbcompkttt h e k t ofthe theorem, try ~ 2 4 .

Wt:Put 5 - , a ,

i 1, ...n, and use (i))

Now kt us consider m applicadon of Tchebychev's inequalities. &t.rmrlc7:Sbowthllt

Taking thetsquare root on 'both sides, we get

I -8

Now you can try some exercises. E25) Show it$ \

(Hint: First apply Tchebychev's inquality to the n-tuples

E26) If a, b, c, E%+, then show that.

(Hlnt: See if it is possible to apply Tbeomn7 to b+c, cta,a+b and their inversk.) b

With this ine;1uaIity we come to &e end of this unit. This d-t mean that we've e x h a d all the inequalities, or even all the important ones. We have just exposedyou to a few elementary ones and some of their wlications. As you study more mathematics you will come across these and several others. Now let us quickly go through what we have covered in this unit.

3.4 SUMMARY . We have discussed several inequalities and their applications in this unit. Let us list themone by one,

1) The inequdlity of the means: The AM of any fimie.ct of elements of R+is greater than or equal to their GM, which is greater than or equal to their HM. 2)

Ifx,, ...,x,,~Rsu~hthatnotall~fthemarequal,andm~Q,mtO,l.then -

3) The triangle inequality :For x,, 3...., x, E R #

The inequalit)' is strict in case all the non-zero xis don't have the same sign. 4)

Cauchy-Schwarz (or Bunyakovaskii) inequality: If a,, ...; a,,, b,, ...,b, ER;then

and Inequalltkr

--.

;:c::pi;*-?t,is j?

,137i3.%9x:31 5; y!,

with

??!i4b~:?*bd~~~!~$~~1: n

incquhties: For 0 < a,, ....an< l'and S, =

5) Weiaa-'

...I*.*.I*

. ly.L%.

.)-

?,<,!

%v=PQ.

- ?

irl

....a,), (b,, ...,bd) E R" such that

6) Tchebychev's i n w t i e s : If (a,,

a, S a , S

..:Sa,, bi9 b,S ...S b,, ,then

I .

I \

n(albl+a,b2+ ...+ anbd)2(a,+a,+....+an)(bl+bp+...+bn) n(a,b, + a,b, + ...+ anbn)S (a, +

+ ...+an)(bl '+bp+ ...+*n)

' i :

As usual, are suggest &t you go back to the beginning of & d >hddsdt'ifyou'hi&

With . this we have cqma tb %e. end of. ?this,course. We , hope , ii yov .. -*,have,enj,oied > and wgl, I;;: . . .; ., . I f . Lit, , :.:!.A ,

,,Z,?[

3, 01 gj;:?

;;.3&<>f!.9m.

,,.

1.'..

i,,!';

:;.~

; ;>

.:<:;

..,.: 4 ; ; :

:,.,~!,I;:,;;. . ,cm.:;:

2ij.-

. .; . . ..dt:;:t::,:,*>(F:

;L)..

;

!.&

Note that equality tiolds iff ab = xy and ax = by e a = y and b = x. r

E2) a) Applying the inequality A 2 G to the n+ 1 nup.lyrs X, Y, Y,

..:, Y (ntimes), we get

n r l

" +

Note that equality holds iff x = y.

-s

I '

, ,:1 I

*-;*

> 1

C) k t m, n ci N and.m n. Then, by fi), r " .

.., . . ..

.,,;.*

-:.,:

1:. > > , r

." ,

Hence the result,

3 : r. .<

.,I

E 3) No, since the GM of negative numbers is not defintd.,:E~qg, 2 H does not remain true any longer. . ... h r ~ 1 e , & e t h e & n ~ ~ - 2 , 1 . n d 3 b - . ;; . : :,; ':'.-,: :.:,;%i,',,.,,' . ,,,,.. . 6.; _ .,

. ,

:,,'

..%..:

2 3 18 Their AIM is - and their HM is -1 1 = 3 -+I+5 2 3

, 2 -5 . ,A

.' ''p..: I

E4) a) Let x, y E R+such that x + =~c, a constant. Then, xy hmaximum when , xy IS mxiqum, that is,when x = y. ,,;;,, ';,, <

;,,>. : .,

,;, ...;: *.: S ' : ' : *

. .. .

.'!>j

.: ,.: '..

+,,: . b) N~.PO* example, let i,y E R such that x i 4. *,

: :?.

' .

.: ,;, ,a.jI : ? .

. !

x + y, since x =5 and y =

-5.1 fa-k;give

. ' ,,~ f , .-.,:. ,':.,.

. $

a @rgk.vdtieof x +y,

E5) Since-5<~<5,5+x>Oand5-x >0.

The volume of the cuboid is xyz.

.. , >. .-.. .. . . .

:%! 5 >,, j4.>;.,) 2t:,,Liv;.

..:

Thus, the cube is a cuboid.wjfh.mU@umum~&,.&the.gj&& ,C . .. ,i,,

%, , !,, 7

E 7) Let x, y and z be dimensions of the cuboid, where xyz = c, a constant. NO?, the surface area bf the cuboid is 2(xy

is mininnun, that when

Now, xyz=c, and the HM of x, y and z is i!!: '>,,

+ yz +hi.. T ~&L-' .,u:l

." . . :>" . b

! :: ;.

,L I" ..

-.r

.,.: if!,3:jJ.,,.?a.:;:,

3 .. . ;". . 1 + --1 + .-1 ' . X y X.; ; - I .,. ,,.','! .

xy +yl +a

!?&

,::;j:

: :

;, . ;. ... ,. x

i t . ; .

;liijj

, ! ; = : "<

; {c{':j. .

minimumwhen x = y = p. Thus the surface area of a cuboid with fixed volume j i

minimum when the cuboid is a cube. - -.-

I '

E8) We apply Theorem 2 to then numbers 2,4,6, ...,2n.

m.1 I.cq.mlltla

We get

Ji+fi+...+& n. Hence the result. t

E 10)Putrnz -.ThenO~rn< l.Nowwe&pply'lheorem2tothinpositivenumbers P a!,

at, ....a!. w e get

Q a:

+a: +...+a! s (nH)l1q (a! + a t +...+a!)q1p

<d7 (a: +&:+...+a:),

1 '9 since - < 1 md < 1. P . P

E l l ) Letx,,..., x,benmt&em. 'Ihm their AM, A =

.-.\A\ =

I'+ X,

+ X2 +...+X,

I I+

x2J...+)L,

:AMoftx,

(XI

Ix~(+..-+Ix,( by theorem 3. n 4

1.13I ,.-.,I \ I .

Hence the statement is true.

'~12)F a l s e . , F o r ~ l e , ~ e xl,y=-3. = Thenlx-yl={ 4 1=4,

md( x 1-1 y 1-1-31-2.

E13) I x 1-1 ( ~ - Y ) + Y IClx-Y l + l Y I.

..,(lo)

l x 1-1 Y 151 x-Y 1 #

Similarly,[ y 1-1 x+(y-x) IS1 il + l y - x

l = l x l + l x-y I,

with equality iffaibj= ajbi v i, j = 1, ...,n, i #j,

that is, iffa, =cbl \I i = 1, ...,II, where c is a co~utant.

the equation is positive. Thus, the given equation hrrs distinct red . 2 2 fi) and j(l + fi).~ftheyarc rational, they have y lie between roots. ~ h .must

The -of

to be factors of 2. Hence, they can be 1,-I, 2 or -2. Of these, 1 and 2, lie within the given bounds. On substitution we find that they actually arc the mots. Since fhe f equation has only 3 roots, - 1 , l and 2 an its roots. ~ u sapplyins-.Theorem4 t ot the numbers a, b,

Y, we get

(a)We apply the Cauchy-Schwa inquality to the n-tuple&

b) Applying Theorem 4 to the n-tuples (1; 1, ...,1) and (a,, 4, ...,a,,), we get the mdt. C)ApplyingTheorem4tothen-mples(1,1, ..., l)and(&,& the result.

,...,&)&get

Also, by Theorem 4 (ab + xYl2s(a2+ x2)(b2+ y2) 3I ab+xy = /,I-/,.

..(13)

(12)and(13)*(~)~+(x-y)~~

which gives us the desired result.

. .

By the Cauchy-Schwarzinequality applied 'to (x, y, z) and (l,1, I), we have (x+y+z)'

..(14) .

' ~ ~ a iapplying n, W r e m 4 to (

A,J;, & ) and (x312';y" ,z3")

E 20) False. For example, take p = 3 and the pairs (1, O), ( 1. 1 hen {(1)(1)+(0)(-1)}~i ( 1 ~ + 0 ~ ) ( 1 ~ + ( )- 1 ) ~

E21) We use the principk of ind~ctiof"~?$p.,~ ,. >,

.,, .

.,, *

I ,

k F,o r n = 1' a = I ,4ndhence, a . . 1 +s,,s 1 +"::r$<.:, -\: b k 3 , y ;

* , I*

* ;

ver* f

Thcn(l +a,)(l +%):..(I + a J > p + s i .*.

"-.'

TJzq,\<

7 i C ~ f # :*? ) L ,

* ' (1 +=,)(I +%).,.(I + - a ~ ~ b ' a m , ) $ 1 ~ + ~ ~ ; 4 i $ ~ ~ ~:it3~ g :(li ~l''.A

> l+sm,

Also, (1 +a,) (1 +a2)..;.(1+ g,J -< il

.: (1 + a ,)...(l + Q ( 1 +a,,)<

-I- " # V L ? ?&,, s(iiSm)(i~'.m,"i9 . -

'.*

T.&

,It

;

.,3

Thus, (ii) is true fbrn = m+ 1, mehence t/ n E N. 4. r ' -. <.E 22) We apply T h e o m 6 (i) to the n ~ufhbers1 +,,.kla;; ..., 1 - a,. !

1-(1-a++1-%+ 132 Y3J< $

,< ' ,

(Sir , a 1- = a , s n t a i j=l

...+ 1-an)4a,% ...a,. .

, i ,

r ,

.QJ

-, , . ,;;

,iy

i-1

- *:

- ,, * * & >

I , , -

4" rP~ 9::*id( jd .+;dfik ,

i - 2 + -d,;=x-. *i,!>+ ,ryT

E23) ~ o . ~ ~ r e x a m ~ l ~ , l e t u s t=-1 a k em a ~d % = 2 . :i-;- < iF1 j

41,

' . c

Then 1 - (a, + a2) 1(,I a,) .( 1a 0 S2(- 1)=-2,v&ichisfil&.

' ,

. -, , , "

;- ? j . r

' b

,+dt . , ,

EX) Ifweputxi=-4,thenx,sr,s ... 5xn.Alsob,9 ~ , B ' . : J . I - ~ ~ , , ' I

,. . _ _ '

to (L

E25) Applying T'cheb@hev9$-1iify

1 (1, ,; ...,

....

we get

Vi. since T 1 (i-1)i = n {l+(l-;)

+...+(A-g}

=n ( l + l - 2 . *l+

+,..+ - s J = = 2 n

Again, applying Theorem '7 to

...(16)

[&-~>--.*~

[ ~ ~ . and- . * ~

we get

E26) Given a, b and c, they can always be ordered. Let us assume that a 5 b 5 c. 1 1 1 Therefore, -2 2 a+b a+c b+c'

1 1 1 andtoc,b,a. We apply Theorem 7 (i) to a+b a+c9b+c We get

Also, applying T h e o m 7 (ii) to b + c, a+c, a+b and their invmcs, we get

MISCELLANEOUS EXERCISES This section is optional As in the previous block, this section contains some extra problems related to the material covered in this block, Doing them may give you a better understanding of simultaneous linear equations and inequalities. As before, our solutions to the questions follow the list of problems. 1) Solve the following linear system, if possible.

a) If a + b + c = 0,solve the equation

b) Solve the equation

Use Cramer's rule to solve the system

A company produces three products, each of which must be processed through three different departments. In Table 1 we give the number of hours that each unit of each product must stay for in each department. We also give the weekly capacity of each department. Table 1 No. of hours available per week

Product

Department PI

100

What combination of the three products will use up the weekly labour availability in all depamnents?

Prove the following identities:

II I

1 I

6) A fish of species S, consumes 10 gm of food F, and 5 gm. of food F, per day. A fish of species S, consumes 6 gm. of F,and 4 gm. of F, per day. If a given environment has 2.2 kg of F 1 and 1.3 kg of F2 available daily, what population sizes of the species S, and S, will consumeexactly all of the available food? 7) Does the system

x+y+z+w=o x+3y-2z+w=o

have a non-trivial solution?

8) Obtain a solution of the following system, if it exists. x+'2y+4z+t=4 2~-~-3t=4

3 ~ - y - Z - 5t=5 9) Show that the following lintar'system has a two-parameter solution set. 2x, +x3-x4+xs=2

x, tx3-x4+xs=1

10) Use Cramer's rule, ifpossible, for solving the following linear systems: a ) . 3x+y=3

11) If the coordinate axes in a plane are rotated through an angle ,then we can express the old coordinates (x, y) in t e r n of the new coordinates (x', y') as

y=x'sin0+y'cos8 Use Cramer's rule to write (x', y') in terms of (x, y).

Solutions 1) We apply the Gaussian elimination process. We get the solution set

2 a) Adding the 2nd and 3rd columns to the fmt column of the determinant doesn't change its value. On d6ing this and using the fact that a + b + c = 0, we get c b a C-X

-0

~iscellrneousExercises

Equations and Inequalities

1 c o0 ' b -a-c x-c

a(-,)

3 (-X) [(b

=O,applyingP4twice.

c-x-b

- c - X)(C- b - X)- (a - c)] = 0, expanding along the first column.

On solving this equation, we find that x = 0 or

2x2= 3(a2+ b2+ c2),using the condition that a + b + c = 0. Thus, the solution set is

b) Using the properties P1 to P5 of determinants, we see that

3) In matrix notation the system is

Here D =

-5

3 ' 4

-2

=-32~;O.

Thus, we can apply Cramer's rule. Now

D,= -1

-2

4) Let x, y and z be the required quantities ofP,, P, and P,. Then we need to solve the system.

By Gaussian elimination (or Cramer's rule) we get x=S,y=O,z=25.

So, the ideal combination is 5 units of P,, 25 units of P,, and none of P,, per week.

1 a a2 1 b b2 1 c c2

1 a 0 b-a 0 c-a

a2 b2-a2 c2 -a 2

subtracting the f;lrstrow from the second and third rows

I c+a a+b I

-2p + p p+q ,subtracting the second and third columns from the fmt column. -2x z + x x+y

a c b

,subtracting the fmst column from the second and third columns. a b c

,interchanging the second and third columns. 6) Let x and y denote the number of fish of species S, and S2,respectively. We have the information given in Table 2 below.

Table 2 Fish

Food consumed per week (in gms)

Spec~es

2200

1300

Total food available per week Thus, we need to solve the system 10x+6y=2200

solving by any of the methods, we get

Thus, the required sizes are 100 fish of species S, and 200 of species S,.

7) By elimination we find that the system has infrnitely many solutions (x, O,0, - x), where x E R. Thus, for any x # 0, we would get a non-trivial solution.

Mlscellrneous Exerclscr

Equations and Inequalities

By Gaussian elimination, we reach a stage where we get 0 = 817. Hence, the given system is inconsistent.

9) We apply Gaussian elimination. After a few steps we get the following system x, = 1

x* +4x4-3x,=o x3-x,+x,=O.

Thus,x,=l,x,=-4x4+3x,,x3=x4-x,. So, if x, = s and x, = t, then our solution set is ( ( 1 , - 4 ~ + 3 t , ~ - t , s , t ) I ~R). ,t~

Thus, we have expressed the solutions.interms of the two parameters, s and t. 10)a)

In matrix notation, the system is

Since D = 5 2 = 1 0, we can apply Cramer's rule. lI

Thus, the solution is x = 5, y = - 12.

b) Here the coefficient matrix is

Thus, we can apply Crarner's rule.

11) Here we can write the equations as

(

Since D = cOsO 0

= ms20+ sin20= 1

+ 0,

we can apply Cramer's rule.

- sin8 = x cose + y sine, and

,,,e

Miscellaneous, Exercises

UNIT 1

PRELIMINARIES IN PLANE GEOMETRY

Structure Introduction

I. 1

Objectives

1.2

Equations of a Line

1.3

Symmetry

1.4

Change of Axes

1.4.1 Translating the Axes 1.4.2 Rotating the Axes

1.5

Polar Coordinates

1.6

Summary

INTRODUCTION

1.1

In this short unit, our aim is to re-acquaint you with some essent~alelements of twodimensional geonleh-y. We will briefly touch upon the distance formula and various ways of representing a straight line algebraically. Then we shall look at the polar representation of a point in the plane. Next, we will talk about symmetry with respect to origin or a coordinate axis is. Finally, we shall consider some ways in which a coordinate system can be transformed. This collection of topics may seem random to you. But we have picked them according to our need. We will be using whatever we cover here, in the rest of the block. So, in. later units we will often refer to a section, an equation or a formula from this unit. You are probably familiar with the material covered in this unit. But please make sure to go through the following list of objectives and the exercises covered in the unit. Otherwise you may have some trouble in later units.

Objectives After studying this unit you should be able to :

./. find the distance~betweenany two points, or a point and a line, in two-dimensional , space. obtain the equation of a line in slope-intercept form, point-slope form, two-point form, intercept form or normal form; check if a curve is symmetric with respect to either coordinate axis or the origin; effect a change of coordinates with a shift in origin. or with a rotation of the axes; relate the polar coordinates and Cartesian coordinates of a point; obtain the polai form of an equation.

1.2

EQUATIONS OF A STRAIGHT LINE

In this section we aim to refresh your memory about the ways of representing points and lines algebraically in two-dimensional space. Since we expect you to be familiar with the matter, we shall cover the ground quickly.

Firstly, as you know, two-dimensional space can be represented by the Cartes~an coordinate system. Thls is because there is a 1-1 correspondence between the points in a plane and ordered pairs of real numbers. If a point P is represented by (x, y) under this correspondence, then x is called the abscissa (or x-coordinate) of P and y is called the

Conics

Acmrding to the pythagoras theorem, in the right-angled triangle ABC,

ordinate (or y-coordinate) of P. If P(x,, y,) and Q(x,, y,) are two points in the plane, then the distance between them IS

...( 1)

PQ = J(x1 - ~ 2 +) (YI ~ -~2)*

From Fig. 1, and by applying the Pythagoras thkorem, you can see how we get (I).

( A B ) ~+ ( B C ~ = (ACP

Q(x2, ~ 2 )

Fig. I: Distance between two points.

(1) is called the distance formula. Another formula that you must be familiar with is the following: if the point R(x, y) divides the line segment joining P(x,, y,) and Q(x2, y2) in the ratio m:n (see Fig. 2), then

Y Qx = nx, + mx,- and y m+n

P Fig. 2 : R divides the segment

~ Y +I my,

(2) is called the -section formula. TO regain practice in using (1) and (2), you can try the following exercises.

PQ in the ratlo m:n

E 1) What are the coordinates if the midpoint of the line segment with endpoints a) A(5, -4) and B(-3, 2) ? b) A@,, a,) and B(b,, b,) ?

E 2) Check if the triangle PQR, where P, Q and R are represented by (1, O), (-2, 3) and (1. 3), is an equilateral triangle.

Fig. 3 : y * a is parallel to the x-axis. The plural of 'r*ls* is 'axes'

Let us now write down the various ways of representing a straight line algebraically. We start with lines parallel to either of the axes. A line parallel to the x-axis is given by the equation ...(3) Y = a, where a is some constant. This is because any point on the line will have the same ordinate (see Fig. 3). What do you expect the equation of a line parallel to the y-axis to be ? It will be x = b, for some constant b.

1 r I

Now let us obtain four forms of the equation of a line which is not parallel to either of the axes. Firstly, suppose we know that the line makes an angle a with the positive direction of the x-axis, and cuts the y-axis in (0. c). 'Then its equation will be

where m = tan a, and m is called its slope and c is its intercept on the y-axis. From Fig. 4 you should be able to derive (5). wh~chis called the slope-intercept form of the equation of a line. Now. suppose we know the slope m ofia line and that the point (x,, y,) lies on the line. Then. can we obtain the line's equation '? We can use (5) to get the point-slope form. y - Y, = m(x - x,) of the equation of the line.

...

We can also find the equation of a ,line that is not parallel to either axis if we know two d ~ s t ~ ~polnts l c t ly~ngon it. If P(xl. y,) and Q(x,,- y,) - are the points on the line (see Fig. 5). then 11s equatlon In the two-point form will be

Fig. 1 : I. is given by y = r tan a + c.

Note that both the ternls in the equation alc \\.ell-defined since the denominators are not zero.

you can see that its slope IS

Yr, - Y, 7 . and its intercept on the y-axls IS the constant term. x2 I

Why don't you try some exercises now ? E 3) What are the eqitiltlons of the coordlnatc axes !'

E 4) Find the equatlon of the line that cuts off an Intercept of I from the negative direct~onof tllr y-axls. and IS 1ncl11ic.dat 120" to thc x-axls.

E 5 ) What is the cquatlon of a I I I I ~ ' passillg through the ~ I I ~ I anti I I 1nak111ga11 anglc t) with the x-axis? E 6) a) Suppose wc know that the intercept o f a liac on'the s-axis 1s ? 3nti on the y-axis IS -3. Then sho\v that its equation is

(Hint: See ~f yo11 can use ( 7 ) j b) More generally, if a line 1. cuts off 311 Intercept a ( # 0 ) on the x-axls alld bt* 0 ) on the y-axis (sce Fig. 6). then show tbat its equatlon 1s

(8) is called the intercept form of the equation of L. We can obta~nthe equation of a l ~ n eIn yet another form. Suppose we know the length p of the perpendicular (or the normal) from the or~ginto a l ~ n eL, and the angle a that the perpend~cularmakes with the x-axis (see Flg. 7).

Fig. 5: Thc rlopc uf PQ is tan U

Conics

Fig. 7 : x cos a + y sin a = P is the normal form o f AB.

Then, using (8) we can obtain the equation of L in the normal form x cos a + y sin a = p ..(9) where p is the length of perpendicular from origin to the line and a is the angle which the perpendicular makes with the positive direction of x-axis. Further, from Fig. 7. A = OA = p sec a and b = OB = p cosec a using in (8), we get For example, the line which is at a distance of 4 units from (0, O), and for which X

a = 135", has equation - - + - = 4, that is, x - y + 4

JZ JZ

The distance o f a line from a point is the length of the perpendicular from the point to the line.

JZ = 0.

Here's a small remark about the form (9).

Remark 1 : In (9) p is positive and the coefficients of x and y are "normalised, that is, the sum of their squares is 1. Using these facts we can easily find the distance of any line from the origin. For example, let us find the distance of the origin frpm the line you got in E4. We rewrite its equation as -J5 x - y = 1. Then we divide throughout by

,/KT to get

- X - - yI = - . 1 This is in the form ax + by = c, where a2 + b2 .= 1 and c 2 0. Thus, 2

1 the required distance is c, which is 2 ' Now, have you noticed a characteristics that is common to the equations (3) to (8)? They are all linear in two variables, that is, are of the form ax + by + c = 0, where a, b, c E R and at least one of a and b is non-zero. This is not a coincidence, as the following theorem tells us.

Theorem 1: A linear equation in two variables represents a straight line in twodimensional space. Conversely, the equation of a straight line in the plane is a linear equation in two variables. So, for example, 2x + 3y - 1 = 0 represents a line. What is its slope? We rewrite ~t as 2 y = --x

3' I 1 and y axes are -and -, respectively? And what is its distance from the origin? To find 2 3 this. we "normalise" the coefficients of x and y, that is, we divide the equation 3

Fig. 8: PD is the distance from P to the line L.

+ -, to find that its slope is - - Do you agree that its intercepts on the x 3

throughout by 3 2 ---x+-y=-

J13

= f i . Wk get

which is in the form (9). Thus, the required distance is

J13'

In general the distance.of a point P(xl, yl) from a line ax + by + c = 0 (see Fig. 8) is given by

You may like to try some exercises now.

1 E 7) Find the distance of (I, 1) from the line which has slope -1 and intercept - on 2 the y-axis. E 8) What is the distance of x ' + c froni (0, O)? a) y = m b) x = 5 from (1, I)? c) x cos a + y sin a = p form (cos a,sin a)? d) (0, 0) from 2x + 3y = O? E 9) Prove the equation (9). Let us now see what the angle between two lines is. Suppose the slope-intercept fonna of the lines are y = mlx + c, and y = q x + c2 (see Fig. 9). Then the angle 0 between them is given by

Tan 0 can be positive or negative. If it is positive, 0 is acute. If tan 8 < 0, then 0 is the obtuse angle between the lines (which would be n - 0 in Fig. 9). .

Note that the constant terns in the equaiions of the lines play no role in finding the angle between them.

Now, from (1 1) can you say when two lines are parallel or perpendicular? The conditions n follow immediately if you remember what 0 and tan - are. Thus, the lines 2 y.= m,x + c, and y = q x + c,. are parallel if m, = q,and are perpendicular if m,l% = - 1.

ii)

...( 12) ...( 13)

For example, y = 2x + 3 and x :- 2y = 5 are perpendicular to each other, and y = 2x + 3 is parallel to y = 2x + c y c G R. Why not try an exercise now?

E 10) a) Find the equation of the line parallel to y + x + 1 = 0 and passing through (0, 0). b) What is the equation'of the line perpendicular to the line obtained in (a), and passing through (2, I)?

c) What is the angle between the line obtained in (b) and 2x = y?

Let us now step our discussion on lines, and move on to more general equations. We shall discuss a concept that will help us to trace the conics in the next unit.

1.3 . SYMMETRY '

. M i l t studying this block you will come across several equations in x and y. Their geometric representations are called curves. For example, a line is repwsented by the equation ax + by + c = 0, and a circle with radius a and .centre (0, 0) .is represented by 2 the equation x2 + y2 - a = 0.

Conics

Note that these equations are of the form F(x, y) = 0, where F(x, y) denotes their left hand sides. 'Now suppose the curve C, represented by an equation F(x, y) = 0, is such that when (x, y) lies on it, then so does (x, -y). ~ h e n~ ( x y) , = o a ~ ( x -y) , = 0. 2 (For example, x2 + y = a2 * x' + (- y2) = a'). In this situation we say that C is symmetric about the x-axis. Similarly, C will be symmetric about the y-axis if F(x, y) = 0 * F (-x, y) = 0. We say that C is symmetric about ihe origin (0, 0) if F(x, y) = 0 F(-x, -y) = 0.

Let us look at an example. The circle x' + y2 = 9 is synunetric about both the axes and the origin. On the other hand, the line y = x is not symmetric about any of the axes, but it is symmetric about the origin. Fig. 10: The curve C i s symmetric about the x-axis.

Geometrically, if a curve is symmetric about the x-axis, it means that the portion of the curve below the x-axis is the mirror image of the portion above the x-axis (see Fig. 10). A similar visual interpretation is true for symmetry about the, y-axis. And what does symmetry about the origin mean geometrically? It means that the mirror image of the portion of the curve in the first quadrant is the portion in the third quadrant, and the mirror image of the portion in the second quadrant is the portion in the fourth quadrant (see Fig. 11). .. Why don't you try some exercises on symmetry to see if you have grasped the concept?

E 11)

Which axis is the curve y2 = 2x symmetric.about? Is it symmetric about the origin?

Fig. I I :The line y = x is symmetric about the origin.

Fig. I2

E 12)

Discuss the symmetries of the line y = 2.

E 13)

Which of the curves in Fig. 12 are symmetric about the x-axis? And which ones are symmetric with respect to the origin?

E14) a) Show that if F(x, y) = 0 is symmetric about the x-axis, then F (x, y) = 0 iff F(x, - y) = 0. b)Show that if f (x, y) = 0 is symmetric about both the axes, then it is symmetric about the origin. Is the converse true? There is another concept that you will need while studying Units 2 and 3, which we shall now take up.

1.4

CHANGE OF AXES

In the next unit you will see that the general equation of a circle is x2 + y2 + 2ux + 2vy + c = 0. But we can always choose a coordinate system in which the equation simplifies to x2 + Y2 = r2, where r is the radius of the circle. To see why this happens, we need to see how to choose an appropriate set of coordinate axes. We also need to know how the coordinates of a point get affected by the transformations to a new set of axes. This is what we will discuss in this section. There are several ways in which axes can be changed. We shall see how the coordinates of a point in a rectangular Cartesian coordinate system are affected by two types of changes, namely, translation and rotation.

1.4.1 Translating the Axes The first type of change of axes that we consider is a shift in the origin without changing the direction of the axes. Let XOY be a rectangular Cartesian coordinate system. Suppose a point 0' has the coordinstes (a, b) in this system, what happens if we shift the origin to Of? Let O'X', parallel to OX, be the new x-axis. Similarly let O'Y', parallel to OY, be the new y-axis (see Fig. 13). Now, suppose a point P has the coordinates (x, y) and (x', y') with respect to the old and the new coordinate systems, respectively.

Fig. 13: Translation of axes through (a, b)

How are they related? From Fig. 13 you can see that x = x' + a and y = y' + b. r

Preliminaries in Plane Geometry

Conics

Thus, the new coordinates are given by x' = x - a and y ' = y - b.

...( 15)

For example, if we shift the origin to (-1, 2), the new (or current) coordinates (x', y') of a point P(x, y) will be given by x' = x + 1, y' = y - 2. When we shift the origin, keeping the axes parallel, we say that we are translating the axes. So, whenever we translate the axes to a point (a, b), we are transforming the coordinate system to a system with parallel axes through (a, b). We can write this briefly a s transforming to parallel axes through (a, b).

Now, if we translate the axes to a point (a, b), what will the resultant change in any equation be? Just replace x by x' + a and y by y' + b In the equation and you get the new equation. For example, the straight line x + 2y = 1 becomes (x' + a) + 2(y' + b) = 1, that is, x' + 2y' + a + 2b = 1 in the new system. Now for some exercises!

E15) If we translate the axes to (-1, 3), what are the new coordinates of the origin of the previous system? Check your answer with the help of a diagram. E16) Transform the quadratic equation 5x' + 3y' + 20x - 12y + 17 = 0 to parallel axes a) through the point (-2, 2). and b) through the point (1, 1). If you've done E16, you would have realised how much simplification can be achieved by an appropriate shift of the origin. Over here we would like to make an important observation.

Note: When you apply a translation of axes to a curve, the shape of the curve doesn't change. For example, a line remains a line and a circle remains a circle of the same radius. Such a transformation is called a rigid body motion. Now let us consider another kind of change of axes.

1.4.2 Rotating the Axes '

Let us now see what happens if we change the direction of the coordinate axes without shifting the origin. That is, we shall consider the transfc o?ation of coord~nateswhen the rectangular Cartesian system is rotated about the origin ~ l ~ o u ganh angle 0. Let the coordinate system XOY be rotated through an angle 0 m the anticlockw~sedirect~on

I I

Fig. 14: The axes OX' and OY' are obtained by rotating the axes O X and O Y through an angle 0.

about 0 in the XOY plane. Let OX' and OY' be the new axes (see Fig. 14). Let P be a point with coordinates (x, y) in the XOY system, and (x', y') in the X'OY' system. Drop perpendiculars PA and PB from P to OX' and OX, respectively. Also draw AC perpendicular to OX, and AD perpendicular to PB. Then x = OB, y = PB, x' = OA, y' = PA. Also L DAO = L AOC = 8. Therefore, L DPA = 8. Thus, x = OB = OC - AD = OA cos 8 - PA sin 8 = x' cos 8 - y' sin 8 and y = PB = AC + PD = x' sin 8 + y' cos 8 (16) and (17) give us x and y in terms of the new coordinate x' and y' Now, how can we get x' and y' in terms of x and y? Note that the xy-system can be obtained from the x'y'-system by rotating through angle (- 8). Thus, if we substitute - 8 for 8, x' by x and y' by y in (16) and (17), we get x' and y' in terms of x and y. Thus, x' = x cos 8 + y sin 8 ... (18) and y' = - x sin 8 + y cos 8 For example, the coordinates of a point P with current co-ordinates (x, y), when the rectangular axes are rotaied in the anticlockwise direction through 45", become 1

X' = x cos 45"

+ y sin 45O = JZ (x + y)

y' = - x sin 45" + y cos 4.5" =

1 JZ (y - x)

Now, what happens if we shift the origin and rotate the axes? We will need to apply all the transformations (14) - (17) to get the current coordinates. For example, suppose we transform to axes inclined at 30' to the original axes, the eqvation 1lx2 + 2 ,6 xy + 9y2 = 12 (x& + y + I), and then translate the system through

(f 0) , what do we get? We first apply (16) and (17), to get 9

Il(x'- & - y?2 + 2 , 6 ( x t ,6 - y') ( x ' + y' A ) + 9 ( x t + y' multiplied both sides by 4 = 12 { , 6 ( x '

we have

,6 - y ' ) + ( x t + y ' & ) + I}, that is,

Now if we shift the origin to

(f 0) and use (14). we find that the new coordinates 9

1 (X, Y) are related to (x', y') by x' = X + - , y' = Y + 0. 2 Thus, the equation will become

Isn't this an easier equation to handle than the one we started with? In fact, both the translation and rotation have been carefully chosen so as to simplify the equation at each stage. Note: The rotation of axes is a rigid body.motion. Thus, when such a transformation is applied to a curve, its position may change but its shape remains the same. Try these exercises now.

PreIiminalSiesin Plane Geometry

Conics

E 17)

Write the equation of the straight line x + y = 1 when the axes are rotated through 60".

E 18)

a) Suppose the origin is shifted to (-2, 1) and the rectangular Cartesian axes are rotated through 45". Find the resultant transformation of the equation x2 + Y2 + 4x - 2y + 4 = 0. b) Now, first rotate the axes through 45" and then shift the origin to (-2, 1). What is the resulting transformation of the equation in (a)? c) From (a) and (b) what do you learn about interchanging the transformations of axes, (You can study more about this in our course 'Linear Algebra').

So far we have been working with Cartesian coordinates. But is'there any other coordinate system that we can use? Let's see.

1.5

In the late 17th century the mathematician Bernoulli invented a coordinate system which is different from, but intimately related to, the Cartesian system. This is the polar coordinate system, and was used extensively by Newton. You will realise the utility of this system when you study conics in Unit 2. Now, let us see what polar coordinates are.

r '/': ,

,/ /

POLAR COORDINATES

:Ae

Fig. 15: Polar coordinates.

/n,4

To defme them, we first fix a pole 0 and a polar axis OA, as shown in Fig.' 15. Then we can locate any point P in the plane, if we know the distance OP, say r, and the angle AOP, say. 8 radians. (Does this remind you of the geometric representation of complex numbers?) Thus, given a poiilt P in the plane, we can represent it by a pair (r, 8), where r is the "directed distance" of P from 0 and 8 is L AOP, measured in radians in the anticlockwise direction. We use the term "directed distancev'&cause r can be negative

- also. For instance, the point P in Fig. 16 can be represented by (59 F)

(-59

a)..

Note that by this method the point 0 corresponds to (0, 8), for any angle 8. Fig. 16: P's polar coordinates are ( - 5 -

A point has many coordinates.

polar

Thus, for any point P, we have a pair of real numbers (r, 8) that corresponds to it. They are called the polar coordinates. . Now, if we keep 8 fixed, say 8 = a,and let r take on all real "slues, we get the line OP (see Fig. 17), where L AOP = a. Similarly, keeping r fixed, say r = a, and allowing 8 to take all real values, the point P(r, 8) traces a circle of radius a, with centre at the pole (Fig. 18). Here note that a negative value of 8 means that the angle has magnitude ( 8 1,

): is also

but is taken in the clockwise direction. Thus, for example the point (2. -

L /

represented by (2.

As you have probably guessed, the Cartesian and polar coordinates are very closely related. Can you find the relationship? From Fig. 19 you would agree that the relationship is

* A

211)

Pig. 17: The line L is given by 0 = -13.

1 Jw= I

X = r COS 8, y = r sin 8, or

tan-'

T3 Fig. 18: circle r = 1.

Note that the origin and the pole are coinciding here. This is usually the situation. -b

We use this relationship often while dealing with equations. For exampl,e, the Cartesian equation of the circle x2 + y2 = 25, reduces to the simple polar form r = 5. So we may prefer to use this simpler form rather than the Cartesian one. As 8 is not mentioned, this means 8 varies from 0 to 2x to 4x and so on. Doing the following exercises will help you get used to polar coordinates.

Prellmlnarles in Plane

Ccomdry

Fig. 19: Polar and Cartesian mrdinates.

E 19)

From (9) and (19), show that the polar tquation of the line AB in Fig. 7 is r cos (0 - a) = p.

E 20)

Draw the graph of the curve r cos (0 -

E 21)

Find the Cartesian forms of the equations a) rZ = 3r sin 8 b) r = a (1 - cos 0), where a is a constant.

= 0, as r and 0 vary.

Apart from the polar coordinate system, we have another method of representing points on a curve. This is the representation in terms of a parameter. You will come across this simple method in the next unit, when we discuss each conic separately. Let us now surnmarise that we have done in this unit

In this unit we have b-;?fly run through certain elementary concepts of two-dimensional analytical geometry. In pa~icular,we have covered the following points :

The distance between (x,, y,) and the line ax + by + c = 0 is

Any line parallel to the x-axis is y = a, and parallel to the y-axis is x = b, for some constants a and b.

The equation of a line in i) slope-intercept form is y = rnx + c, ii) point-slope form is y -yl = m(x - x,), Y - Y I = x-Xl iii) two-point form is Y2 'Yl x2 - X I

iv) intercept form is - + - = 1 , a b v) normal form is x, cos a + y sin a = p. 5)

The angle between two lines with slopes m, and IT+ is

Also see Unit 9 of MTE-01 (Calculus) for more about tracing of curves.

The lines are parallel if m, = q,and perpendicular if m , q

Conics

= -1.

Symmetry about the coordinate axes and the origin.

i) If we translate the axes to (a, b), keeping the directions of the axes unchanged, the new coordinates x' and y' are given by x' = x - a and y' = y - b. ii) If we rotate the axes through an angle 8, keeping the origin unchanged, the new coordinates x' and y' are given by x' = x cos 8 + y sin 8 y' = - x sin 6 + y cos 8.

A point P in a plane can be represented by a pair of real'numbers (r, 8), where r is the directed distance of P from the pole 0 and 8 is the angle that OP makes with the polar axis, measured in radians in the anticlockwise direction. These are the polar coordinates of P. They are related to the Cartesian coordinates (x, y) of P by rZ = x2 + yZ and -1

8 = tan

In the next unit we shall start our study of ellipses and other conics. But before going to it, please make sure that you have achieved the unit objectives listed in Sec. 1.1. One way of checking is to ensure that you have done all the exlrcises in the unit. Our solutions to these exercises are given in the following section.

Thus, the sides of the triangle are not equal in length. Hence, APQR is not equilateral.

E 3) The x and y-axis are y = 0 and x = 0, respectively.

-+ X

E 4) In Fig. 20 we have drawn the line. Its equation is y = mx + c, where c = -1 and m = tan 120' = - JI. Thus, the required equation is y = - ( J l x + 1). E 5) Here c = 0. Thus, the equation is y = x tan 6.

E 6) a) (2, 0) and (0, -3) lie on the line. Thus, its two-point form is y-0 x-2 -that is, 2y = 3(x - 2). 0-2' -3-0 b) (a, 0) and (0, b) lie on the line. Thus, its equation is y - 0 --x-a *-+-=I. x y -b-0 0-a a b 1 E 7) The equation of the line is y = - x + - ,that is, 2x + 2y - 1 = 0. 2 The distance of (1, 1) from this line is 3 2.1 + 2.1 - 1

I m I=z

Prdiminrria In .Plane Geometry

E 9) Using the intercept form (9) and Fig. 7, we see that the equation of the line is

Now, LOAC

- -a

OA = oc cosec

and LOBC = a. Thus, '

:( - a) p sec a (J-] , and cos a =

P OB = OC cosec a = sin. x c o s a y sin a -1 Thus, (20) 3 -+ -P P 3 x cos a + y sin a = p.

E 10)

a) Any line parallel to y + x + 1 = 0 is of the form y + x + c = 0. where c E R. Since (0, 0) lies on it, 0 + 0 + c = 0, that is, c = 0. Thus, the required line is y + x = 0. b) The slope of the line y + x = 0 is -1. Thus, the slope of any line perpendicular to it is 1, by (13). Thus, the equation of the required line is of the form y = x + c, where c E R. Since (2, 1) lies on it, 1 = 2 + c 3 c = -1. Thus, the required line is y = x - 1. c)

In this case m, = 1, m2 = 2. Thus, the angle between the lines is

Note that born -tan-]

(- i) and tan-' (- i) are angles between the

given lines.

E 11)

If we substitute y by (-y) in the given equation, it remains unchanged. Thus, the curve is symmetric about the x-axis. If we substitute x by (-x), the curve changes to y2 = -2x, Thus, it is not symmetric about the y-axis. If we substitute (-x) and (-y) for x and y, respectively, in the equation, it changes to y2 = -2x. Thus, it is not symmetric about the origin.

E 12)

It is not symmetric about either axis or the origin.

E 13)

(a) is symmetric with respect to the x-axis. (a) and (d) are symmetric with respect to the y-axis. (a) and (c) are symmetric with respect to the origin.

E 14)

a) The curve is symmetric about the x-axis. Thus, f(x, y) = 0 3 f(x, - y) = 0 t/ x, y E R. A F(x, - y) = 0 *Y;(x, -(-y)) = 0 3 F(x, y) = 0 v X, y E R. Hence, the equivalence. b) The curve is symmetric about both the axes. Now, F(x, y) = 0 3 F (x, -y) = 0, because of symmetry about the x-axis. 3 F (-x, -y) = 0, because of symmetry about the y-axis.

Conics

3 F is symmetric about the origin. The converse is clearly not true, as you can see from Fig. 11

E15) In Fig. 21 we show the new and old systems.

Fig. 21: The coordinates of 0 are (1, -3) with respect to X'O'Y'.

E16) a) If the new coordinates are x' and y', then x = x' - 2, y ='y' + 2. Thus, the equation becomes

b) The equation becomes 5(x' + 1)' + 3(yf + 1)' + 20 (x' + 1) - 12 (y' + 1) + 1,7 = 0

Thus, x + y = I , becomes y'J3

XI&

+(l+T] = 1, that is.

x'(1 + J3) + y'(1 - A ) = 2. E18) a) By shifting the origin, the new coordmates x' and y' are related to x and y by x = x ' - 2 , y = y ' + 1. Thus, the equation becomes (x' - 212 + (f + 112+ 4(xf - 2) - 2(yf + 1) + 4 = 0

Now, rotating the.axes through 45O, we get coordinates x and y given by, X-Y

= -

and y' =

Thus, (2 1) becomes

X+Y JZ

b) If we first rotate the axes, the given equation becomes

Preliminrrics in Plane Geometry

Now, shift the origin to (-2. I), the equation (2) becomes

c) From (a) and (b) you can see that a change in the order of transforniations makes a difference. That is, if T , and T, are two transformations, then T, followed by T, need not be the same as^, followed by T I . Diagrammatically, the circles C , and C, in Fig. 22 correspond to the final equations in (a) and (b), respectively.

Fig. 22

E19) From (9), the equation of L is x cos or + y sin or = p. Using (19), this becomes r (COS8 cos or + sin 8 sin a ) = p. r cos (8 - or) = p.

Fig. 11 : The line r ros (0 1

0. 1

E21) a) Since r2 = x- + y- and ); = 4 sin 8, the equation becomes x- + y- = 3y. b) The equation becomes

UNIT 2 THE STANDARD CONICS Structure 2.1 2.2 2.3

Introduction Objectives Focus Directrix Property Parabola

Description o f Standard Forms Tangents and Normals

2.4

Ellipse Descr~pt~on o f Standard Form String Property Tangents and Normals

2.5

2.6

2.7

2.1

Hyperbola Description of Standard Form String Property Tangents and Narmals Polar Equation of Conics Summary

INTRODUCTION

In this unit you will be studying some curves which may be-familiar to you. They were first studied systematically by the Greek astronomer Apollonius (approximately 225 B.C.). These curves are the parabola, ellipse and hyperbola. They are called conics (or conic sections) because, as you will see in this course, they can be formed by taking the .intersection of a plane and a cone. We start next unit by defimng conics as cw-es that satisfy the 'focus-directrix' property. From this definition, we will come to the particular cases of standard forms of a parabola, ellipse and a hyperbola. The standard forms are so called because any conic can be reduced to one of these forms, and then the various properties of the conic under consideration can be studied easily. We will trace the standard forms and look at their tangents and normals. We will also discuss some other characteristics, along with some of their applications in astronomy, military science, physics, etc. In the next unit we will discuss conics in general. And then, what you study in this unit will certainly be of help. If you achieve the following unii objectives, then you can be sure that you have grasped the contents of this unit.

QBJECTIVES After studying this unit, you should be able to obtain the equation of a conic if you know its focus and directrix; obtain the standard forms of the Cartesian and polar equations of a parabola, an ellipse or a hyperbola;

prove and apply the string property of an ellipse or a hyperbola; obtain the tangent or normal to a standard conic at a given point lying on it;

check whether a given line is a tangent to a given standard conic or not;

find the asymptotes ~f a hyperbola in standard form.

Let us now start ,our discussion on conics.

2.2 FOCUS-DIRECTRIX PROPERTY Suppose you toss a ball to your friend. What path will the ball trace'? It will be similar to the curve in Fig. 1, which is a parabola. With this section, we begin to take a close look at curves like a parabola, an ellipse or a hyperbola. Such curves are called conic sections. or conics. These curves satisfy a geometric property, which other curves do not satisfy. We treat this property as the definition of a conic section. Definition : A conic section, or a conic, is the set of all those points in two-dimensional space for which the distance from a fixed point F is a constant (say,e) times the distance from a fixed straight line L (see Fig.2).

T h e Standard Conics

8y---,

Fig. I: T h e ball. when thr?wn. traces a parabola.

The fixed point F is called a focus of the conic. The line L is known as a directrix of the conic. The number e is called the eccentricity of the conic. Since there are infinitely many lines and points in a plane, you may think that there are infinitely many types of conics. This is not so. In the rest of this block we will list the types of conics that there are and discuss them in detail. As a first step in this direction. lkt us see what the definition means in algebraic terms. We will obtain the equation of a conic section in the Cartesian coordinate system. Let F(a,b) be a focus of the conic. and px + qy + r = 0 be the directrix L (see Fig. 2). Let e be the eccentricity of the conic. Then a point P (x, y) lies on the conic iff J(x - a12 + (y - b12 = e

1 7%

, by Formulae (1) and ( l o ) of Unit 1.

Thus, (1) is the equation of the conic with a focus at (a, b), a directrix px + qy + r = 0 and eccentricity e. For example. the equation of the conic with a eccentricity 112, a focus at (1, 1) and a directrix x + y = 1 is

Why don't you try an exercise now?

.El) Find the equation of the conic section with a) eccentricity 1, (2, 0) as its focus x = y as its directrix, b) eccentricity 112, 2x + y = 1 as its direckix and (0. 1) as its focus (Note that in this case the focus lies on the directrix). In E l you have seen the two different possibilities that the focus may or may not lie on the directrix. Let us first consider the case when the focus does not lie on the directrix. In this.case the conics we get are called non-degenerate conics. Ylere are three types of such conlcs, depending on whether e < 1, e = 1 or e > 1. When e < 1, the conic is an ellipse; when e = 1, we get a parabola; and when e > 1, we get a hyperbola. We shall discuss each of these conics in detail in the following sections. Let us start with the non-degenerate conics with eccentricity 1.

2.3 PARABOLA In this section we will discuss the equation and properties of a parabola. Let us first define a parabola.

Fig. 2: T h e set of all points P, where P F = e P M is a conic.

Conics

Definition : A parabola is the set of all those points in two-dimensional space that are equidistant from a line L, and a point F not on L. L is its directrix and F is its focus. Let us use (1) to obtain the equation of a parabola. To start with let us assume F is (0, 0) and L is the straight line x + c = 0, where c > 0. (Thus, L is parallel to the y-axis and lies to the left of F). Then, using (I), we see that the equation of the parabola is 7

+ y- = (x + c)', that is, ... (2)

y' = c(2x + c).

(- 7.0).If we put 2 C

Now, to simplify the equation let us shift the origin to

= a, then

we are shifting the origin to (- a, 0). From Sec. 1.4.1 you know that the new coordinates x' and y' are given by x = x' - a and y = y' Thus, (2) becomes y'2 = 4ax'. .

This parabola has a focus at (a, 0) (in the X'Y' - system) and the equation of the directrix is x' + a = 0. So, what we have found is that the equation y2 = 4ax

... (3)

represents a parabola with x + a = 0 as its directrtx and (a, 0) as its focus. This is one of the standard forms of the equation of a parabola.. There are three other standard forms of the equation of a ptrabola. They are

= - 4?x, and

x2 = - 4ay

where a > 0. These equations are called standard forms because, as you will see in Unit 3, we can transform the equation of any parabola into one of these forms. The transfoimation th'at we use are the rigid body motions given in Sec. 1.4. So they do not affect the geometric properties of the curve that is being transformed. And, as you will see in the following sub-sections, the geometry of the standard forms are very easy to study. So, once we have the equation of a parabola, we transform it to a standard form and study its properties. And these properties will be the same as the properties of the parabola we started with. Now let us see what the standard,forms look like.

2.3.1 Description of Standard Forms Let us now see what a parabola look's like. We stait with tracing (3). For this, let us see what information we can get from the equation. Firstly, the curve intersects each of the axes in (0, 0) only. Next, we find that for the points (x, y) of the parabola, x 2 0, since y2 2 0. Thus, the curve lies in the first and fourth quadrants. Further, as x increases, y also increases in magnitude. And finally, the parabola (3) is symmetric about the x-axis, but not about the y-axis or the origin (see Sec. 1.3). Thus, the portions of the curve in the first quadrant and the fourth quadrant are mirror images of each other. Using all this information about the curve y2 = 4ax, we trace it in Fig. 3.

The Standard Conics

I Fig. 3 : y2 = 4nx, a > 0.

The line through the focus and perpendicular to the directrix is called the axis of the parabola. Thus, in this case the x-axis is the axis of the parabola.

\ I /

The point at which the parabola cuts its axis is called its vertex. Thus, (0, 0) is the vertex (plural 'vertices') of the parabola in Fig. 3.

\ I /

Now, what happens if we interchange x and y in (31? We will get (4). This is also a parabola. Its focus is at (0, a), and directrix is y + a =, 0. If we study the symmetry and other geometrical aspects of the curve, we find that its geometrical representation is as in Fig. 4. Its vertex is also at (0, 0) but its axis is not the same as that of (3). Its axis is x = 0, that is, the y-axis. Why don't you trace some parabolas yourself now? E2) Trace the standard forms (5) and ( 6 ) of a parabola. Explicitly state the coordinates of their vertices and foci. So far we have considered parabolas whose vertices are at (0, 0) and foci lie on one of the coordinate axes. I11 Unit 3 you will see that by applying the changes in axes that we havediscussed in Sec. 1. 4, we can always obtain the equations of a parabola in one of these standard forms. In this section we shall keep our discussion to parabolas in standard form. Now let us look at a simple 1.1echanical method of tracing a parabola. On a sheet of paper draw a s&aight line L and fix a point F not on L. Then, as in Fig. 5, fix one end of a piece of string with a drawing pin to the vertex A of a set-square. The length of the string should be the length of the side AD of the set-square. Fix the other end of the string with a drawing pin at the point F. Now slide the other leg of the set-square along a ruler placed on the line L (as in Fig. 5), and keep a pencil point P pressed to the side AD so that the string stays taut. Then, PQ = PF. This easily 2. follows from the fact that AD = AP + PF = length of string But AD = AP + PD Thus, as P moves, the curve that you draw will be part of a parabola with focus F and directrix L. Why don't you try this method for yourself?.Instead of a set-square you could simpry cut out a right-angled triangle from a piece of cardboard. E3) Use the mechanical method to trace the parabola x' + 8y = 0 So far we have expressed any point on a parabola in terms of its Cartesian coordinates x and y. But sometimes it is convenient to express it in terms of only one variable, or parameter, say t. you can check that the point (at2, 2at) lies on the parabola y2 = 4ax, for all t E R. Further, any.point (x, y) on this parabola is of the form (at2, 2at) where t = y/2a

ly+a=O Fig. 4 : x'

4ay. a > 0.

'Foci' is the plural 'of 'focus'

R. Thus, a point lies on the parabola yZ = 4ax if it can be represented by (atz, 2at) for some t E R. In other words.

Conics

Fig. 5 : ~ e c h a n i c a lmethod for t r a c i n g a parabola.

the parametric representation of any point on the parabola yZ = 4ax is x =. at2 , y = 2at, where t E R.

And now let us look at the intersection of a line and a parabola.

<. 'le llne segment PQ

1 chord

I S called

o f the parabola

P ( x I ,Y I )

2.3.2 Tangents and Normals

Let us consider the parabola y2 = 4ax. What is the equation of the line joining two distinkt points P(x,, y,) and Q(x,, y,) on it (see Fig. 6)? From Unit 1 we know that the equation of PQ is

-'")

Fig. 6 : PQ is a chord of the parabola.

Y-YI --

_-XI

Y2 -YI

x2 - X I

Since P and Q lie on the parabola,

= 4axl and

= 4ax2.

So we can write the equation of PQ as

e (y - y,) (y, +y2) = 4a(x - x ) (Note that y, # y,, since P and Q are distinct.)

e y(y, + yz) = 4ax + y,yz , since y:

= 4ax,.

...(7)

This is the equation of any line passing through two distinct points on the parabola. In particular, the equation of the line joining A(a, 2a)\and B(a, - 2a) is x = a, which is parallel to the directrix of the parabola. The chord AB has special significance. it is called the latus rectum o'f the parabola yZ = 4ax, and its length is 4a. Note that the focus lies on the latus rectum. Thus, the latus rectum is the chord of the parabola which corresponds to the line through its focus and perpendicular to its axis (see Fig. 7) Note that the length of the latus rectum is the coefficient of x in the equation of the pirabola. Similarly, the length of the latus rectum of xZ= 4ay is the coefficient of y.

T h e Standard Conics

Now for an exercise. 24) Find the equation of the latus rectum of xZ + 2y = 0. Wow, let us go back to Equation (7). .Suppose we take the point Q closer and closer to P, in Fig. 6 that is, Q tends to P. Then x2 tends x, and y2 to y,. In this limiting case the line PQ is given a special name. Definition : Let P and Q be any two points on a curve C which are close to each other. Then the line segment PQ is called a secant of C. The position of the line PQ when the point Q is taken closer and closer to P, and ultimately coincides with P, is called the tangent to the curve C a t P. P is called the point of contact or point of tangency. Thus, in Fig. 6, as Q moves along the curve towards P, the line PQ becomes a tangent to the parabola at the point P(x,, y,) (see Fig. 8). So, from (7) we see that the equation of the tangent at P is

Fig. 7: AB is the latus rectum of the parabola.

y . 2y, = 4 a x + y 2, = 4a (x + x,), since

= 'lax,.

w yy, = 2a (x + x,).

So. (8) is the equation of the tangent to the parabola y2 = 4ax at (x,, y,).

Fig. 8: The line L is the tangent at P to the parabola.

For example, the tangent to (3) at its vertex will be the y-axis, i.e., x = 0. And what will the equation of the tangent to yZ = x at (4, 2) be? It will be \

yy, =

X+X,

where x, = 4 and y, = 2, that is, 4y = (x + 4).

Have you noticed how we obtained (8) from (3)? We give you a rule of thumb that we follow in the remark below. Remark 1 : To get the equation of the tangent to the parabola y2 = 4ax at the point 1 (x,. y,), we replace yZby yy, and x by - (x + x, ). Similarly, the equation of the tangent 2 to xZ= 4ay at a point (x,, y,) l y q on it will be XX, = 2a (y + y,). You may like to try the following exercise now. E5) Find the equation of the tangent to a) xZ + 2y = 0 at its vertex, and b) yz + 4x = 0 at the end of its latus rectum.

E 6) Give an example of a line that intersects yZ = 4ax in only one point, but is not a tangent to the parabola. So, given a parabola and point on it, you have seen how to find the tangent at that point. But, given a line, can we tell whether it is a tangent to a given parabola? Let us see under what conditions the line y = mx + c is a tangent to y2 = 4ax. If y = mx + c meets the parabola at (x,, y,), then y:

= 4ax, and y , = mx, + c.

SO (mx, + c ) = ~ 4ax,, that is, mZx i + (2mc - 4a) x, + cZ= 0.

...(9)

Conics

Now, there are two possibilities m = 0 and m # 0. Can the first one arise? Can the line y = c be a tangent to y2 = 4ax? Suppose it is a tangent at a point (x,, y, ). Then y = c is the same as yy, = 2a (x + x,). This is not possible, sillce a # 0. Thus, for y = mx + c to be a tangent to y2 = 4ax, we must have m # 0. Then (9) is a quadratic equation in x,. So it has two roots. Corresponding to each root, we will get a point of intersection of the line and the parabola. Thus, a line can intersect a parabola in at most two points. If the roots of (9) are real and distinct, the line and parabola have two distinct common points. If the roots of (9) are real and coincide, the line will meet the parabola in exactly one point. And if the roots of (9) are imaginary, the line will not.intersect the parabola at all. So, if y = mx + c is a tangent to y2 = 4ax, the discriminant of (9) must be zero, that is,

a c = - , since m # 0. m

Thus,

y2 = 4ax if m # 0 and c = - . And then, what will the point of contact be? Since (9) has coincident roots, we see that a 4a-2m a 4a 2mc x, = m = - ; and then 2m2 2m2 m

a. Thus, y = mx + - will be a tangent to y2 = 4ax at the point m

TJsing the condition for tangency, we can say, for example, that the line 3x +2y = 5 is a tangent to yZ + 15x = 0, but not to y2 = 15x. And now for an exercise.

E 7) Under what conditions on m gnd c, will y = mx + c be a tangent to x2 = 4ay? A tangent to a parabola has several properties, but one of them in particular has many practical applications. This is the reflecting property. According to this, suppose a line L, parallel to the axis of a parabola, meets the parabola at a point P (see Fig. 9). Then the tangent to the parabola at P makes equal angles with L and with the focal radius PF. That is, a = P in Fig. 9. The reason this property is called the reflecting property is the following application: Take a mirror shaped like a parabola, that is, a parabolic mirror (see Fig. 10). If a ray of light parallel to the parabola's axis falls on the mirror, then the reflected ray will pass through the focus of the parabola. Thus, a beam of light, parallel to the axis converges to the focus, after reflection. Similarly, the rays of light that are emitted from a source at the focus will be reflected as a beam parallel to the axis. This is why parabolic mirrors are used in car headlights and searchlights.

The Standard Conics

The CQcai radius of a point P on a parabola with focus F is the line segment PF.

Fig. 9 ; Reflecting property ef a parabola.

It is also because of this property that the ancient Greek mathematician Archimedes could use parabolic reflectors to set fire to enemy vessels in the harbour! How did he manage this? Archimedes ingeniously thought of applying the following fact: If a parabolic reflector is turned towards the sun, then the rays of the sun will reflect and converge to the focus and create heat at this point. This is also the basis for solar-energy collectors like solar cookers.

L a

Fig. 10: A parabolic mirro.

'Focus' is Latid for 'fireplace'

The reflecting property is also the basis for using parabolic radio and visual telescopes, radars, etc. ' The following exercise is about the reflecting property. I

E8) A parabolic mirror for a searchlight is to be constructed with width 1 metre and depth Q.2 metres. Where should the light source be placed? In Fig. 11 we have given a cross section of the mirror. ~ i g 11 .

(Hint: The parabola is yZ= 4ax, and (0.2, 0.5) lies on it.) Now let us consider certain lines that are often spoken of along with tangent lines. These are the normals.

Definition : The normal to a curve at a point P on the curve is a straight line which is perpendicular to the tangent at P, and which passes through P (see Fig. 12). I

For example, a parabola's axis is the normal at its vertex.

' Now, let P (x, y,) be a point on y2 = 4ax. Then, you know that the equation of the

tangent at P is

.. ,

Fig. 12: L is the normal to the parabola at P.

yy, = 2a(x + x,). IS y, = 0, then x, = O.and the normal at (0, 0) is y = 0, the axis of the parabola. 2a On the other hand, if y, # 0, then the slope of the tangent at (x,, y,) is - . Y, Yl

So the slope of the normal will be - - (see Equation (13) of unit 1). Then, from Unit 1 2a you knpw that the equation of the normal at (x,, y,) is

Conics

Y-YI

= -

since y:

YI 2a

- (x - x ) ,

that is,

= 4ax1,

Note that (10) is valid even when y, = 0. So, for example, what will the equation of the normal to y2 = x at (1, 1) be? 1

Here a = - , x, = y, = 1. So, by (10) we find that the required equation is. 4

We end this section with some easy exercises. E9) Find the equation of the tangent and normal at (1, 1) to the parabola xZ= 4y. E1O)What is the normal at the point of contact of the tangent a

y = mx + - to the curve yz = 4ax? m With this we end our rather long discussion on the standard forms of parabolas. Now let us consider a conic whose focus doesn't lie on the directrix, and whose eccentricity is less than 1.

As the title of this section suggests, in it we shall study an ellipse and its properties. Let . us s'tart with a definition.

Definition: An ellipse is a set of points whose distance from a point F is e ( < I ) times its distance from a line L which does not pass through F. Let us find its Cartesian equation. For this we shall return to Equation ( I ) in Sec. 2.2 As in the case of a parabola, let us start by assuming that F is the origin and L is x + c = 0, for some constant c. Then (1) becomes

which is equivalent to

we now shift the origin to (I__ O) , the equation in the new X 1 Y1-system

becomes

This is of the form

e22

where a =

The Standard Conics

(ec12

and b2 =

= a'

(1 - e').

In the X'Y' - systems, the focus is (-ae. 0) and the directrix is x' + ae + c = 0, a that is, x' + - = 0. e Not that b2 = a2 (1 - e2) and e < 1. Thus, b1 < a'. So, if we simply retrace the steps we have taken above, and find the equation of an ellipse with focus (-ae, 0 ) and directrix a x + - = 0, we will get e

where b2 = a2 (1 - el). (1 1) is the standard form of the equation of an ellipse. As in the case of a parabola, we can always rotate and translate the axes so that the equation of any ellipse can be put in this form, for some a and b. We call it the standard form because it is a convenient form for checking any geometrical properties of an ellipse, or for solving problems related to an ellipse. Let us now study ( 1 1) carefully, and try to trace it.

2.4.1 Description of Standard Form Let us start by studying the sy~nnletryof the curve (see Sec. 1.3). Do you agree that the curve 1s symmetric with respect to the origin, as well as both the coordinate axes? Because of this, it 1s enough to draw the ellipse in the first quadrant. Why is this so? Well, the portion in the second quadrant will then be its reflection in the y-axis; and the rest of the curve will be the reflection in the x-axis of the portion in these two quadrants. Next, let us see where (1 1 ) intersects the coordinate axes. Putting y = 0 in (1 l), we get x = -+ a; and putting x = 0, we get y = b. So, (1 1) cuts the axes in the four points (a, 01, (- a, 01, (0, b), (0, - b).

Thirdly, let us see in which area of the plane. the ellipse (1 1) is defined. You can see that if I x i > a, y IS imaginary. Thus, the ellipse must.lie between x = - a and x = a. Similarly, it must lie between y = b and y = -.b. This infolmation helps us to trace the curve, which we have given in Fig. 13.

F (-ae, 0)

A' (-,:

Fig. 13: 'The ellipse

]A (a. O )

= 1

Conics

Looking at the symmetry of the curve, do you expect (ae, 0) to be a focus also? If you do, then you are on the right track. (1 1) has another focus at F' (ae, 0) with I a corresponding directrix x = - . Thus, (1 1) has two foci, namely, F(- ae, 0) and F' (ae, 0) ; e and it has two directrices (plural of 'directrix'), namely,

The chord of an ellipse which passes through the foci is called the major axis of the ellipse. The end points of the major axis are the vertices of the ellipse. Thus, in Fig. 13. A and A' are the vertices and the chord A' A. is the major axis. Its length is 2a. . The midpoint of the major axis is called the centre of the ellipse. You can see that the centre of the ellipse (1 1) is (0, 0). The chord of an ellipse which passes through its centre and is perpendicular to its majc axis is called the minor axis of the ellipse. In Fig. 13, the minor axis is the line segment B'B. Its leng,th is 2b. c'

Let us look at an example. Example 1: Find the eccentricity, foci and centre for the ellipse 2x2 + 3y2 = 1.

Solution: The given equation is

-x2 + - = I . y2

1 1 Comparing with (1 1), we get a = - b = -. Since.

1 1 b2 = a2(1- el), - = - ( 1 - e 2 ) ,

that is, e = - . I

The foci are given by (* ae, 0)

01.

And of course, the centre is (0, 0).

Now let us sketch an ellipse whose major axis is along the y-axis. In this case a and b in (1 1) get interchanged. x 2 Y2 Example 2: Sketch the ellipse -+ -= 1. . 9 25

Solution: This ellipse intersects the x-axis in (i3, O), and* y-axis in (0, 5). Thus, its major axis lies along the y-axis, and the minor axis lies along the x-axis. Thus, a and b of (1 1) have become interchanged. Note that (0, 0) is the centre of this ellipse too. Also, if e is the eccentricity of this ellipse, then 9 = 25 (1 - e2). Therefore,

e = - Thus, the foci lie at (0, 4) and (0, -4) (Remember that in this case the major axis

lies along the y-axis.) The directrices of this ellipse are y =

* -.4

We sketch the ellipse in Fig. 14. iIere are some exercises now. Ell)

Find the length of the major and minor axes, the eccentricity, the coordinates of the vertices and the foci of 3x2 + 4y2 =12. Hence sketch it. '-

El2)

Find the eauation of the elli~sewith center (0, 0). vertices at (* a, 0)

The Standard Conies

and eccentricity 0. Sketch this ellipse. Does the figure you obtain ltave another name?

x2 y2 Fig. 14: T h c ellipse -i - = 1 9 25

E13)

El4

The astronomer Johann Kepler discovered in 1609 that the earth and other planets travel in approximately elliptical orbits with the sun at one focus. If the ratio of the shortest to the longest distance of the earth from the sun is 29 to ' 30, find the eccentricity of the earth's orbit. Y x2 Consider the ellipse -+ ----?- = 1. where e is 'its eccentricity. 4(1-e-) 4 1 1 . 3 Sketch the ellipses that you get when e = -, e = - and e = -. Can 4 4 2 you find a relationship between the magnitude of e and the flatness of the ellipse?

What E 12 shows you is that a c;.-cle is a particular case of an ellipse, and the ...(12) equation of a circle with centre (0, 0) and radius a is x2 + y2 = a2. ,

You may be wondering about the directrices of a circle. In the following note we make an observatioil about them. Note : As the eccentricity of an ellipse gets smaller and smaller its directrices get farther and farther away from the centre. Ultimately, when e = 0, the directrices become lines at infmity. At this point let us mention the'parametric representation of an ellipse. As in the case x2 Y 2 of a parabola, we can express any point on the ellipse -+ y= 1. In terms of a a2 b parameter t. In this case, you can check that any point (x, y,) on the ellipse is given by x = a cos t, y = b sin t, where 0 5 t < 2 x . Note that the vertices will correspond to

f Let us now look at some important properties of an ellipse.

In this'section we derive a property that characterises an ellipse. Let us go back to Equation (1 1). Its foci are F (ae, 0) and F' (-ae, 0). Now take any point P (x, y) on the

Conics

ellipse. The focal distances of P are PF and PF'. What is their sum? If you apply the distance formula, you will find that PF + PF' = 2a, which is a constant, and is the length of the major axis. This property is true for any ellipse. Let us state it formally. Theorem 1 a) The sum of the focal distances of any point P on an ellipse is the length of the major axis of the ellipse. b) Conversely, the set of all points P in a plane such that the sum of distances of P fiom two fixed points F and F' in the plane is a constant, is an ellipse.

\ "\

.-_

F , ,, , - / -

Fig. 15

Proof: We have already proved (a). Let us prove (b). We can rotate and translate our coordinate system so that F and F' lie on the x-axis and (0, 0) 1s the midpoint of the line segment F' F. Then if F has coordinates (c, 0), F' will be glven by (- c, 0). Let P(x; y) be an arbitrary point, such that PF + PF' = 2a, where a is a constant (see Fig. 15). Then, by the distance formula we get

On squaring, simplifying again squaring and simplifying, we get (a2 - c2) x2 + ai y2 = a2(a2- c2). Now, since PFF' forms a triangle, FF' < PF + PF'. Therefore, 2c < 2a, that is, c < a. So we can rewrite the above equation as

-+-

= 1. where b =

Jn.

Comparing this with (1 1) we see that the set of all points (x, y) that satisfy the given condition is an ellipse with foci F and F', and major axis of length 2a. Mathematicians often use Theorem 1 as the definition of an ellipse. That is, an ellipse is the set .of all points in a plane for which the sum of the distances from two fixed points in the plane is constant. B

This property of an ellipse is also called the string property, because it is the basis for

, the following construction of an ellipse. A mechanical method for drawing an ellipse.

Flg. 16: Sketching an ellipse using a string.

Take a piece of string of length 2a and fix its ends at the points F and F' (where FF' < 2a) on a sheet of paper (see Fig. 16). Then, wlth the point of a pencil P, stretch the string into two segments. Now, rotate the pencil point all around on the paper while sliding it along the string. Make sure that the string is taut all the time. By doing this the point P will trace an ellipse with foci F and F' and major axis of length 2a. Why don't you try this method now? 1 ~ 1f u5s e the method we have just given to draw an ellipse with eccentricity - and a _2 string of length 4 units. What will the coordinates of its vertices and foci be?

There is another property o f an ellipse which makes it very us'eful in engineering. We shall tell you about it in our discussion on tangents.

2.4.3 Tangents and Normals In Sec. 2.3.2 you studied about a tangent of a parabola. We will discuss the tangents of an ellipse in the same manner.

Let P (x, , y, ) and Q(x2, y 2 ) be two distinct points on the -eH$se

T h e Standard Conics

2+ -= 1. a b2

If x, = x, = c, say, then the equation of PQ is

Similarly, if y, = y2 =A, say then the equation of PQ is y = d. If x, ;t x2 and y, ;t y 2 , then the equation of PQ iS

. x2 +y: - - (x-x1) (XI+ x 2 ) , since A - = >x2+ -y; z ( Y - Y , ) ( Y I+Y2) -

Fig. 17: L is tangent to the ellipse at P.

.(I51 since ( x,, y, ) lies on the ellipse. So, (13), (14) and (15) are the various possibilities for the equation of the line joining P and Q. Now, to get the equation of the tangent at P, we see what happens to the equation of PQ as Q tends to P (see Fig. 17). In this case, can you see from Fig. 13 that we need to consider (15) only? This is because as Q nears P, the line PQ can't be parallel to either axis. Now, as x2tends to X, and y2approaches y, (15) becomes, in.the limiting case, 2(% a

+ a) = 2, that is. b

x2 y2 Thuse, (16) is the equation of the tangent to -T + 7 ; .=~ 1 at I x l , y l ) . a

Remark 1 n;ay have already suggested this equation to you. The same rule of thumb works here too: thai is, replace x2 by xxi and y2 by yy,. 2

For example; the tangent to the ellipse in Example 1, i.e., to 2x + 3y,2 = 1 at

Now try this exercise on tangents. E 16)

Find the equations of the tangent's at the vertices and ends of the minor axis of x2 Y 2 the ellipse 7+ - = 1. a b2

Let us now find the equation of the normal to (1 1) at any point (x,, y,). If y, = 0, from

Conics

E 15 you know that the slope of the tangent will be n 12; and hence, at these points the; I normal is just the x-axis, that is, y = 0. Similarly, you can see that the normal at the points at which x, = 0 is the y-axis. Now suppose x, + 0, y, + 0.What is the slope of the normal at ( x,, Y, )? By (16) b2x, you know that the slope of ths: tangent is - - Thus, the slope of the normal at a2yI

, ,

( x ,y ) to the ellipse is

b'r, (See (13) of Unit I). Thus, the equation of the normal at

aLyl Y - Y I =( x - X I ) that is, b xl A diameter of an ell~pseis a chord that passes through the centre.

Why don't you try these exercises now? E17) Find the equation of the tangent and normal at (2, 1) to x2 + 4y2 = 8. E18) Show that the tangents at the extremity 0f.a diameter of an ellipse are parallel. Now, in Sec. 2.3.2, we discussed the reflecting property of a parabola. Do you expect it to hold for an ellipse too? The same property is not satisfied, but something like that is. Reflecting property: The tangent to an ellipse at a point makes equal angles with the focal radii from that point. That is, if you take the tangent PT at a point P on an ellipse (see Fig. 18), then it makes equal angles with the lines PF and PF' we' shall not prove this here. We leave the proof to you as an exercise (see Miscellaneous Exercises). Because of the reflecting property, a ray of light (or sound, or any other type of wave) that is emitted from one focus of a polished elliptical surface is reflected back to the other focus (see Fig. 19). One of the applications of this fact is its use for making whispering galleries.

. Fig. 18: a = P

Why don't you try and apply this property now? E19) An elliptic reflector is to be designed so as to concentrate all the light radiated from a point source on to another point 6 metres away. If the width of the reflector is 10 metres, how high should it be? Fig. 19: Reflected wave property.

Let us now see under what conditions a given line will be a tangent to the ellipse

Let the line be y = mx + c. I'

Substituting for y in'the equation of the ellipse we get x2 -+

(IILX+C)~ = 1, that is, b2

y = mx + c will be a tangent to the ellipse if this quadratic equation in x has equal roots. This will happen if its discriminant is zero, that is, 4 m 2 c 2 a 4= 4(b2+a'rn2) a 2 ( c 2 - b 2 ) 3 c -7= a -7m 2 + b 2 . So (18) is the condition that y = rnx + c is a tangent to

- Here's an opportunity for you to use this condition.

E 20) Check whether y = x + 5 touches the ellipse 2xZ +. 3y2 = 1 We shall now stop our discussion on ellipses, and shift our focus to another standard conic.

Let us now consider the coilic we get if e > 1 in Equation (I), namely, a hyperbola. Let us define it explicitly. Definition : A hyperbola is tbe set of points whose distance from a fixed point F is e ( > I ) times its distance from a fixed line L which doesn't pass through F. There is a similarity between the derivation of the standard equation of an ellipse and that of a hyperbola. In the following exercise we ask you to obtain this equation for a hyperbola. E 21) a) Show that the equation of a conic with focus at (0. 0), directrix x + c = 0 and eccentricity e > 1 is

Shift the origin suitably so as to get the equation

where a = ec c)

.b = a J K

What are the coordinates of the focus and the equation of the directrix in the

X'Y '-system? What is the equation of the conic with a focus at (-ae, 0) and directrix a x = - - , where e ( > I ) is the eccentricity? e

E 22)

As iq the case of the other conics, we can always translate and rotate our coordinate axes so as to get the focus of any given hyperbola as (-ae, O), and its directrix as x = - - a ~h us, we can always reduce, the equation of any hyperbola. to the equation e that you otitained in E 21, namely,

The Standard Cul~ics

where b2 = a2(eZ- 1).

Conics

So (19) is the standard form of the equation of a hyperbola. Let us now trace this curve.

2.5.1 Description of Standard Form Let us study (19) for symmetry and other properties.

Firstly, if - a < x < a, then there is no real value of y which satisfies 7 a -b?=' --

Thus, no part of the curve lies between the lilies x = - a and x = a. Secondly, it is symmetric about both the axes, as well as (0, 0). So it is enough to trace it in the first quadrant. Thirdly, the points (* a, 0) lie on it, and it does not intersect the y-axis. Finally, since x y2 , = 1+7 as x increases so does y. Thus, the hyperbola extends to infinity in both a b the x and y directions. We have sketched this conic in Fig. 20. You can see that it has two disjoint branches, unlike the other conics.

Fig. 20: The heperbola

y2 -= 1. b2

Looking at the curve's symmetry, do you feel that it has another focus and directrix? You can check that (ae, 0) is another focus with corresponding directrix

A hyperbola intersects the line joining its foci in two points. These points are called its vertices. The line segment joining its vertices is called its transverse axis. (Some people call the line joining the vertices the transverse axis.) Thus, the points A ' (- a, 0) and A (a, 0) are the vertices of the hyperbola in Fig. 20 and the line segment AA' is its transverse axis. The length of this transverse axis is 2a. The midpoint of the transverse axis is the Centre of the hyperbola. in Fig. 20, the line segment BB', where B is (0, b) and B' is (0, - b) is called the conjugate axis of the given hyperbola.

Note that it is perpendicular to the transverse axis and its midpoint is the centre of the hyperbola. The reason it is called the conjugate axis is because it becomes the transverse a

axis of the conjugate hyperbola of (19), 7- 7= 1 . (We shall not discuss conjugate b a hyperbolas in this course. If you would like to know more about them, you cac look L, 'A Textbook of Coordinate Geometry' by Ramesh Kumar). Let us consider an example of a hyperbola.

Example 3: For the hyperbola 4x2 - 9y2 = 36 find the vertices, eccentricity, foci and the axes. x7 -Y? Solution: We can write the equation in the standard form as = 1. 9 4

Comparing this with (19), we find that a = 3, b = 2. Therefore, the vertices are (* 3. 0). 13 Ji3 Now, since b2 = aZ (e2 - l), we find that eZ = Thus, the eccentricity is --- . Then the 9 ' 3 foci are (* ae, 0), that is, (5 f i ,0) The transverse axis is the line segment joining (3, 0) and (- 3, O), and the conjugate axis is the line segment joining (0, 2) and (0, - 2). Why don't you try some exercises now?

E 23) '

E 24)

Find the standard equation of the hyperbola with eccentricity & . (Such a hyperbola is called a rectangular hyperbola. Find the equation of the hyperbola with centre (0, 0), axes along the coordinate axes, and for which a) a vertex is at (0, 3) and the transverse axis is twice the length of the conjugate axis: b) a vertex is at (2. 0) and focus at F

(a, 0)

E 25 a) 'show that the lengths of the focal radii from any point P(x, y) on the hyperbola (19) are I ex + 9 ( and I ex - a 1. x2 y2 b) What is the analogue of (a) for the ellipse 7 + - = I? a- b'

E 26)

The more eccentric a hyperbola, the more its branches open out from its transverse axis. Ture or false:' Why?

As in the case of the other conics, we can give a parametric representation of any point on a hyperbola. What do you expect it to be? Does the equation Sec2t - tan2t = 1, v t E R help? Using this, we can give the parametric form of any point on (19) by x = a sec t, y = b tan t, for t E R such that 0 I t < 2x. Let us now look at some properties of a hyperbola.

2.5.2 String Property In Theorem 1 you saw that an ellipse is the path traced by a point, the sum of the distances of which fonp two fixed points is a constant. A similar property is true of a * hyperbola. Only, in this case, we look at the difference of the distances. Theorem 2 : a) The difference of the focal distances of any point on a hyperbola is equal to the length of its. transverse axis. b) Conversely, the set of points P such that I PF, - PF2 ( = 2a, where FI and F2 are two fixed points, a is a constant and FlF2> 2a, is a hyperbola. xz y 2 Proof : a) As you know, we can always assume that the hyperbola is a2 - ,2 = 1. Let P(x, y) be a point on it, and let its foci be F, and F,. Further, let D l and Dz be the feet of the perpend~cularsfrom P on the two directrices (see Fig. 21). In the figure you can see both the. cases - when P is on one branch of the hyperbola or the other. Now, by definition,.

PF,'= e PD2 and PF2 = ePD2. Therefore,

T h e Standard'Conics

Conics

):(

I PF, - PF, I = e I PD, - PD, I = e

Fig. 21:

= 2a, the length of the transverse axis.

1 PFI - PFI ( is constant

(You can also prove this by using E 25 (a)) We ask you to prove (b) in the following exercise. -

E 27)

Prove Theorem 2 (b). Where is the condition F, F2 > 2 a used?

Theorem 2 is called the string property, .for a reason that you may have guessed by now. We can use it to mechanically construct a hyperbola with a string. Since this construction is more elaborate than that of an ellipse, we shall not give it here. h

The string property is also the basis for hyperbolic navigation - a system developed during the World Wars for range finding and navigation. And now we shall see how to find the tangent to a hyperbola.

2.5.3 Tangents and Normals You must have noticed the \similarity between the characteristics of an ellipse and a hyperbola. The derivation of the equation of a chord joining two points on a hyperbola and of the equation of a tangent are also obtained as in Sec. 2.4.3. We shall not give the details here. Suffice it to say that all these equations can be obtained from the elliptic case by substituting -b2 for b2. x2

Thus, the equation of the tangent to 7- 7= 1 at a point ( XI, y l ) lying on it is a b

Also, the equation of the normal at a point (x,, y,) to

Similarly, the condition for the straight line y = mx + c to be a tangent tb

The Standard Conics

Kow for a short exercise. x2

E 28)

a ) Find the tangent and normal --II- = I at each of its vertices. 4 9 b) Is 3y = 2x a tangent to this hyperbola? If so, find the point of tangency.

We will now introduce you to some special tangents to a hyperbola. Consider the x2 Y 2 b hyperbo1,a 7- 7= 1 and the lines y = f - x (see Fig. 22). These lines satisfy the a b a condition for tangency. They are the pair of tangents to the hyperbola which pass through its centre. Such tangents are called the asymptotes of the hyperbola. Now, let P (x, y) be a point of the branch of the hyperbola in the first quadrant. b

Then its distance from y = - x is a lay- bxi - la'y'-

b'x'~

- J-r7 a + b- (ay+ bx)

b Since b2x2 -a2Y' = a 2 b 2and y = a a'b

Fig. 22: y =

d m

x2 y2 *-b x are the asymptotes of --.2 b2 ' ''

As x increases, this distance gets smaller and smaller. Thus, as P tends to infinitely alongb the branch y of the hyperbola, its distance from y = - x tends to zero. a a But the asynlptote never actually intersects the curve. So we say that its point of contact is at an infinite distance.

- 'm

b You can check that the same is true of y = - - x. In fact, tangents with their points of a contact 'at infinity' are called asymptotes. Try these exercises now. 41

Conics

E29) Find the asymptotes of the rectangular hyperbola x2 - y2 = aZ.Are they the same for any value of a ? Asymptotes are d~scussedtn the course MTE-01 ~n d e t a ~ l

E30) Under what conditions on a and b will the asymptotes of the hyperbda (19) be pelpendicular to each other? So far we have discussed the Cartesian and parametric equations of the conics. But, in some applications the polar equation (see Sec. 1.5) of a conic is more useful. So let us see what this equation is.

2.6

POLAR EOUATION OF CONICS

Consider a conic with eccentricity e. Take a focus F as the pole. We can always rotate the conic so that the corresponding directrix L lies to the left of the pole, as in Fig. 23. Let the line FA, perpendicular to the directrix, be the polar axis and d the distance between F and L. Let P (r, ) be any point on the conic. Then, if D and E are the feet of the perpe~idicularsfrom P onto L and FA, we have

3 r = e(d - EF) = e (d - r cos ( z - Q ) ) = e(d + r cos

,g ).

Fig. 23: Obtaining the polar equation of a conic

which is the polar equation of a conic. Can you find the .polar equations of the standard conics from this?

For instance, the polar equation of the parabola (3) is r =

d 1 - e cose '

where d = 2a. Now, suppose you try to derive the polar equation of a conic by taking the directrix L corresponding to a focus F, to the right of F. Will you get (22)? You can check that the equation will now be

Let us consider an application of the polar form.

I.:ranlple 4: In Fig. 24 we show the elliptical orbit of the earth around the sun, which is at a focus F. The point A on the ellipse closest to the sun is called the perihelion; and the point A ' farthest from the sun is called the aphelion.

Fig. 24: A p l i r l i o ~ iancl p e r i l i r l i o ~ iO I I the o r b i t of the e a r t h a r o u n d t.he sun.

Show that the perihelion distance FA and the aphelion distance FA' are given by ed ed and FA' = -- where d is as given in (22) and (23). FA = -I+e 1-e'

n ). Thus, from (23)

5olution: The polar coordinates of A are (FA, 0) and of A' are (FA', we find that

You can do the following exercises to see if you have understood what we have done in this section.

E 31)

Let 2a be the length o f t h e major axis of the ellipse ed ed r = show that a = 7 . I + e cos 8 I -e-

E 32)

A cornet is travelling in a parabolic course. A polar coordinate system is introduced id the plane of the parabola so that the sun lies at the focus and the polar axis is along the axis of the parabola, drawn in the direction in which the curve opens. When the comet is 3.0. lo7 km from the centre of the sun a ray from the sun to the comet makes an angle of ~c / 3 with the polar axis. Find a) an equation for this parabolic path, b) the miilimum distance of the comet from the sun, 7-l

c) the distance between the comet and the sun when 0 = -

So in this unit you have seen the Cartesian, parametric and polar representations of the various conics for which the foci do not lie on the corresponding directrices. Such conics are called non-degenerate conic$.In case a focus of a conic. lies on the directrix corresponding to it, the conic we get is callcd a degenerate conic. We will not go into details about them. But let us list the possible types that there are. A degenerate conic can be of 5 types: a point, a pair of intersecting lines, a pair of distinct parallel lines, a pair of coincident 1ines.and the empty set. Now let us do a brief run-through of what we have covered in this unit.

In this unit we have discussed the following points : 1)

The focus-directrix defining property of conics.

The Standard C o ~ ~ i c s

Conics

A standard form of a parabola is y2 = 4ax. it's focus is at (a, O), and directrix is x = - a. Its eccentricity is 1. The other standard forms are xZ = 4ay, x2 = - 4ay and y2 = - 4ax, a > 0.

The tangent to y2 = 4ax at a point (x,, y,) lying on it is yy, = 2a(x + x l ).

a y = mx + c is a tangent to yZ = 4ax if c = - . .

yif The normal at ( X I , yl ) to y2 = 4ax is y = - Y l x + y , +2a 8a2

The standard form of the equation of an ellipse with eccentricity e (< 1) is

a Its foci are (* ae, 0) and directrices are x = - .

The sum of the focal distances of any point on an ellipse equals the length of' the major axis of the ellipse.

xi yL YY =I x+ The tangent to the ellipse --i+ - = I at ( x,, X, ) is a b2 ab

xi yL y = mx + c is a tangent to -+ - = 1 if c2 = a2m2+ b2. a2 b2

x y2 10) The normal to ,+--5-=1 a- b-

a' b at(x,,y,)is-(y-y,)=-(x-x,) YI XI

11) The standard form of the equation of a hyperbola with eccentricity e ( > I ) is x2 Y 2 1 where b' = a* (e2 - 1). Its foci are (* ae, 0) and directrices are a 2 b2 -

12) The difference of the focal distances of any point on a hyperbola equals the length of its transverse axis. 13) The tangent to the hyperbola

14) y

mx + c is a tangent to

15) The normal

x2 a

XXl

YYl

7- -y= 1 at ( x l , y 1 ) is 7 7 x2 a2

= 1.

y2 - 1 if b2

-b2.

x2 y 2 a' b' = I at ( X I . Y ) ) is y ( X - X ~ ) + - ( Y - Y I ) = ~ a 2 b2 -

x2 Y 2 16) The asymptotes of the hyperbola 7- - = lare y = a b2

* -ax .

17) The parametric representation of any polnt on

a) the parabola y2 = 4ax is (at2, 2at). where t E R; x2 y 2 b) theellipse - + T = I a2 b

i s ( a c o s e , b s i n e ) w h e r e O I0 < 271.

x2 y 2 c) the hyperbola - - - --1 i s ( a s e c e , a 2 b2

b t a n e ) , w h e r e O I 0 < 271;

18) The polar equation of a conic with eccentricity e is r =

I - e cose

The Standard C o ~ ~ i c s

or.

ed depending on whether the directrix being considered is to the I + e cose ' left or to the right of the corresponding focus.

r =

Here. d is the distance of the focus from the directrix. 19) The list of possible conics is

Table 1: Standard Forms of Conics. Conic

Standard Equation

Ellipse

xz Y 2 -+T=l,a,b>O a2 b

Sketch

Circle

x2+y2=a2, a t 0

Hyperbola

x2 a:

Parabola

y' =4px, p > O

y2 = I , a, b > O b2

.+.

x2 Y 2 -0, a,b t o Pair of intersecting l ~ n e s -- a 2 b2

$,-

Pair of parallel lines

y =a2, a > 0

Pair of coincident lines

y' = 0

Point conic

x2 y2 --i-+;l=O.a,b a b

* *+-

t 0

. *

Conics

And now you may like to check whether you have achieved the objectives of thisunit (see Sec. 2.1). If 'you'd like to see our solutions to the exercises in this unit, we have given them in the following section.

2.8 SOLUTIONS 1 ANSWERS 4y 2 (x-Y)~

E 1) a) The required equation is (x- 2)2 + y 2 = 1

a x 2 + y 2 +2xy-8x+8=0. 1 (2x+y-I)' b) The required equation is X' + ( Y- 1 )2 = 4 5 a 1 6 x 2 - 4 ~ ~ + +4&38y+19=0. 1 9 ~ ~

Fig. 25: y' = - 4ax, a > 0.

E2) In Fig. 25 we have traced the parabola y2 = - 4ax, a > 0. Its vertex (0, 0) and focus is (-a, 0). In Fig. 26 we have drawn x2 ' = - 4ay, a > 0. Its vertex 1s (0, 0) and focus is (0, -a). E3) The parabola is similar to the one in Fig. 26. E4) The parabola is x2= - 2y. Thus, its fmus is (0, - I!?),

and its latus rectum is

1 =

-2.

Fig. 26: x' = - 4ay, a > 0.

E5) a) The equation is XXl + 2

(

~ 0, )where", =YI = 0, that is, y = 0.

b) The ends of the latus rectum are-+ 1, 2) and (- 1, -2). The tangents at these points are x + y - 1 = 0 and x - y - 1 = 0. E6) The axis of the parabola intersects it at the vertex only, but it is not a tangent at the vertex. E7) The first point to note is that no tangent line can be parallel to the axis of the parabola. For any other m, the line will be a tangent at ( x , , y,) if

x: = 4ay,, y,= mx, + c, and xf 4a (mx, + c) has coincident roots. Thus, y = r n x + c willbeatangentifm # t a n n / 2 a n d c = - a m 2 . E8) We have to find the focus of the parabola. We know thaf (0.2, 0.5) lies on it. Therefore, a = 0.3125. 0.25 = 4a(0.2) = 0.8a

1 E9) The tangent is x = 2(y + 1). Its slope is 2

Thus, the normal is y - 1 = - 2 (x - 1). E1O)The point of contact is

(

) ;2a .

1 The slope of the normal is - - .

n u s , its equation is Y

=.-;(x ; - +) m 2a

T h e Standard Conics

x2 y2 The equation can be rewritten as - + - = 1 . 4 3 The major axis is of length 4 and lies along the x-axis. The minor axis is of length 2 & . :.

(A)'= 2'(.1 - e 2 ) , where e is the eccentricity.

The vertices are (* 2, 0) and the foci are (*I, 0). We trace the curve in Fig. 27.

Fig. 27

x2 y 2 The equation is -+T = 1 , that is, x2 + = a 2 . The foci colncide with the a2 a cente (0, 0). In this case the ellipse becomes a circle, given in Fig. 28.

a '

The shortest and longest distances will be the distances of the vertices from the focus at which the sun lies. So, suppose the orbit is

x2 . y2 7 a2( - e2) = 1 and the sun is at (ae, 0). +

The vertices lie at (- a, 0) and (- a, 0). Then

Fig. 28: T h e circle x' + yz = a'.

Y, -

a-ae 29 1 - =e=-. a + a e 30 . 59. As e grows larger the nlinor axis becomes smaller, and the ellipse grows ,flatter. Thus, the ecceiltricity of an ellipse is a measure of its flatness.

Your ellipse should be similar to the one in Fig. 27 Its vertices will be (+ 2,O). Its foci will be (* 1,O). xa y.0 The tangent at (a, 0) is 7+ -- 1 X = a . Similarly, the tangents at (- a, 0), a b2 (0, b) and (0, -b) are x = :a , y = b and y = - b, respectively. I

The tangent is 2x +. 4y = 8, that is, x + 2y = 4. Therefore, the slope of the normal is 2. Thus, its equation is y - 1 = 2 (x - 2) 2 y = 2x - 3. \

x2 y 2 Let the ellipse be -+ - = 1 . The equatlon of any diameter will be y = mx, a2 b2 for some m, since it passes through (0, 0). Further, if one end of the diameter is ( x , , y , ), then y, = m x , . Thus, (- x , - y, ) also lies on the ellipse and the

Conics

line y = mx. Thus, it is the other end of the diameter. So, we need to find the tangents at ( x,, y , ) and (-x, - y,). They are XXI YYI ---i-+-=l anda b3

-+(yl 7)

= I , respectively. Since both their

b2xl slopes are -, they are parallel. a2y1 \>'-.,;,'<\ 2 ,,., . ;--:/ py, h~,&,+\,,

,,,

y'--

:, ,

E 19)

3 ellipse is 10 metres and its foci lie at (5 3, 0). Thus, its eccentricity, e = 5

.We have shown a cross-section of the reflector in Fig. 29. The major axis of thiy

Fig. 29

So, if h is its height, then h = 5

.. C2 + E 21)

= 4 metres.

+ b2. SO the line is not a tangent to the given ellipse.

a) Using (I), we see that

( 21) x+-

- e2c2

e2-1

(e2-l)2

as. in Sec. 2.4.

(-=* ce

Shifting the origin to

=---= "' Y" a2

O) the equation in (a) becomes

ec 1 , where a = T e -1

a c) In the X'Y' -system the focus is (- ae, 0) and directrix is x + e = 0.

E 22)

The required equation is

E 23)

The required equation is

E 24)

a) The transverse axis lies along the y:wis in this case. Thus, we interchange ' x and y in (19):Also, the length of' the transverse axis is 6. So, the required equation is

b) Here, the transverse axis lies along the x-axis, and a = 2 and ae =

1 :. e = 2

The Standard Conics

J13.

Thus, the required equation is

E 25) a) The foci lie at F(- ae, 0) and F' (ae, 0). Thus

But, since P lies'on the hyperbola.

Similarly, PF' = ( ex - a 1. b) In this case PF = ex + a and PF' = a - ex. Remember, from Sec. 2.4.2, that PF + PF' = 2a.

E 26)

Suppose you fix the transverse axis and increase the eccentricity of a hyperbola. You will see that the lengths of its latera recta (plural of 'latus rectum') increase. Thus, the given statement is true.

E 27)

Let F, F2 = 2c and let (0, 0) bisect F, F2 . Then the coordinates of F, will be (- c, 0) and of F2 will be (c, 0). If P is given by (x, y), then IP F, - P F2 I = 2a

= 1 ( ~ ~ + ~ ~ + =~ ~ ~( X- 2~ -a C~ ~ ) ) ~~ + ~ ~ ~ ( C ~ squaring + X ~ ) and + ~ ~ . O ~ simplifying. 3 (c2 - a2) x2 - aZy2= aZ (c2 - a2), again squaring and simplifying. Now, since 2

X y2 c > a, we can rewrite this equation as -- m= 1 , which is a hyperbola. a2 C - a .

E 28) a) The vertices are (2, 0) and (-2, 0). The tangents at these points are x = 2 and x = - 2, respectively. The normals at both these points is the x-axis. 2 2 b) Here a2 = 4 , b = 9 , ,TI=--,c = O . 3

:. c 2 # a 2 m 2 - b 2 .

:. 3y = 2x is not a tangent to the given hyperbola.

E 29)

y = *x, which are independent of a . Thus, these are the asymptotes of any rectangular hyperbola.

E 30)

b y = -X a

(3(-:)

b and y = - - x will be mutual]y perpendicular iff a = -1,

that is, iff a = b, that is, iff the hyperbola is rectangular.

Conics

E 31)

ed ed As in Example 5, you can show that FA = - and FA' = I -e l+e' where A and A' are the vertices of the ellipse and F is a focus. 2ed Then 2a = AA' = FA + FA' = 7 , I -e-

E 32) a) Since the curve is a parabola, its equation is r= . We also h o w that (3.0 x l o 7 .n 1 3) lies on it. 1- C O S ~

1.5 l o 7 Thus, the required equation is r = I-COS~ b) The minimum distance will be when the comet is at the vertex of thr parabola, that is, when 0 = x . 1.5 x 10' km. Thus, the minimum distance = 2

c) The required distance is 1.5 x l o 7 km.

GENERAL THEORY OF CONICS

UNIT 3 Structure

Introduction Objectives

General Second Degree Equation Central and Non-central Conics Tracing a Conic 3.4.1 Central Conics 3.4.2 Parabola

Tangents Intersection of Conics Summary Solutions/Answers

3.1

INTRODUCTION

So far you have studied the standard equations of a parabola, an ellipse and a hyperbola. We defined these curves and other conics by the focus-directrix property of a conic. This defining property was discovered by Pappus (approx. 320 AD) long after the definition of conic sections by the ancient Greeks. In his book "Conics", the ancient Greek mathematician Apollonius defined these curves to be the intersection of a plane and a cone. You will study cones later, in unit 6, but let us show you how conics are planar sections of a cone, with the help of diagrams (see Fig.1)

Fig. 1: A planar section o f a cone can b e (a) an ellipse, (b) a parabola, (c) a hyperbola (d) a p a i r o f lines, (e) a point.

In this unit we will prove a result that may surprise you. According to this result, the general second degree equation ax2 + hxy + by2 + gx + fy + c = 0 always represents a conic section. You will see how to identify it with the various conics, depending on the conditions satisfied by the coefficients. In Unit 2 you saw one way of classifying conics. There is another way of doing so; which you wilI study in Sec. 3.3. We shall discuss the geometric properties of the different types of conics, and see how to trace them. After that, we shall discuss the tangents of a conic. And finally we shall see what curves can be obtained when two conics intersect. With this unit we end our discussion on conics. But in the next two blocks you will be coming across them again. So, the rest of the course will be easier for you to grasp if .you' ensure that you have achieved the unit objectives given below.

Objectives After studying this unit you should be able to identify the conic represented by a quadratic expression; .-

c-a *I.-

---&..-

/:r :4 -.,:"*"\

..-A

--.-- r., - -,.-:-.

Conics

trace any given conic; find the tangent and normal to a given conic at a given point; obtain the equations of conics which pass through the points o f intersection of two given conics.

3.2 GENERAL SECOND DEGREE EQUATION In U~lit2 you must have noticed that the standard equation of each conic is a second degree equation of the fornl ax' + hxy + by2 + gx + fy + c = o for some a, b, c,.f, g, h E R and where at least one of a,h,b is non-zero. We write the coefficient; of xy, x and y as 2h. 26. and 2f to have simpler expressions later on. as you will see.

In this section we will show you that the converse is also true. That is, we will prove that the general second degree equation

where at least one of a,h,b is non-zero, can be transformed into a standard equation of a conic. We achieve this by translating and rotating the coordinate axes. Let us see how. We first get rid of the term containing xy by rotating the XY-system through a "suitable" angle 0 about origin. You will see how we choose8 a little further on. Now, by (16) and (17) of Unit 1. we see that (1) becomes a(x' cos 0 - y' sin O)z + 2h(x'cos 0 - y'sin 0) ( x ' s ~ n0 + y'cos 0) + b(x'sin 0 + y'cos 0)2 + ~ ~ ( X ' C0O-Sy'sin 0) + 2f(x'sin 0 + y'cos 0) + c = 0 3 (a cos2 0 + 2h cos 0 sin 0 2 b sinZ 0) x r Z- 2 ((a - b) sin 0 cos 0) h(cosz 0 - sin2 0)) x'y' + (a sin2 0 - 2h sin 0 cos 0 + b cosZ 0 ) ~ +' ~ (2g cos 0 + 2f sin 0) x'4 (2f cos 0- 2gsin 0) y' + c = 0. The x'y' term will disappear if (a - b) sin 0 cos 0 = h(cosz 0 - sin2 0), that is, I -(a - b) sin 20= h cos 20. 2 Tt

So, to get rid of the x'y' term, if a = b we can choose 0 = - ; otherwise we, 4

sin 28 = 2 sin 8 cos 8 cos 28 = cos28 - sin28

1 call choose 0 = - tan-' 2 (a!b) '

( We can always choose such a 0 lying between 4 and ): . X

For this choice of 0 the x'y'

term becomes zero.

So, if we rotate the axes through an angle 0 = - tan-i 2 the second degree equation

AX'^ +

(&), then ( I ) tra~isformsinto

+ 2Gx' + 2Fy' + C = 0,

...(2)

where A = a cosZ 0 + 2h cos 0 sin 0 + b sinZ 0 and

B = a sinZ 0 - 2h sin 8 cos .0 + b cos' 0. Thus, A + B = a + b. Also, with a bit of computation, you can check that ab - hZ = AB. Now various situations can arise.

Case 1 (ab - h2 = 0): 111this case we see that either A = 0 or B = 0. So, let us assume

that A = 0. Then we claim that B must be non-zero.. Do you agree? What would happen if A = 0 and B = O? In this case we would get a = 0, b = 0 and h = 0, which contradicts our assumption that ( I ) is a quadratic equation. ,So, let A = 0 and B # O . Then (2) can be written as I

Now, if G = 0, then the above equation is

F F'-BC ( Y P + ~ ) =T

, that is,

F IF'-BC y+B=+ljB'.

This represents a pair of parallel lines if F' 2 BC, and the empty set if F' < BC. On the other hand, if G # 0, then we write (3) as 2C

2BG

Now if we shift the origin to

2BG -

-F

ZG7T)

. then the equation becomes

where X, Y are the current coordinates. From Sec. 2.3. You know that this represents.a G G parabola with (-%. 0) and directrix X = -. 2B

Now let us look at the other case. Case 2

(ab - h'

* 0) : Now both A and B are non-zero. We can write (2) zis , which is a constant K, say.

Let us shift the origin to

-i\Y-B F, . Then this equation becomes

where X and Y are the current coordinates. Now, what happens if K = O ? Well, if both A and B have the same sign, that is, If AB = ab - hZ > 0, then (4) represents the point (0, 0). And. if ab - hZ < 0, then (4) represents the pair of lines, -

And,'what happens if K # O'? Then we'can write (4)

Does this equatich look familiar'? From Sec. 2. 4. You can see that this represents an K K ellipse if both - and - are. positive, that is, if K > '0 and AB = ab - hZ > 0. A B But, what if WA < 0 and KIB < O ? In this case K < 0 and ab - hZ > 0. And then (5) represents the empty set.

General Theory of Conics.

Conics

And, if -- and - are of opposite signs, that is. if AB = ab - h2 < 0, then what will A B (5) represent? A hyperbola So we have covered all the possibilities for ab - h2, and hence for ( 1 ) . Thus, we have proved the following result. Theoren1 1: The general second degree equation ax2 + 2hxy + by2 7 2gx 4 2fy + c = 0 represents a conic. While proving this theorem you nlust have noticed the inlportailce we gavc expression ab - h2 . Let us tabulate the various types of non-degenerate ailif di :nerate conics that ax2 + 211xy + by2 + 2gx + 2fy + c = 0 represents, according to the u a y ab - h2 behaves. (Recall from Unit 2 that a degenerate conic is a conic whose focus lies on the corresponding directix.) Table 1 : Classification of Conics. I

Types of Con~cs

Condition

Non-degenerate

~egenerate

ab - hZ = 0

parabola

pair of parallel lines, or empty set

ab - h2 > 0

ellipse

point, or einptv set

ab - hZ < 0

- hyperbola

pair of intersecting lines

Table 1 tells us about all the possible conics that exist. This is what the following exercise is about.

El) a) b)

Write down all the possible types of conics there are. Which of them are degenerate? If ( 1 ) represents a circle, will ab - h2 = O'?

Now let use the procedure in the proof above in some examples. Example 1: Find the conic represented by 9x2 - 24 xy + 16y2- 1 2 4 + ~ 132y + 324 = 0. Solution: The given equation is of the form ( l ) , where a = 9, b = 16, h = - 12. Now let us rotate the axes through an angle 8, where tan 28

2h a-b

24 7

= -= - , that

2tan9 - 24 is, -- -- , that is, 1 - tan% 7

So we can take tane' = - , and then sin0 4

3 4 and cos0 = 7 . 5

?'hen, in the new coordillate system the given equation becomes

. Now let us ihift the origin to o()~ !-

. Then the equation becomes

where X and Y are the current coordinates.

c e n e r h ~Theory of Conics

Can you recognise the conic represented by this equation? From Unit 2 you know that this is a parabola. Since the tl-ansformations we have applied do not alter the curve, the original equation also represents a parabola.

Example 2 : Identify the conic x2 - 2xy + y2 = 2. Solution: Over here, since a = b = 1, we choose 8 = 45" So, let us rotate the axes . through 45".The new coordinates x ' and y' are given by

[

x =-(XI-

y') and y = - ( x l +

y')

I Then, in the new coordinate system.

I I

xZ - 2xy + y2 = 2 transforms to Y'2 = 1

which represents the pair of straight lines y' = 1 and y' = -1. You can do the following exercise on the same lines.

E 2) Identify the conic

So far you have seen that any second degree equation represents one of the following conics:

a parabola, an ellipse, a hyperbola, a pair of straight lines, a point, the empty set.

But, from Table 1 you can w e that even if we know the value of ab - h2, we can't irnmediately.say what the conic is. So, each time we have to go through the whole procedure of Theorem 1 to identify the conic represented by a given equation. Is there a short cut? Yes, there is. We have a simple condition for (1) to represent a pair of lines. It can be obtained from the proof of Theorem 1 after some calculations, or independently.. We shall only state it, and then see how to use it to cut short our method for identifying a given conic. .

Theorem 2: The quadratic equation

represents a pair of straight lines if and only if abc + 2fgh - af2 - bg2 - ch2 = 0, that is, the determinant

Further, if the condition is satisfied, then the angle between the lines is

The 3 X . 3 determinant given above is called the discriminant of the given conic. You can see that the discriminant looks neater if we take 2h, 2g and 2f as coefficients, instead of h, g and f. Let us consider some exa.mples of the use of Theorem 2. I

Recall the definit~ono f a determinant from Unit 5 o f MTE-04.

Conics

Example 3: Show that x2 - 5xy + 6y2 = 0 represents a pair of straight lines. Find the angle between these lines. 5 Solution: With reference to Theorem 2, in this case a = 1, h = - - , 2 b = 6, g = 0 = f T C. Thus, the related discriminant is

O Thus, the given equation represents a pair of lines.

The angle between thim is tan-'

Js)

= tan-'

Example 4: Find the'conic represented by 2x2 + 5xy + y2 = 1. Solution : In this case ab - h2 = -23 < 0. So, from Table 1 we know that the equation represents a hyperbola or a pair of lines. Further, in this case the discriminant becomes.

So, by Theorem 2 we know that the given equation doesn't represent a pair of lines. Hence, it represents a hyperbola. Why don't you do these exercises now? -

E3) Check whether 3x2 + 7xy + 2y2 + 5x +5y + 2 = 0 represents a pair of lines.

W) Show that the real quadratic equation ax2 + 2hxy + by2 = 0 represents a pair of lines

E5) Under what conditions on a, b and h, will the equation in Theorem 2 represent a) a pair of parallel lines? b) a pair of perpendicular lines? So far we have studied all the conics in a unified manner. Now we will categorise them according to the property of centrality.

3.3 CENTRAL AND NON - CENTRAL CONICS In our discussion on the ellipse in unit 2, we said that the midpoint of the major axis was the centre of the ellipse. The reason that this point is called the centre is because of a property that we ask you to prove in the following exercise. 2

x2 ' Y Ed) Consider the equation + 7- 1. Let (x, , y , ) be a point on this ellipse and 0 . , a2 b

be (0, 0). Show that the line PO also meets the ellipse in P' (-x,, - y , ) . What you have just proved is that O(0, 0) is bisects every chord df the ellipse

x2. Y ? x2 Y 2 - 1 7+ 7= 1 that passes through it. Similarly, any chord of the hyperbola 7- - a b a b2 through O(0, 0) is bisected by 0 . Hence, according to the following definition, 0 is the centre of the ellipse and hyperbola given above.

Definition: The centre of a conic C is a point which bisects any chord of C that passes through it. Not all conics have centres, as you will see. A conic that has a centre is called a central conic. For example, an ellipse and a hyperbola are central conics. Now, can a central conic 'have nlore than one centre? Suppose, it has two centres C , and C,. Then the chord of the conic intercepted by the line C , C, must be bisected by both C; and C,, which is not possible. Thus, a central conic has a unique centre. Let us see how we call locate this point. Considerothe conic (1). Suppose it is central with centre at the origin. Then we have the following result, which we will give without proof. I

Theorem 3: A central .conic with centre at (0; 0) is of the form ax2 + 2hxy + by2 = 1, for some a, h, b in R. This result is used to prove the following theorem about any central conic. We shall not prove the theorem in this course but we will apply it very often.

Theorem 4: Let ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 be a central conic. Then its centre is the intersectioil of the lines. ax + hy + g =

and hx + by + f = 0.

What this theorem.tells us is that if ax + hy + g = 0 and hx + by + f = 0 intersect, then the conic is central; and the point of intersection of these straight lines is the centre of the conic. But what if the lines don't intersecta?Then the conic under consideration can't be central: that IS, non-central. Thus, the conic IS non-central if the slopes of these lines are equal, that is, if ab = h2. So, we have the following result:

central if ab # h2, and ii) non-central if ab = h2. Does this result and Table 1 tell you which conics are non-central? You can immediately tell that a parabola doesn't have a centre. 1,et t s see how we can apply the above results on centres of conics.

Example 5: Is the conic 17x2- 12xy + 8yZ+ 46x - 28y + 17 = 0 central? If it is, find its centre. Solution : In this case a = 17, b = 8, h = -6, g = 23, f = -14. So, ab # h2. Hence, the conic is central. Its centre is the intersection of the lines 17x - 6y + 23 = Q and 3x - 4y + 7 = 0, which is (-1, 1).

Why don't you try some exercises now? E7) Is the conic in h3 cenrral? If yes, find its centre. ES) Identity thc conic x2 - 3xy + y2 + lox - 10y + 21 = 0. If it is central, find its centre. E9) Which degel!erate conics ,are central, and which are not?

C c n e r a l T h e o r y o f Conics

Conics

r l ~ a yUL.

au ILL

u3 JCC LLUW

,LILLU

LIIL

~ A L J . vv L J 1 L a i L LU~~JLULL

LLLL

LLLLLI~I

allu LIUII-LLIILI~I

cases separately.

3.4.1 Central Conics Suppose we are given the equation of a central conlc. By translating the axes, if necessary, we can assume that its centre lies at (0, 0). Then, by theorem 3, its equation IS ax2 + 2hxy + by2 = 1 where a, h, b E R.

..(6)

In theorem 1 you saw that if we rotate the coordinate axes through an angle 1 I 2h €)=-tan- - then the axes of the conic lie along the coordinate axes. 2 a-b' Therefore, the axes of the conic are. inclined at the angle Q t o the coordinate axes. (Here if a = b, we take Q = 45") Now, 2h tan 28 = a-b

2 tan8 - 211 -1-tan28 a - b

3 tan2 8

a-b

lane -

This is a quadratic equation in tan 8, and hence IS satisfied by two values of 8, say 8, and 0 2 . Then the slopes of the axes of the conic are tan 8, and tan 0 2 . Note that the axes are mutually perpendicular, since (tan 8, ) (tan O2 ) = - 1. Now , to find the lengths of the axes of the conic, we write (6) in polar form (see Sec. 1.5). For this we substitute x = r cos 8, y = r sin 8 in (6). Then we get

3 r 2=

A semi-axis is half the axis.

cos' 8 + sin28 writing 1 = cos28+sin2@ a cos' 8 + 2h cos 8 sin8 + b sin28 '

If we substitute tan 8, and tan O2 in (7), we will get the corresponding values of r, which will give the lengths of the corresponding semi-axes. Let us use what we have just done to &ace the conic in Example 5. Since ab - hZ > 0, from Theorem 1 we know that the conic is an ellipse. You have already seen that its centre lies at (-1, 1). Now, we need to shift the axes to the centre (-1, I), to get the equation in the form (6). The equation becomes

G e n e r a l Theory e l Conics

Now we can obtain the directions of the axes from 3 tan 2 0--tan€)-l.=O. 2. 1

This gives us tan 8 = 2, - - . 2 Therefore, we can take 8,=tan-' 2 = 63.43" (approximately), and

8, =

=+ tan-' 2. 5[

I The lengths of the semi-axes, r, and r, , are given by s~bstitutingthese values in (7). So = $ a=q 2 . and 17 - 6 + 8 2 0 5 5 1 I+ 2 rl = 1 7 + 3 + 1 = i a r 2 = 1 .

r,2 = '

Thus, the length of the major axis is 4, and that of the minor axis is 2. So now we can trace the conic. We first draw a line 0' X' through 0' (-1, 1) at an angle of tan-l 2 to the x-axis (see Fig. 2). Then we draw O'Y' perpendicular to O'X'. Now we mark off A' and A on 0' X: such that A'O' = 2 and O'A = 2. Similarly, we mark off B and B' on O'Y' such that OB =. 1 and O'B' = 1. The required ellipse has AA' and BB' as its axes. For further help in tracing the curve, we can check where it cuts the x and y axes. It cuts the x-axis in (-4, O), (-2.2, O), and the y-axis in (0, 2.7) and (0, .8). So the curve is what we have drawn in Fig. 2. Now why don't you see if you've understood what has been done in this section'?

.E 10)

Trace the conic in E8.

E 11)

Under what conditions on the coefficients, will xZ+ 2hxy + yZ + 2fy = 0 be central ? And then, find its centre and axes.

So fir you have seen how to trace a central conic. But what about a non-central conic? Let us look at 'this case now. .

3.4.2 Parabola In this sub-section we shall look at a method for finding the axis of a parabola, and hence tracing it. We will use the fact that if (1) is a parabola it can be written in the form. Ax+By+C

A'x + B'y + C'

C = 0 is the axis of the parabola and A'x + B'y + C' = 0 is the tangent at the vertex, nd hence they are perpendicular to each other.The vertex ( x , , y ) ,

f this parabola is the intersection of Ax + By + C = 0 and A-( x

+ B-(y + c-(

,I Fig. 2: T h e ellipsc 17x2 12xy + 8y2 + 46x -, 28y + 17 = 0.

Conics

A'x + B'y + C' = 0,k is the length of its latus rectum, and F

is its focus, where tan 6 is the slope of the axis. Let us see the method with the help of an illustration.

Example 6: Show that the conic x2 + 2xy + y2 - 2x - 1 = 0 is a parabola. Find its axis and trace it. Solution: H e r e a = l , b = l , h = l . :. ab-h 2 =O.

Further, the-discriminant of the conic is

0 =-1#0.

-1

Hnece, by Theorem 2, the equation does not represent a pair of straight lines. Thus, by Theorem 1, we know that the given conic is a parabola. We can write the given equation as (x + Y ) =~ 2x + 1. Now we will introduce a constant c so that we can write the equation in the form (8). So, let us rewrite the equation as (x + y + c)' = 2x + 1 + 2cx + 2cy + c2, that is, (x + y + c y = 2 ( 1 + c ) x + 2cy + c Z + 1 . We all choose c in such a way that the lines x + y + c = U and 2(1 + c)x + 2cy + c2 + 1 = 0 are perpendicular. From Equation (13) of Unit 1, you know that the condition is

Then (9) becomes

This is in the form (8). 1 Thus, the axis of the parabola is x + y - - = 0, and the tangent at the vertex is 2 5 x - y + - =o. 4 3

The vertex is the intersection of these two lines, that is,

(-g9

The length of the latus rectum of the parabola is

Thus, the focus is at

- - 8+ C O 0). 4&S 8 v ' + L S8 i n 4&

makes with the x-axis, tha is, 8 = tan-' (.-1).

3~ where 8 is the angle that 'the axis

General Theory of Conics

Therefore, the focus F I

What are the points of intersection of the parabola and the coordinate axes? They are (1

t I

3 z).

+JT,0) (1 -JT,0) (0, 11, (0 -1). 9

So, we can trace the parabola as in Fig. 3.

1 Fig. 3 : The parabola x2 + 2xy + y2 - 2x - 1 = 0

Has the example helped you to understand the method for tracing a parabola? The following exercise will help you to find out. E 12)

Trace the conic 4x2 - 4xy + y2 - 8x - 6y + 5 = 0. -

Let us now see how to obtain the tangents of a general conic.

3.5 TANGENTS .

In Unit I you studied the equations of tangents to the conics in standard form. Now we will discuss the equation of a tangent to the general conic (1) So, consider two distinct points P (x

,, y ,) and Q (x2, y ) on the conic

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 I

If x, = x, = a , say, then t h e l i n e P Q i s x = a . Similarly, if y , = y 2 = a , say, then the line. PQ is y = a . Otherwise, the line PQ is

Since P and Q lie on the conic,

andax; +2hx2y2+by; +2gx2 +2fy2 + c = 0 . Then (12) - (1 1) a(x: -x:)+2h(x2y2 -x1yI) + b(y; -y:)+2g(x2 - ~ I ) + ~ ~ ( Y ~ - Y I ) = O a a ( x ; -x:)+2h(x2~2 -x1y2 + ~ 1 - x~I Y 2I ) + ~ ( Y-Y:)+ :

Conics

Putting this in (lo), we get

As (x2, y2 ) tends to ( x l , y, ), (13) gives us the equation of the tangent to the given

Thus, the equation of the tangent at P(x,, y, ) is (Y-YI)(~Y +hx, I +f)+(x-x,)(ax, +hy, +g) = 0 ~ x ( a x , + h y , + g ) + y ( b y l + h x I + f+) ( g x l + f y , + c ) = O , u s i n g ( l l ) .

Thus, (14) is the equation of the tangent to the conic ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 at the point ( x , , y, ) lying on the conic. From (14) you can see that we can use the following rule of thumb to obtain the equation of a conic. In the equation of the conic, replace x2 by xx,, y 2 by yy,, 2x by(x+x,), ~ Y ~ Y (~Y1 )+a n d 2 x y b ~ ( x ~ , + to ~ xget , ) ,the equation of the tangent at (x,, yl ) For instance, the tangent to the parabola y2 - 4xy = 0, at a point ( x,, y, ) is yyI- 2a (x + x l ) = 0. We have already derived this in Sec, 2.3.2.

In fact, the equations of tangents to the ellipse and hyperbola given in standard form are also special cases of (14), as you can verify from Unit 2. Now you may like to .try your hand at finding tangents at some points. E13)

Obtain the equations of the tangent and the.norma1 to the conic in E8 at the points where it cuts the y-axis.

In Unit 2 you have seen that not every line can be a tangent to a given standard conic. Let us now see which lines qualify for being tangents to the general conic ax2 + 2hxy + by2 + 2gx + 2fy + c = @.With your experience in Unit 2, can you tell the conditions under which the line px + qy + r = 0 will be a tangent to this conic? Suppose it is a tangent at a point ( x , , yl ) to the conic, Now, either p i 0 or q + 0. ( 9 +~r) Let us suppose p # 0. Then we can substitute x = -in the equation of

the conic, to get 1

The roots oftthis, quadratic equation in y give us the y-coordinates of the points of intersection of the given line and conic. The line will be .a tangent if these points

General T h e o r y o f Conics

comcide, that is, if the quadratic equation has coincident roots, that is, if (prh + pqg - aqr - p2f)l = (aq2 - 2hpq + bp2) (ar2 - 2gpr + cpZ). In terms of determinants (CS-60, Block 6), we can write this condition as a

= 0

...(16)

Thus, (159 or the determinant condition (16) tell us if px + qy + r = 0 is a tangent to the general conic or not. For example, the line y = rnx + c will touch the parabola y2 = 4ax, if 0 -2a m 0 0

-1

-2a

c-;m-2a

-1

-2a

= 0

s(-2a)

= 0, expanding along the first row.

s (-2a) (cm - 2a) - m(2ac) = 0 a sc=-. m This is the same condition that we derived in Sec. 2.3.2. Why don't you try these exercises now?

E 14)

Is x + 4y = 0 a tangent to the conic x2 + 4xy + 3y2 - 5x - 6y + 3 = O? Find all the tangents to this conic that are parallel to the given line.

E 15) a) Prwe that the condition for ax + by + 1 = 0 to touch x2 + y2 + 2gx + 2fy + c = 0 is (ag + bf - 1)' = (a2 + b2) (g2 + P - c). b) In particular, under what conditions on C will y = Mx + C touch xZ+ yZ= A3? In this section you saw that a line and a conic intersect in at most two points. Now let us see what we get when two conics intersect.

3.6 INTERSECTION OF CONICS

Consider the intersection of an ellipse and a circle (Fig. 4*(a)) or of an ellipse and a hyperbola (Fig. 4 (b)).

/-----\, ,A-

I---\

' , ' I

y L-

(a)

(b) Fig. 4: Intersecting conics.

, ,

Conies

You can see that these conics intersect in four points. But, do any tyo conics intersect in four points? The following result answers this question. Theorem 5 : In general, two conics intersect in four points. \

Proof : Let the equations of the two conics be ax2 + 2 (hy + g) x + by2 + 2fy +c = 0, and .

These equations can be considered as quadratic equations in x. If we e l ~ n ~ ~ nxa tfrom e them, we will get a fourth degree equation in y. This will have four roots. Cl [responding to each of these roots, we will get a root of x. So there are, in general, four points of intersection for the two conics. Since a fourth degree equation with real coefficients may have two or four complex roots (see CS-60, BlockJ), two conics can intersect in i) Four real points, ii) two real and two imaginary points, or iii) Four imaginary points. These points of intersection can be distinct, or some may coincide, or all of them may coincide. ~ e us f consider an example. Example 7 : Find the points of intersection of the parahola y2 = 2x and the circle x2 + y2 = 1 (see Fig.5). ~ o l u t i o:i If ( x , , y, ) is a point of intersection, then x: + y:

= 1 and

= 2 x,

Eliminating y, from these equations, we get Fig. 5: y' = 2x and x' + y' 1 Intersect i n the points and Q.

x:+2x1

= 1, that is, ( x , + 1 ) l = 2 .

Then y:

= 2x

gives us

YI = + - f i ( ~ 2 - 1 ) ' i f x, = - I + & ,

and

y, = +_JT;(JZTi)/' ifx, = - I - A , Thus, there are only two real points of intersection, namely, ( J Z - I , J Z ( J Z - ~ ~ ~and ) (A-I,-JS(JZ-~)'~).

This is why you see only two

points of intersection in Fig. 4. Here is.an exercise for you now.

E 16)

x2 Y 2 x2 y 2 7 + 7- 1 . Find the points of intersection of -j- + 7= 1 and a b b a

You have seen that two conics intersect in four real or imaginary points. Now we will find the equation of any conic that passes through these points. Let ax2 + 2hxy + by2 + 2gx +2fy + c = 0, and a,x2 + 2 h l x y + b I y 2+2glx+2f,y+c, = O be the equations of two conics.

* Let us briefly denote them by S = 0 and S , = 0, respectively. Then, for each k E R, S + k S, = 0 is a second degree equation in x and y. So it is a conic, for each value of k.

On the other hand, any point of intersection of the two conics satisfies both the equations S = 0 and S, = 0. Hence it satisfies S + kS, = 0. Thus, the conic S + kS, = 0 passes through all the points of intersection of S = 0 and S,= 0. So we have proved. Theorem : The equation of any conic passing through the intersection of two conics = O and S, = 0 is ofthe form S + kS, = 0 , w h e r e k ~ R .

For different values of k, we get different conics passing through the points of intersection of S = 0 and S, = 0. But, will all these conics be of the same type? If you do the following exercises, you may answer this question. If S = 0 and S,= 0 are rectangular hyperbolas, then show that S + kS,= 0 is a rectangular hyperbola, for all real k.

E17)

(Hint: Recall that axZ+ 2hxy / byz + 2gx +2fy + c = 0 is a rectangular hyperbola if a + b = 0.) 8

E18)

x2 Y 2 Let S ~ - + - - l = o

ands, z x y - 9 = 0 .

Under what conditions' on k will S + kS,= 0 be a) an ellipse? b) a parabola? C) a hyperbola? Now we have come to the end of our discussion on conics. Let us see what we have covered in this unit:

3.7

SUMMARY

In this unit we discussed the following points: 1)

The general second degree equation axZ+ 2hxy + by2 + 2gx + 2fy + c = 0 represents a conic. It is

. i) a pair of straight lines iff

Further, if the condition is satisfied, then the angle between the lines is

ii) a parabola if ab - h2 = 0, and the determinant condition in (i) is not satisfied;

iii) an ellipse if ab -h2 > 0; iv) a hyperbola if ab - h2 < 0. 2)

An ellipse and

hyperbola are central conics; a parahla is a non-central conic.

3) A central conic with centre at the origin is of the form ax2 + 2hxy + by2 = 1, where a, h, b F ??. 4)

ax2 + '2hxy + by2 + 2gx + 2fy + c = 0 represents a central conic if ax + hy + g = 0 and hx + by + f = 0 intersect. And then the centre of the conic is the point of intersection of these lines. The slopes of the axes of this conic are the roots of the equation.

General Theory of Conics

a-b

tan2e+(--)

tane-I = o ,

5) Tracing a conic. 6)

The tangent to the conic axZ+ 2hxy + byZ + 2gx + 2 f y + c = 0 at the point I

( X I ,Yl) is axx,+h(xy, +x,y)+byyl + g ( x + x , ) + f ( y + y l ) + c = O . Further, a line px + qy + r = 0 is tangent to the given conic if a

r = O

0 .

7) Two conics intersect in four points. which can be real or imaginary.

8) The equation of a conic passing through the four points of intersection of the conics S = 0 and S, = 0 is S + kS, = 0, k ER.

3.8

SOLUTIONSIANSWERS

El) a) There are 3 types of non-degenerate conics: parabola, ellipse, hyperbola. There are 5 types of'degenerate conics: point. pair of intersecting lines, pair of distinct parallel lines, pair of coincident lines. empty set. b) A circle is a particular case of an ellipse. Thus, if (1) represents a circle then ab - h2 > 0. E2) a)

x2-2xy+y 2 + f i x - 2 = 0 . H e r e a = l = b , h = - 1 . If we rotate the axes through n14, then the new coordinates x' and y' are given by

1 x = -(XI--y')

1 and Y = -h - ( X I + y')

Thus, the given equation becomes 2y'2 + x l - y 1 = 2 .

Now, if we shift the origin to

(ylf]

the equation becomes the parabola

1 y1 = - - X, where X and Y are the new coordinates. 2 b) 9x2 - 6xy + yZ- 40x - 20y + 75 = 0. Here a = 9, b = 1, h =-3. So, let us rotate the axes through 8, where

3 1 So we can take tan 8 = 3 so that sin 8 = - ~ c o s ~ =fi - . T h e n t h e p n l l a t i n n in t h e

Y'V' - c ~ r c t e mh p r n r n i = c

\ \

General Theory o f Conics

which can be transformed and seen to be the equation of a parabola.

5 7 E3) In this case a = 3 , b = 2 , c = 2 , f = - = g , h = 2 2

Hence the given equation represents a pair of lines. E4) Here the discriminant concerned is

Thus, the given equation represents a pair of lines.

E 5) a) The lines will be parallel if ,/hz

= 0, that is, ab - h1 = 0.

b) The lines will be perpendicular if a + b = 0. E6) Since P lies on the ellipse, os does P'. Y-Yl - x-x, The equation of PO is _yl , that is, X ~ ( Y - ~ ~ ) = ~ , ( X - X ~ ) . -xi

P' also lies on this since (-x,, - y , ) satisfies this equation. Hence, we have shown the result. E7) In this caseab #hZ. So the conic is central. Its centre is the intersection of

3 E8) In this case a = 1 = b, h = - 2 So the given equation is central, and can be a hyperbola or a pair of intersecting lines. 1 3 2 5

-Since

2 1

-5

#O ,

using Theorem 2 we can say that the equation represents a hyperbola. Its centre is the intersection of

3 x--y+5=0 2

3 and - - x + ~ - 5 = 0 , that is, (-2, 2). 2

Conics

E9)

The central degenerate conics: point, pair of intersecting lines. The non-central degenerate conics : pair of distinct parallel lines, pair of coincident lines. The empty set is both central and non-central.

E 10)

The equation represents a hyperbola with centre (-2, 2). If we shift the origin to (-2, 2), the equation becomes _XI*

+3Xryf- y'* = 1.

n -

Thus, the axes of the conic are at an angle of - and - to the x-axis, So, putting these values of 9 in (7), we get the lengths r, and r 2 , of the semi-axes, on solving r:

Thus, r, =

and r, =

= 2 and :r

32 .

Note that over here, though r:

is negative, we only want its magnitude to

compute the length of the axis. Now, you know that if e is the eccentricity of the hyperbola then

r2 = r,

JE that is,

df JS ,/= =

Now let us also see where the hyperbola cuts the x and y axes. Putting y = 0 in the given equation, we get

So, the hyperbola intersects the x-axis in (-3, 0) and (-7, 0). Similarly, putting x = 0 in the given equation and solving for y, we see that the hyperbola intersects the y-axis in (0, 3) and (0, 7). With all this idonnation the curve is as given in Fig. 6.

Fig. 6 .

It will be central if hz# 1. And then its centre will be the intersection of x + hy - 0 and hx + y + f = 0, which is

G e n e r a l Theory o f Conies

If we shift the origin to this point,'the given equation is transfom-led to f2 Xz - 2hXY + YZ = 1-h2 1-h2 1-h2 y2 = 1. x2- 2h(l-h2) XY + ---f2 f2 f2

This is in the standard form AXZ + 2HXY + BYZ=I of a central conic. 1-h2 ~. Therefore, the axes of the conic are at an angle of 45" f' and 135" to the x-axis. Since they pass through the centre, their equations are

HereA = B =

1-h

hf 1-h2

f y + ---T = X-.-

and

The conic is a parabola since ab = hZ, and the determinant condition for it to represent a pair of lines is not satisfied. We can rewrite the equation as (2x - y)' = 8x + 6y - 5. We inttoduce a constant c to the equation, to get (2x - y + c ) = ~ 8x + 6y -- 5 + 4cx - 2cy + c2. o ( 2 -~y + c)' = 4(2 + C) x + 2(3 - C) y + C' - 5

We choose c in such a way that

Then the equation of the curve becomes ( 2 ~ - ~ - 1 =) 4~( ~ + 2 y - 1 )

The vertex of this parabola is the intersection of 2x - y - 1 = 0 and. x + 2y '- 1= 0, that is,

('.5 '15. The focus lies at , where tan 0 = 2.

2 :. sin8 =-,cosQ

=-.

:. The focus lies at The curve intersects the y-axis in (0, 1) and (0, 5). It doesn't intersect the .x-axis. Thus, the shape of the parabola is as given in Fig. 7.

Conics

Fig.

E 13)

The conic's equation is x2-3xy + y 2 + lox- lOy+21 = o . From E l 0 you know it intersects the axes in (-3, O), (-7, O), (0, 3), (0, 7). The tangent at (-3, 0) is

Its slope is 4. 1

Thus, the slope of the normal at (-3, 0) is -- . Hence, its equation is 4 1 y = - -4x( +

3).

You can similarly check that the tangents at (-7, O), (0, 3)' and (0, 7) are . respectively,

4x- l l y + 2 8 = 0 , ' x - 4 y + 12=0, llx-4y+28=0. The normals at these points are respectively,

1 y-3= qx,

.E 14)

x + 4y = 0 will be a tangent to the given conic if

General T h e o r y o f conics

a 4.0 = 0, which is false. Thus, the given line is not a tangent to the given conic. Any line parallel to the given line is of the form x + 4y + c = 0. This will be a tangent to the given conic if (15) is satisfied, that is, (Sc + 2 8 ) l = 3(3cZ+ 24c + 48) a c = - 5 o r - 8. Thus, the required tangents are x+4y-5=Oandx+4y-8=0.

E 15)

Using (IS), we see that the condition is 1

a .

o b ( f L b c ) - ( 1 - bf) + g { a f + b(bg-af)} + a { ( a c - g ) + f(bg-af)} = O a bzgz + aZP +2bf - 2abfg - bzc - 1 + 2ag - a2c = 0. Adding azgZon both sides and simplifying, we get the given condition. a 1 b ) . In (a) we put g = 0 = f, c = - A Z , -- = M,--= C. b b So the condition for y = Mx + C to touch xZ + yZ = AZ is C2 = AZ(Mz+ 1)

E 16)

Thus, C = A

Substituting

x' = a 2 I - Y

Then x = a

'[I--- a 2 y b 2 ] = m

(

;: f

= 1, we get

a2b2

ab Fig. 8

Thus the 4 points of intersection are

and We have drawn the situation in Fig. 8.

E 17)

Let S r axz + 2hxy + by2 +2gx + 2fy + c = 0 and S , = a,x' + 2 h l x y + b l y 2+ 2 g l x + 2 f , y + c ,

Conics

be rectangular hyperbolas. Then a + b = O a n d a,+b,=O. :.(a+b)+k(a, +b,)=OVk ER o(a+ka,)+(b+kb,)=OVk~R

a S + kS1= 0 is a rectangular hyperbola t/ k E R

a) This conic will be an ellipse if

(

(i)

k' 1 -- > 0,that is, k2 c -. 9

b) The conic will be a parabola if 1 -

--k

9 k 7 1 k-=-and-9 2

2 1

0 0

# 0,

But this can't be true. So the conic can't be a parabola 1

But it will be a pair of lines if k = - . 3 1

be a hyperbola if k2 > -. c) The conic will . . 9

that is,

Miscellaneous Exereises

MISCELLANEOUS EXERCISES (This section is optional) In this section we have gathered some problems related to the contents of this block. You may like to do them to get a better understanding of conics. Our solutions to the questions follow the list of problems, in case you'd like to counter check your answers. 1) Find the equation of the path traced by a point P, the sum of the squares of the distances from (1, 0) and (-1, 0) of which is 8.

A path Raced by a moving point i s called its loeus.

2) Find the equation of the circle which passes through (1, O), (0, -6) and (1,'4). (Hint: The general equation of a circle is x2 + y2 + 2gx + 2fy + c = 0). 3) Prove the reflecting property for a parabola. (Hint: Show that a = P in Fig. 9, Unit 2.) 4) Prove Theorem 2 of Unit 3. 5)

A circle cuts the parabola y2 = 4ax in the points ( at:, 2at ) for i = 1, 2, 3, 4. Prove that t, + t 2 + t 3 + t 4 = 0. (Hint: t, + t2 + t3 + t, are the solutions of the quadric equation obtained by putting x = O.), y = 2at in the equation of a circle.

6). Trace the curves xy = 0 and xy - 4x - 5y + 20 = 0.

7) What relations must hold between the coefficients of ax2 + by2 + cx + cy = 0 for it to represent a pair of straight lines? 8) Find the angle through whlch the axes should be rotated so that the equation Ax + By + C = 0 is reduced to the form x = constant, and find the-value of the

constant. 9) Prove that yZ+ 2Ax + 2By + C = 0 represents a parabola whose axis is parallel to the x-axis. Find its vertex and the equation of its latus rectum. 10) Prove that the set of midpoints of all chords of yz = 4ax which are drawn through its vertex is the parabola y2 = 2ax. x: Y: 11) a) Prove that 7 + -- 1 is negative, zero or positive, according as the point a b2 x2 y2 ( x, , x2 ) lies inside, on or outside the ellipse 7+ - =l. a b2 b) Is the point (4, - 3) inside, or outside the ellipse 5x2 +7yz = 1l? 12) A line segment of fixed length a + b moves so that its ends are always on two fixed perpendicular lines (see Fig. 1). Prove that the path traced by a point which divides this segment in the ratio a :'b is an ellipse. 13) Find the equation of the common tangent to the hyperbolas

x2 to -- Y = 1 meets the x and y axes in M zad N, respectively. The lines a2 b2 through M and N drawn perpendicular to the x and y-axes, respectively, meet in the point P. h ~ v '?at a type locus of P is the hyperbola a2x2- bZy2= (a2.+b2)2.

14) A no-1

15) Consider the hyperbola in Fig. 20 of Unit 2. Through A and A' draw parallels to the conjugate axis, and through B and B' draw parallels to the transverse axis. Show that the diagonals of the rectangle so formed. lie along the asymptotes of the hyperbola.

A method for draw~ng asymptotes of a hyperbola.

Conics

16) Which conlcs are represented . - by- the following equat~ons'? ' a) (X - y)2 -t (X - a)Z= 0, b) r sin20 = 2a c o s 0 ,

17) Trace the conics a) 9x2- 24xy +16y2 - 18x - lOly + 19 = 0 b) xy - y2 = a2 C) ( 3 ~ - 4 y +1 ) ( 4 x + 3 y + 1 ) = 1. 18) Find the kquation to the conic which passes through (1, 1) and the intersection of x2 + 2xy + 5y2 - 7x -8y + 6 = 0 with the pair of straight lines 2x-y-5=Oand3x+y-11=0.

Solutions 1)

Let P be (x, y). Then

o xZ + y2 = 3, which is a circle with centre (0, 0) and radius & 2)

Let the equation be x2 + y2 + 2gx +2fy + c = 0. Since (1, 0), (0, -6) and (3, 4) lie on it, 1 +2g + c = 0, 3 6 - 12f + c =0, 9+ 16+6g+8f+c=O. Solving these three linear equations in g, f and c, we get

71 47 69 Thus the equation is X ' + ~ ~ - - X+-y+-=O. 2 4 2 3) The parabola is yZ= 4ax. The tangent T at a point P(x,, y , ) is

yy, = 2a(x+ x,).

2a So tana = YI

Y-Y1 x-Xl The line PF, where F(a, 0) is the focus, is -= ----. -Y, a-XI

Its slope is

Thus, tan0 =

xI -a

2a &, using (1 1) of Unit 1.

2a = -,using

the fact that yf = 4ax,.

Thus tan a = tan p and a and P are both less than or equal to 90'. 4) We want to show that axZ+ 2hxy + byZ+ 2gx + 2fy + c = 0 can be written as a product of two linear factors iff its discriminant is 0. if a # 0, we multiply (1) throughout by a and arrange it in decreasing powers of x. We get a2x2+ 2ax (hy +g) = -aby2 - 2afy - ac. O n cnrnnletincr the p n ~ ~ a ron p the left hand. we pet

...( 1)

From this we can obtain x in terms of y, only involving the first degree iff the quantity under the square root sign is a perfect square, that is, iff (gh - af)t = (h2 - ab) (g2 - ac), o a b c + 2 f g h - a f ' -bg2 -ch2 =O. l a I

g J

Let the circle's equation be xZ+ yZ+ 2gx + 2fy + c = 0. Substituting x = at2, y = 2at in this, we get a2t4 +4aY2 + 2agtz + 4aft + c = 0. We know that it has 4 roots t i , t 2 , t,: t, . So, from (CS-60/Block 5 & 6) you know that the 1

sum of the roots will be --;-(coefficient of t3) = 0. a-

:.t,+ t , + t , + t , = 0 . ky = 0 is the pair of lines x = 0 and y = 0. We have traced it in Fig. 2. xy - 4x 5y + 20 = 0 is a pair of lines since its discriminant is 0. In fact, we can easily factorise it as (x - 5) (y - 4) = 0.

Fig. 2

Thus, it represents the pair of lines x = 5 and y = 4, which we have traced in Fig. 3. ax2 + by2 + cx + cy = 0 represents a pair of lines iff

Let us rotate the axes through 8. Thendhe equation becomes A(x1cos9- y1sin9)+B(x'sin9+ y1cos9)+C=O. ox1(Acos9-Bsin9)+ y1(Bcos8-AsinO)+C = 0. This will reduce to the form.x'= constant iff B cos8 =A sine, B that is, 9 = tan-' A'

And then the equation becomes I

-C Thus, the constant is

J X F

We rewrite the given equation as y2 = AX + 2By + c) ( + k)2 = -

2 - 2~~ ~ ~- c + 2ky + k2, where k is a constant.

o ( y + k ) 2 = - 2 ~ x + 2 ( k - ~ ) ~ -+Ck ~ We choose k so that

Fig. 3

Conics

c-p'

Ax + (B - k)y + -- 0 is parallel to the y-axis, that is, 2 k = B. Then the equation becomes

(*:A

Its axis is y + B = 0, vertex is -,-B) and the equation of its latps rectum is x =

B~ - A ~- C 2A

10) The midpoint of any chord through P (x,, y,) and O(0, 0) is

Thus, the set of all such Q is y2 = 2ax. 'I)

11) a) Firstly, if (x, ,y ,) lies on the ellipse, then clearly

Y2 b2 Now, if (x, ,y,) lies outside the ellipse (see Fig. 4) then either Ixll > a or a2

-a

IYIl>b

I Fig. 4

similarly, you can show that if (x,,y,) lies inside the ellipse,

b) Since 5(16) + 7(9) = 143 > 11, the point lies outside the ellipse. 12) Let the perpendicular lines be the coordinate axes. Let the segment intersect the axes in (x, 0) and (0, y). Then the coordinates of the point P are

Now, since (x2 + y2) = (a + b12

x2 = 1. Thus, the path traced by P is the ellipse 3 7+ a b2 x2 -1 isy=mx+ , / ~ , a n d a n Y 13) Any tangent to -- -a2 b2 tangent to

m.a-':l\

y2 x2 -= 1is x = m,y + a b2

1 For these two lines to be the same, we must have - = m and m1

Miscellaneous Exercises

- b2 = a 2 - m2b2

m2 = 1, as+ b2. Thus the common tangents are y = x + ,/= and y=-x+

422.

xL ' =1 14) See Fig. 5 for a diagram of the situation. The normal to- a2 b2

Thus, M is

(

(a2 + b2)x1

.o) and N is(09

(a2 + b 2)YI b2

Thus, the coordinates of P are ( ( y ) x , ,(y]yl].

Now, since ( xl, y, ) lies on the hyperbola,

a2 + b2 aZ+b2 o a 2 x 2- b 2 Y2 = ( a 2 +bZ)2,WhereX=--X I and Y=Yl a a Now, as P varies, X and Y vary; but always satisfy the relationship a2x2

- b2 Y 2 = (a2 + b 2 )2. Thus, this is the locus of the point P.

15) The lines meet in (a, b) (a, - b), (-a, b) and (-a, -b). Thus the diagonals of the b b rectangles lie along y = - x and y = -- x, which are the asymptotes of the a a hyperbola.

-1 .

-I

16) a) ( ~ - ~ ) ~ + ( x -=aO) ~ a 2 x 2 - 2 x ~ + ~ ~ - 2 a x =o. +a~ Here a = 2, b = 1, h =-I, g =-a, f = 0, c =a2. ... ab - h2 ,o. Thus, the conic is an ellipse.

Changing to Cartesian coordinates, this equation is y2 = 2ax, which is a parabola. c) -1= l + c o s e + f i s i n e

0 2 y 2 +2&+2x.+J;jy+l=0 Here ab - h1 < 0 and its discriminant is

J;j

J?=l#O.

Thus, the curve represents a hyperbola.

Fig. 5

17) a)

You can check that ab - = 0 and the discriminant is non-zero Thus, t' equation represents a parabola. We can write it as ( 3 -~4 ~ ) = ' 1 8 +~lOly - 19. ~ ( 3 ~ - 4 y + c=(6c+18)x+ )~ y(101-&)+c2 -19, where we Choose the constant c so that 3(6~ +18) - 4 (101-8~)= 0 0c=7. Then the given equation becomes (3x - 4y + 7)' = 15(4x + 3y + 2)

Thus, the axis of the parabola is 4x + 3y + 2 = 0. The vertex is the intersection of 3x - 4y + 7 = 0 and 4x + 3y + 2 = 0; 29 22 that is,(-25,13)The length of its latus rectum is 3. Its focus F lies

:. F is (-0.71, 0.28). The curve intersects the y-axis in

49 3 that is, approximately, -and 8 16 It doesn't intersect the x-axis. We have traced it in Fig. 6.

-D

Fig. 6

b) xy - y2 = 1 This is a hyperbola. Its centre is the intersection of 1 1 x + y = 0, that is, (0, 0).Its axes are inclined to the 2 coordinate 'axes at an angle of 8,where tan 2 8 =l. Thus, the slope of :' I [ . 5~ transverse axis is =- , and of the conjugate axis is e2= - . Since 8 8 - - Y = 0 and -

Miscellaneous Exercises

tan 8, = .4 1, the length of the transverse axisi r, ,:is given by

:. r, = 1.24, approximately. '

We similarly find r2 =.91. Thus its eccentricity is 1.24. It does't intersect the axes. With all this information we have traced the curve in Fig.'7.

Fig. 7 c) The equation is .ahyperbola whose centre is the intersection of 3x - 4y + 1 = 0 and 4x + 3y + 1 = 0,that is,

(-&,A).

7 Here a = 12, b = -12, h = -- . 2 Thus, its axes are inclined at angles 8, andO2to the coordinate axes, where tanel and tan

ez are roots.of the equation

a-b tan2@+-tang-1 h

otan@,= 7 and tan 9, = - 7

@ I =8 1.9O (approx.) and

6 = 171.9" (approx.). The length of its axes are r, and r, , where

The curve intersects the coordinate axes in (0, 0) (-12y So, its curve is as given in Fig. 8.

0) , (0,-

A)'

Conics

Fig. 8

18) Let S, = x 2 + 2 ~ ~ + 5 ~ ~ - 7 x - 8 ~ +and 6=0

SZ'(2~-y-5)-(3x+y-11)=0. Then the required conic is S, + kS2 = 0, where we choose k so that (1, 1) lies on the curve. Thus, the curve is (1 + 6k) x2 + (2 - k) xy + (5 - k) y2 - (7 +37k) x - (8 -6k) y + (6 +55k) = 0. 1 Since (1, 1) lies on it, k = - . 28 Thus, the conic 's 34x2+55xy+ 13 yZ-233x-218x-218y+22-3=0.

UNIT 1

PKELIMINARIES IN THREEDIMENSIONAL GEOMETRY

Structure 1.1

Introduction Ohject~ves

1.2

Points

1.3

Lines 1 .?. 1 1 .!.2

L)~~.ectton Cos~nes Eq~iatlonso f a Straight Line

I . 3 7 Anglc Betwccn Two I.ines

1.4

Planes 1.4.1 Equations of a Plane

1.4.2 lntersect~ngPlanes and Lines

INTRODUCTION

1 .

With this unit we start our discussion of analytical geometry in three-dimensional space, or 3-space. The aim of this unit is to acquaint you with some basic facts about points, lines and planes in 3-space. We start with a short introduction to the cartesian coordinate system. Then we discuss various ways of repremting a line and a plane algebraically. We also discuss angles between lines, between planes and between a line and a plane. The facts'covered in this unit will be used constantly in the rest of the course. Therefore, we suggest that you do all the exercises in the unit as you come to them. Further, please do not go to the next unit till you are sure that you have achieved the following objectives.

Objectives After studying this.qnit you should be able to : find the distance bktween any two points in 3-space; obtain the direction cosines and direction ratios of a line; I

obtain the equations of a line in canonical form or in two-point form; obtain the equation of a plane in three-point form, in intercept form or in normal form; find the distance between a point and a plane; find the angle between two lines, or between two planes, or between a line and a plane; find the point (or points) of intersection of two lines or of a line and a plane.

Now let us start discussion on points in 3-space.

1.2

POINTS

Let us start by genqralising the two-dimensional coordinate system to three dimensions. You know that any poi$ in two-dimensional space is given by two real numbers. To locate the

'..-

T h e Sphere, Cone and Cylinder

position of a point in three-dimensional space, we have to give three numbers. To do this, we take three mutually perpendicular lines (axes) in space which intersect in a point 0 (see Fig. 1 (a)). 0 is called the origin. The positive directions OX, OY and OZ on these lines are so chosen that if a right-handed screw (Fig. 1 (b)) placed at 0 is rotated from OX to OY, it moves in the direction of OZ.

Fig. 1 : The Cartesian coordinate axes i n three dimensions.

To find the coordinates of any point P in space, we take the foot of the perpendicular from P on the plane XOY (see Fig. 2). Call it M. Let the coordinates of M in the plane XOY be (x, y) and the length of MP be I z 1. Then the coordinates of P are (x, y, z). z is positive or negative according as MP is in the positive direction OZ or not.

Fig. 2

So, for each point P in space, there is an ordered triple (x, y, z) of real numbers, i.e., an element of R3. Conversely, given an ordered triple of real numbers, we can easily find a point P in space whose coordinates are the given triple. So there is a one-one correspondence between the space and the set R3. For this reason, three-dimensional space is often denoted by the symbol R3. For a similar reason a plane is denoted by R2,and a. line by R.

,/m

Now, in two-dimensional space, the distance of any point P (x, y) from the origin is . Using Fig. 2, can you extend this expression to three dimensions? By Pythagoras's theorem, we see that

Preliminaries in ThreeDimensional Geometry

Thus, the distance of P (x,y, z) from the origin is Jx i +7 y + z7 . And then, what will the distance between any two points P (x,, y ,,2,)and Q (x*,y2,z2)be? This is the distance formula

as you may expect from Equation (1) of Unit 1. Using (I), we can obtain the coordinates ofa point R (x, y, z) that divides the join of P ,y ,2,) and Q (x?,y2,z2) in the ratio m:n. They are

(x, ,

For example, to obtain a point A that trisects the join of P (1,0,0) and Q (1, 1, I), we take

1 1 m = 1 and n = 2 (2) . Then the coordinates of A are (1. j v . Note that if we had taken m = 2, n = 1 in (2), we would have the other point that trisects PQ,

Why don't you try some exercises now?

E 1) Find the distance between P (1, 1, -1) and Q (-1, 1, 1). What are the coordinates of the point R that divides PQ in the ratio 3:4? E 2) Find the midpoint of the join of P (a, b. c) and Q (r, s, t).

Now let us shift our attention to lines.

1.3

LINES i

In Unit 1 we took a quick look at lines in 2-space. In this section we will show you how to represent lines in 3-space algebraically. You will see that in this case a line is determined by a set of two linear equations, and not one linear equation, as in 2-space. Let us start by looking at a triplet of angles which uniquely detenn~nethe direction of a line in 3-space.

1.3.1 Direction Cosines Let us consider a Cartesian coordinate system with 0 as the origin and OX, OY, OZ as the axes. Now take a directed line L in space, which passes through 0(see Fig. 3). Let L make angle a, $ and y with the positrve directions of the x, y and z-axes, respectively. Then we define cos a, . cos $ and cos y to be the direction cosines of L. ,

For example, the direction cosines of the x-axis are cos 0, cos ~ 1 2cos , ~ 1 2that , is, 1.0,O. Note that the direction cosines depend on the frame of reference, or coordinate system, that we choose.

The Sphere, Cone a n d

Cylinder I

Fig. 3 : Cos a,cos B and cos y are the direction cosines of the line L.

Now take aay directed line L in space. How can we find its direction cosines with respect to a given coordinate system? They will clearly be the direction cosines of the line through 0 which has the same direction as L. For example, the direction cosines of the line through ( 1, 1 , l ) and parallel to the x-axis are 1,0,0. Now let us consider some simple properties satisfied by the direction cosines of a line. Let the direction cosines of a line L be cos a, cos p and cos y with respect to a given coordinate system.We Can assume that the origin 0 lies on L. Let P (x, y, z) be any point on L. Then you can see f r o m ~ i g4: that

x=0Pcosa,y=OPcos~andz=OPcosy. Since 0 p 2 = x2 + y2 + z2 = 0 p 2 (cos2 a + cos2 P + cos" )),we find that Fig. 4

This simple property of the direction cosines of a line is useful in several ways, as you'll see later in the course. Let us consider an example of its use. 7r

E x a m ~ l e1: If a line makes angles 7and y with the x and y axes, respectively, then what is the angle that it makes with the z-axis?

Thus, there will be two lines that satisfy our hypothesis. (Don't L, surprised ! See Fig.

Prellminrries lo ThreeDimensional Geometry

Ir 2x They will make angles T and - ,respectively, with the z-axis. 3 3

There is another number triple that is related to the direction cosines of a line.

I !'

Definition :Three numbers a, b, c are called direction ratios of a line with direction cosines 1, mand n,ifa=kl,b=km,c=kn,forsomek~R. Thus, any triple that is proportional to thk direction cosines of a line are its direction ratios. For example,

JZ , 1, 1 are direction ratios of a line with direction cosines -JZ'2'2.

You can try these exercises now.

E3) If cos a,cos p and cos yare the direction cosines of a line, show that sin2a + sin2p + sin2y -- 2.

E4) Find the direction cosines of

a) the y and z axes,

b) the line y = mx + c in the XY-plane. E5) Let L be a line passing through the origin, and let P (a, b, c) be a point on it. Show that a, b, c are direction ratios of L. E6) Suppose we change the direction of the line L in Fig. 3 to the opposite direction. What will the direction cosines of L be now? Let us now see how the direction cosines or ratios can be used to find the equation of a line.

1.3.2 i

Equations of a straight Line

We will now find the equations of a line in different forms. Let us assume that the direction cosines of a line are 1, m and n, and that the point P (a, b, c) lies on it. Then, if Q (x, y, z) is any other point on it, let us complete thecuboid with PQ as one of its diagonals (see Fig. 6).

Flg. 6 : (x

- a), (y - b) and (z - c) are direction tatios of PQ.

Then PA = x - a, PB = y - b and PC =z - c. Now, if PQ = r, you can . see. that cosa=

x-a .r ,that is,

x -a y-b z-c 1- -. Similarly,m = - *=-. r' r ' r

Thus, any point on the line satisfies the equations.

The Sphere, Cone and Cyllnder

x - a --=y-b --

z-c n

Note that (4) consists of pairs of equations,

x-ay-band -=I m

y-b m

- z - c , o r =x - a n

z-c n

y - b --Z - c and --

Conversely, and pair of equations of the f o m (4) represent a ~traightline passing through (a, b, c) and having direction ratios I, m and n. (4) is called the canonical form of the equations of a straight line. For example, the equations of the straight line passing through (1, 1, 1) and having direction cosines

1 -1 1 (zl K p3?) are.

Note that this is in the form (4), but 1, -1,l are direction ratios of this line, and not its direction cosines.

Remark 1 :By (4) we can see that the equations of the line passing through (a, b, c) and having direction cosines 1, m, n are x = a + h , y = b + m r , z = c + n r , w h e r e r ~R.

...(5)

This is a one-parameter form of the equations of a line, in tenns of the parameter r. Let us now use (4) to find another f o m of the equations of a line. Let P (xl, y,, z,) and Q (x,, y,, 3)lie on a 1ine.L. Then, if 1, m and n are its directions cosines, (4) tells us that the equations of L are

Since Q lies on L, we get

Then (6) and (7) give us

(8) is the geneiahsationof Equation (7) of Unit 1, and is called the two-point form ofthe equatio* A a line in 3-space. For exmple, the equations of the line passing through (1,2,3) and (0,1,4) are x - 1 = y - 2 =-(2-3). Note that, while obtaining (8) we have also shown that

i f P ( ~ ~ , y , , z ~ ) a n d Q ( ~ , y ~ , ~ ) l b ~ n a l ~ ex1,y2-y,and ~,then~~2, - z, are direction ratios of L.

Preliminaries in ThreeDimensional Geometry

Now you can try some exercises. E7) Find the equations to the line joining (-1,0, 1) and (1,2,3). E 8) Show that the equations of a line through (2,4,3) and (-3,5,3) are x + 5y = 22, z = 3. Now let us see when two lines are perpendicular.

Angle Between Two Lines

1.3.3

In Unit 1 you saw that the angle between two lines in a plane can be obtained iq terms of their slopes. Now we will find the angle between two lines in 3-space in terms of their direction cosines. Let the lines L, and L2 have direction cosines I,, m,, n, and I,, m,, n,, respectively. Let 0 be the angle between L, and L,. Now let us draw straight lines L,' and L,' through the origin with direction cosines I,, m,, n, and I,, m,, n,, respectively. Then choose P and Q on L,' and L,' , respectively, such that OP = OQ = r. Then the coordinates of P are (l,r, m,r, n,r), and of Q are (12r,m,r, n,r). Also, 8 is the angle between OP and OQ (see Fig. 7). Now Fig. 7

pQ2 = (I, -I2) 2 r 2 + (m, -m,) 2 r2 + (n, - i ~ , ) ~ r ~ = 2(1-1,

1,-m,m2-n, n,) r,,

7 usingl, + m I 2 + n l 2 =1,122 +m,-+n,,= 1.

Also, from Fig. ?and elementary trigonometry, we know that PQ2 = 0 p 2+ 0 Q 2- 2 0 ~OQ . cos 8 = 2 (1 - cos 83 r2.

Therefore, we find that 8 is given by the relation cos 8 = Ill2+ m,m, + n,n2

...(9)

Using (9) can you say when two lines are perpendicular ? They will be perpendicular iff 8 = x/2, that is iff Ill2+ m,m, + n,n2= 0.

...(10) I L ,

Now, suppose we consider dire-.:on ratios a,, b,, c, and a,, b,, c, of L, and L,, instead of their direction cosines. Then, is it true t h a ~L, and L, are perpendicular if ala2-t blb2+ clc2= O? If you just apply the definition of direction ratios, you will see that this is so. I

And when are two lines parallel? Clearly, they are parallel if they have the same or opposite directions.Thus, the lines L, and L, (given above) will be parallel iff 1, = I,, m, = m,, n, = n, or 1, = - I,, m, = - m,, n, = - n,. In particular, this means that if a, b, c and a', b', c' are direction ratios of L, and L2respectively, then

Let us just summarise what we have said. Two lines with direction ratios a,, b,, c, and a,, b,, c, are (i) perpendicular iff a, a, -1- b, b, -t c, c, = 0; (ii) parallel iff,

- bl

- b=

For example, the line - = y = - is not parallel to the x-axis, since 2,1,3 are not proportional

32 The Sphere, Cone and

to 1,0,0. Further, x = y = z is perpendicular to x = - y, z = 0, since 1, 1 , l and 1, - 1,0 are the direction ratio of these two lines, and 1 (1) + 1 (-1) + 1 (0) = 0.

Cylinder

Why don't you try some exercises now?

E 9)

Find the angle between the lines with direction ratios 1,1,2 and ,Fj ,respectively.

,4,

E 10) If 3 lines have direction ratios l , 2 <3; 1, -2,l and 4,1, -2, respectively, show that they aremutually perpendicular. In this section you saw that a line in 3-space is represented by a pair of linear equations. In the next section you will see that this means that a line is the intersection of two planes.

1.4

PLANES

In this section tyou will see that a linear equation representsa plane in 3-space. We will also discuss some aspects of intersecting planes, as well as the intersection of a line and a plane.

L+ ,/"

Let us first look at some algebraic representations of a plane.

1.4.1 Equations of a Plane

Consider the XY-plane in Fig. 1 (a). The z-coordinate of every point in this plane is 0. Co'nverseiy, any point whose z-coordinate is zero will be in the XY-plane. Thus, the equation z = 0 describes the XY-plane.

Fig. 8: The plane z = 3

Similarly, z = 3 describes the plane which is patallel to the XY-plane and which is placed 3 units above it (Fig. 8). And what is thk equationof the YZ-plane? Do you agree that it is x = O? ,.

Note that each ,of these planes satisfies the property that if any two points lie on it, then the line joining the@ also lies on it. This property is the defining property of a plane. Definition :A Piane is a set of points such that whenver P q d Q belong to' it then so does every point on the line joining P and Q. Another point that you may have noticed about the planes mentioned above is that their equations are linear in x, y and z, This fact is true of any plane, a c c o r h ~1 - I he following theorem. Theorem 1:The general linear equation Ax + By + Cz + D = 0, where at least one of A, B, C is non-zero, represents a plane in three-dimensional space.

Further,the coiverse is also tiue. We will not prave this result here, but will always use the fact that a plane is synonymous with a linear equatiop.in 3 variables. Thus, for example, because of Theorem 1we know that 2x + 5z = y represents a plme. . At this point wewould Iike to make an important remark.

Remark 2 :In &space a linear equation represents a line,,whilein 3-space a linear equatiori represents a plane. For example, y = 1 is the line of Fig. 9 (a), as well as the plane of Fig. 9 (b).

Preliminaries in ThreeDimensional Geometry

Fig. 9 : The same equation represents a line in 2-space and a plane in 3-space.

Let us now obtain the equation of a plane in different forms. To start with, we have the following results.

Theorem 2 :Three non-collinearpoints determine a plane. In fact, the unique plane passing through the non-collinear points (x,, y,, z,), (x,, y,, z,) and (x,, y,, 5)is given by the determinant equation.

We will not prove this result here, but we shall use it quite a bit. As an example, let us find the equation of the plane which passes through the points (1, 1, O), (-2,2, -1) and(l,2,1). It will be

=2x+3~-3z=S. Why don't you try some exercises now ?

The Sphere, Cone and Cylinder

tz I

E 11)

Show that the four points (0, -1, -l), (4,5,1), (3,9,4) and (- 4,4,4) are coplanar, that is, lie on the same plane. (Hint : Obtain the equation of the plane passing through any three of the points, and see if the fourth point lies on it.) ,

E 12)

Show that the equation of the plane which makes intercepts 2, -1,5 on the three axes

+ -z= 1 . is -+2 (-1) 5 (Hint :'fie plane makes an intercept 2 on the x-axis means that it intersects the x-axis at (2,0,0.). In E l 2 did you notice-the relationship between the intercepts and the coefficient of the equation?

Fig. 10 : The plane

In general, you can check that the equation of the plane making intercepts a, b and c on the coordinate axes (se~eFig. 10) is

-+-+-=I.

-X+ - + Y- = I Z a

This is because (a, O,, O), (0, b, 0) and (0,O, c) lie on it. (12) is called the intercept form of the equation of aplane.

Let us see how we c:an use this form. Example 2 : Find the intercepts on the coordinate axes by the plane 2x - 3y + 5z = 4. Solution :Rewriting the equation, wle get

4 4 Thus, the intercepts on the axes art: 2, - and - . 3 5 Now here is an exercise on the use of (12). E13)

Show that the planes ax + by + cz + d = 0 and Ax + By + Cz + D = 0 are the same iff a, b, c, d and A, B, C, D are proportional. (Hint :Rewrite the equatir~nsin intercept form. The two planes will be the same iff their 'intercepts on the axe:s are equal).

Let us now c'onsider another form of the equation of a plane. For this, let us drop a perpendicuhir,fromthe origin 0 ointo the given plane (see Fig. 11). Let it meet the plane in the point P. Let cos a , cos p and cos *r be the direction cosines of OP and p = I OP I. Further, let the'plane make intercepts a, b and c on the x, y, and z axes, respectively. Then P P P c o s a = - , c o s p = - andcosy==-. b c a

Fig. I I: Obtaioing tbe normal form of the eq,%tion o f the plaoe.

...(13) X

Now, from (12) we know that the equation of the plane is a +b +c = 1. The, using (13). this equation becomes x c o s a + y cosp+zcosy =-

...(14)

This is caned the PC. rnal form of the equation of the plane For example, let us find the normal form of the equation of tht ,-fie nl Fig. 9 (b). The perpendicular from the origin onto it is of length 1 and 1it:s along the x-axis. Thus, its direction cosines w e 1.0.0. Thus, from (14) we get its equation as x = 1.

Notethat(14)isoftheformAx+By+Cz=D,where)A)5l , I B ) S l , I C ( SlandD20.

Preliminaries in ThreeDlmenslonal Geometry

Now, suppose we are given a plane Ax + By + Cz + D = 0. From its equation, can we find the length of the normal from the origin to it? We will use E 13to do so. Suppose the equation of the plane in the normal form is x c o s a + y c o s p + z c o s y -p=O. Then this is the same as the given equation of the plane. So, by El3 we see that there is a constant k such that cosa=kA,cos~=kB,cosy=kC,p=-kD. Then, by (3) we get

ID I 2 ,where we take the absolute value ofD since p 2 0. $ O , P = - ~J A2 +B2 + C

Thus, the length of the perpendicular from the origin onto the plane Ax + By + Czz 0 is

ID1

jA'.

...(15)

For example, the length of the perpendicular from the origin onto x + y + z = 1 is Now let us go one step further. Let us find the distance between the point (a, b, c) and the plane AX + By + Cz + D = 0,that is, the length of the perpendicular from the point to the plane. To obtain it we simply shift the origin to (a, b, c), without changing the direction of the axes. Then, just as in Sec. 1.4.1, if X, Y, Z are the current coordinates, we get x = X + a. y=Y+b,z=Z+c. So the equation of the plane in current coordinates is A (X + a) + B (Y + b) + C (Z + c) + D = 0.

We have discussed the translation of axes in detail in Unit 7.

Thus, the length of the normal from (a, b, c) to the plane Ax + By + Cz + D = 0 is the same as . the length of the normal from the current origin to

For example, the length of the normal from(4,3,1) to 3 x 1 4 +~ 122+ 14= 0 is

Now you may like to do the following exercises. E 14)

Find the distance of (2,3, -5) from each of the coordinateplanes, as well as from x+y+z=l. Show that if the sum of the squares of the distance of (a, b, c) from the planes ~+~+z=0,x=zandx+z=2~is9,thena~+b~+c~=9.

Now that you are familiar with the various equations of a plane let us talk of the intersection of planes.

.1.4.2

Fig. I2 : A stnlgh; ilne is the intersection of two pirner.

Intersecting Planes and Lines

In Sec., 4.3.2, you saw that a line is represented by two linear equations. Thus, it is the intersection of two planes represented by these equations (see Fig. 12). 15

The Sphere, Cone and Cyllader

In general, we have the following remark.

Remark 3 a A straight line is represented by a linear system of the form ax + by + cz + d = 0, Ax+By+Cz+D=O. Wewritethisinshortasax+by+cz+d=O=Ax+By+Cz+D.For example, 3x + 5y + z - 1 = 0 = 2x + 1 represents the line obtained on intersecting the planes 3x+5y+z-1 and2x+ 1.=0 Now, suppose we are given a line.

This line clqarly lies in both the planes ax + by + cz + d = 0 and Ax + By + Cz + D = 0.In fact, it lies in infinitely many planes given by

where k E R. This is because any point (x, y, z) lies on the line iff it lies on ax + by + cz + d = 0 aswellasAx+By+Cz+D=O. Let us see an example of the use of (17).

Example 3 :Find the equation of the plane passing through the line 1

y - 3 --z +2 x+l =-3 2 1 '

and the poht (0,7, -7).

Solution :The line is the intersection of 2 (x + 1) = -3 (y - 3) and x + 1 = -3 (z + 2), that is, ~X+~Y-TFO=X+~Z+~.

Thus, by (17), any plane passing through it is of the form

Since (0,7, -7) lies on it, we get

Thus, the required plane is 3~+3y+3z=O,thatis,x+y+z=O. You can do the following exercise on the same lines.

E 16).

Find the equation of the plane passing through (1,2,0) and the line x t o s a + y.c o.s p + z c o s y = l , x + y - = z .

Now, give4 a line and a plane, will they always intersect? And, if so, what will their intersection look like? In Fig. 13 we show you the three possibilities.

Fig. 1.3 : (a) A line and a plane can eiiher intersect in a point, or (b) the line can lie in the plane. or (c) they may not intersect at all. I

Let us see some example. x y-1' 2-2 Example 4 :Check whether the plane x + y + z = 1 and the straight line - = -= 1 2, 3 intersect. If they do, then find their point (or points) of intersection.

Solution :By Remark 1 you know that any point on the line can be given by x = t, y = 1 + 2t and z = 2 + 3t, in terms of a parameter t. So, if the line and plane intersect, then (t, 1+ 24 2 + 3t) must lie on the plane x + y + z = 1 for some t. Let us substitute these values in the equation. We 1

get t + 1 + 2t + 2 + 3t = 1, that is, 6t = -2, that is, t =- - . Thus, the line and plane intersect in a 3 ,point, and the point of intersection is

(- j1 1 1) 9 $9

Example 5 :Find the point (or points) of intersection of . ..

Solution :a) Any point on the line is of the form (2k- 2,3k - 3, - 2k + 4), where k E R. 0

Thus, if there is any point of intersection, it will be given by substituting this triple in 3x+2y+6z= 12. So, we have

This is true +j k E R. Thus, for every k E R, the triple (2k - 2,3k - 3, -2k + 4) lies in the plane. This means that the whole line lies in the plane. b) Any point on the line is of the form (2t + 1, t + 2, -2t + 3), where t E R. This lies on x + z = 1 if, some t, (2t + 1) + (-2t + 3) = 1, that is, if 4 = 1, which is false. Thus, the line and plane do not intersect. Yob :an use the same method for finding the point (or points) of intersection of two lines. In Bllowing exercises you can check if you've undestood the method.

E 17)

Find the point of intersection of the line x = y = z and the plane x + 2y + 3z = 3.

NOW; consider any two planes. Can we find the angle between them? We can, once we have

the following definition. Definition : The angle between two planes is the angle between the normals to them from the origin. So, now let us find the angle between two planes. Let the equations of the planes, in the normal form, be 1,x + m, y + nlz = p, and 1,x + m2y+ n,z = p,. Then the angle between the normals is cos-' (1,1, + mlm, + n,n2). Thus, the angle between two planes 1,x + m,y + n,z = p , and l,x + m2y+ n,z = p, is cos-I (/,I, + m,m2 + n,n,). ...(18)

Prclimlnarles im TbrceDimenslomai Geometry

The Sphere, Cone and Cylinder cos 9 = ,/a2 + b" + c2

.J-

...(19)

This is because a, b, c and A, B, C are direction ratios of the normals to the two planes, so

m 7 ,/= ,/KKF9 a

that

and

7- / ,

are their direction cosines.

Thus,

theplanesax+by+cz+d=OrindAx+By+Cz+D=O i)

a b c are parallel iff - = - = - ,and A B C

ii)

are pependicular iff aA + bB t cC = 0.

Let us consider an example of the utility of these conditions.

Example 6 :Find the equation of a plane passing through the line of intersection of the planes 7x - 4y+ 72 + 16 = 0 and 4x + 3x - 22 + 13 = 0, and which is pependicular to the plane 2x-y-2z+5=0. Solution :~ h e ~ ~ e n eequation ral of the plan~$hroughthe line of intersection is given by 7x-4y+7z+ 1 6 + k ( 4 x + 3 ~ - 2 z +13)=0. * ( 7 + 4 k ) x + ( 3 k - 4 ) ~ +(7-2k)z+ 13k+16=0. This will be perpendicular to 2x - y - 22 + 5 = 0, if

Thus, the required equation of the plane is 47x - 48y + 7 l z + 92 = 0. Try these exercises now. -

E 19)

Find the equation of the plane through (1,2,3) and parallel to 3x + 4y - 5z = 0.

E20)

Find the angle between the planes x + 2y + 22 = 5 and 2x + 2y + 3 = 0.

E21)

X-A y - b Z - c Show that the angle between the line = -= -and the plane

a P

(Hind :The required angle is the complement of the angle between the line and the normal to the plane). And now let us end the unit by summarising what we have done in it.

Preliminaries in ThreeDimensional Geometr)

SUMMARY

1.5

In this unit we have covered the following points. 1) Distance formula : The distance between the points (x, y, z) and (a, b, c) is

The coordinates of a point that divides the join of (x,, y,, z,) and (x,, y,, z,) in the ratio m :nare

(

nx, + mx, ny, + my2 nz, + mz2 9 m+,n 9 m + n m+n

If cos a , cos p and cos yare the direction cosines of a line, then cos2a + cos2p + cosZy=1.

The canonical form of the equations of a line passing through the point (a, b, c) and having direction cosines cos a , cos P and cos y is x-a -=-cos a

y - b --z - c cos p cos 7

5) The two-point form of the equations of a line passing through (XI,Y,, 2,) rind (x,, Y,, z,) are X-X 1

x2-x1

Y-Yl z-z, -= -----.

z2-z1

Y2-Yl I

The angle between two lines with direction ratios a,, b,, c, and a?, b2, cl is

Thus, these lines are perpendicular iff a,a2+ bvI6,+ clc, = 0, and parallel iff a, = ka,, b, = kb,, c, =kc2,for some k E R.

The equation of a plane is of the form Ax.+ By + Cz + D = 0; where A, B, C, D'E Rand not all of A, B, Care zero. Conversely, such an equation always represents a plane.

The plane determined by the three points

The equation of the plane which makes intel cepts a, b and c on the x, y and z-axes, X

respectively, is - + - + - = I . a b c 10) The normal form of the equation of a plane is x cos a t y cos $ + z cos y = p, where p is the length of the pependicular from the origin ontothe plane and cos a, cos p, cos yare the direction cosines of the perpendicular. 11) The length of the pependicular from a point (a, b, c) onto the plane Ax+By+Cz+ D=Ois

; - T h e Sphcr:,

Cone and

Cylinder

12) A line is the intersection of two planes. 13) The general equation of a plane passing through the line ax + by + cz + d = 0 = Ax + b B y + C z + D i s ( a x + b y + c z + d ) + k ( A x + B y + C z + D ) = O , w h e r e k ~R. 14) The angle between the planes ax + by + cz + d = 0 and Ax + By + Cz + D = 0 is

And now you may like to go back to Sec. 4.1, and see if you've achieved the unit objectives listed there. As you know by now, one way of checking this is to ensure that you have done all the exercises in the unit. You may like to see our solutions to the exercises. We have given them in the following section.

1.6

SOLUTIONSIANSWERS

E l ) PQ= ,/(I

- ( - I ) ) ~+ (1 - 1)2 + (-1 - 112 = 6

The coordinates of R are

(t, +) 1, -

E 3) sin2a +'sin2p + sin2y= 3 - (cos' a + cos2 p + cos' y) = 2. E4) a) 0, 1, 0 and 0,%, 1, respectively. b) Any line in the XY-plane makes the angle xI2 with the z-axis. Now, if m = tan 0. &err y = mx + c makes an angle 8 with the x-axis, and x/2 - 8 with the y-axis. Thus, its direction cosines are cos 8, sin 8,O. E5) In Fig. 14 we have depicted the situation. a b c Let OP = r. Then you can see that the direction cosines of L are - - ,- . r r r Thus, a, b, c are direction ratios of L.

X Fig. 14

E6) Now, the line L makes angles of x -a,x -P and x -'y with the positive directions of the x, y and z-axes, respectively. Thus, its direction cosines are -cos a , -cos P and - cos y. X+l y 2-1 E 7) --- --- = 2 2 2 ' E 8) The equations are x + 3 -. y - 5 5 ' -1.

2-3 0

-- ----- = - = r, say, that is,

E9) The direction cosines of the line with direction ratios 1, 1 , 2 are I 2 1 1 2

, + + i 12

J3 A 4 Similarly, the direction cosines of the other line are -, - - 5 5 '5' Thus,.if 8 is the angle between them.

(+)($1 +(+) ($1

(2) .6 ()! 5

cos 0

El01

Since1(1)+2(-2)+3(1)=0. 1 ( 4 ) - 2 ( l ) + 1 (-2)=0,and 1 (4)+2(1)+3(-2)=0, the lines are mutually perpendicular.

E 11)

The equation of the plane passing through the first there points is

*5~-7y+ 11~+4=0. Since (-4,4,4) satisfies it, the 4 points are coplanar. E 12)

The points (2,0, O), (0, -1, O), (0,0,5) lie on the plane. Thus, its equation is

E 13)

-I

=0~5~-1Oy+2z=10

d d d The intercepts of the two planes on the axes are -- - - - a ' b' c

D D D and - -,- - - - , respectively. Thus, the planes coincide A B C

i f f -a= - =b- = - c A B C E 14)

d ,that is, iff a, b, c, d and A, B, C, D are proportional. D

The distance of (2,3, -5) from the XY- plane, z = 0, is

Similarly, the distance of (2,3,-5) fromx = O'and y = 0 is 2 and 3, respectively. Its distance from x + y + z = 1 is

E15)

Weknowthat

~ r e l i f n i n a r i e sin ThreeDimensional Geometry

The Sphere, Coae and Cylinder

E 16)

The equation of the plane passing through the given line is (xcqsa+y cosp+zcosy-l)+k(x+y-z)=0, where k E R is chosen so that (1,2,0) lies on the plane.

...(20)

( c o s a + 2 c o s p - 1 ) + 3 k = O d k = - (1-cosa-2cosP). 3

Thus, the required equation is obtained by putting this value of k in (20).

E 17)

Any point on the line is (t, t, t). The line and plane will intersect if, for some t E R.

Thus, the plane and line intersect in only one point

E 18)

1 1 1

(i 2 i) 9

Any point on the first line is given by ( t + 1 , 2 t + 3 , 3 t + s ) , t € R. Any point on the second line is given by ( 3 k - l , S k + 4 , 7 k + 9 ) , k ~R.

Thetwolineswillintersectift+1=3k-l,2t+3=5k+4and3t+5=7k+9for some t and k in R. On solving these equations we find that they are consistent, and k = 5 gives us the common point. Thus, the point of intersection is (14,29,44).

Fig 15 : The llne L makes an angle 0 with the plane 1C and

(7Ut $ with the normal N to r.

E 19)

Any plane parallel to 3x + 4y - 5z = 0 is of the form 3,!+4y-5z+k=O,wherek~ R. Sidce(l,2,3)liesonit,3+8-15+k=Oak=4. Thus, the required plane is 3x + 4y - 5z + 4 = 0.

E20)

If the angle is 8, then

E21)

If 0 is the angle between the line and the plane, then - - is the angle between the 2 line and the normal to the plane (see Fig. 15).Now, A, B, C are the direction ratios of the normal. Thus,

'UNIT 2

THE SPHERE

Structure 2.1

Introduction Objectives

2.2

Equations of a Sphere

2.3

Tangent Lines and Planes 2.3.1 Tangent Lines

2.3.2 Tangent Planes

2.4

Intersection of Spheres Two Intersecting spheres

Spheres Through a Given Circle

2.5

2.1

Summary

INTRODUCTION

With this unit we start our discussion of three-diemensional objects. As the unit title suggests, we shall consider various aspects of a sphere here. A sphere is not new to you. When you were a child you must have played with balls. You must also have eaten several h i t s like limes, oranges and watermelons. All these objects are spherical in shape. But all of them are not spheres from the point of view of analytical geometry. In this unit you will see what a geometry calls a sphere. We shall also obtain the general equation of a sphere. Then we shall discuss linear and planar sections of a sphere. In particular, we shall consider the equations of tangent lines and planes to a sphere. Finally, you will see what the intersection of two spheres is and how many spheres can pass through a given circle. Spheres are an integral part of the study of the structure of crystals of chemical compounds. You find their properties used by architects and engineers also. Thus, a; analytical study of spheres is not merely to satisfy our mathematical curiosity. A sphere is a particular case of an ellipsoid as you will see when you study Block 3. So, if you have grasped the contents of this unit, it will be of help to you while studying the next block. In other words, if you achleve the following objectives, it will be easier for you to understand the contents of Block 3.

Objectives After studying'this unit you should be able to : obtain the equation of a sphere if you know its centre and radius; check whether a given secdnd degree equation in three variables represents a sphere; check whether a given line is tangent to a given sphere; obtain the tangent plane to a given point on a given sphere; obtain the angle of intersection of two intersecting spheres; , 8

find the family of spheres

&rough a given circle.

The Spheye, Cone and Cylinder

Let us now see what a sphere is and how we can represent it algebraically.

2.2

EQUATIONS OF A SPHERE

In 2-space you know that the set of points that are at a fixed distance d from a fixed point is a circle. A sphere is a generalisationof this to 3-space (see Fig. 1).

Definition :The set of all those points in 3-space which are at a fixed distance d from a point C (a, b, c), is a sphere with centre C and radius d.

Fig. 1 :(a) A cireie, (b) a sphere, with origin as centre and radius d.

Spheres are, of course, not new to you. A ball and a plum are spherical in shape. However, whenever we aalk of a sphere in analytical geometry, we mean the surface of a sphere. Thus, for us a hollow ball is a sphere, and a solid cricket ball is not a sphere. Let us find the equation of a sphere with radius d and centre C (a, b, c) now. If P (x, y, z) is any point on the sphere, then, by the distance formula ((1) of Unit 4), we get

which is the required equation. For example, the sphere'with centre (O,O, 0) and radius 1 unit is xZ+ y2 + z2= 1. Now, if we expand (I), we get the second degree equation x2 + y2 + z2 -2ax-2by-2cz+a2+b2 +c2-d2=0. Looking at this you could ask if every equation of the t ) pe a(x2+y 2 + z 2 ) + 2 ~ ~ + 2 v y + 2 w z + d = 0 , where a, u, v, w, d E R, represents a sphere,

It so happens that if a # 0,then (2) represents a sphere. (What happens if a = O? Unit 4 will give you the answer.) Let us rewrite (2) as

u2 v2 w2 Adding 7 + 3 + - on either side of this equation, we obtain a a2

Comparing this with (1), we see that this is a sphere with centre -

( a7 a7 a

and radius

u2 + v 2 + w 2 -ad la l

The following theorem summarises what we have said so far.

Theorem 1 : The general equation of a sphere is ~~+~~+z~+2ux+2v~+2wz+d=0.

Its centre is (-u, -v, -w) and radius is ,/u2 + v2 + w2 - d . Note that the general equation given above will be a real sphere iff u2 + v2 + w2 - d 2 0. Otherwise it will be an imaginary sphere, that is, a sphere with no real points on it. So, what we have seen is that k

a second degree equation in x, y and z represents a sphere iff

ii)

the equation has no terms containing xy, yz or xz.

Why don't you see if you've taken in what has been said so far'? -

E 1) Find the centre and radius of the sphere given by x2+ y2 + z2- 2x + 4y - 62 = 11. -E2) Does 2x2+ 1 + 2y2+ 3 + 2z2+ 5 = 0 represent a sphere? E 3) Determine the centre and radius of the sphere x2 + y2 + z' = 42. Now, if you look at the general equation of a sphere,yi;ru will see that it has 4 arbitrary constants u, v, w, d. Thus, if we know 4 points lying on a sphere, then we can obtain its equation. Let's consider an example.

Example 1 :Find the equation of the sphere through the points (O,O, O), (0,1, -I), (-1,2,0) and (19 293). Solution : Suppose the equation is

Since the 4 given points lie on it, their coordinates must satisfy this equation. So we get

Solving this system of simultaneous linear equations (see Block 2, MTE-04), we get

Thus, the required sphere is 2

7(x + y + z )-15x-25y-llz=O.

T h e Sphere, Cone and Cylinder

Note that it can happen that the system obta~nedby subst~tutingthe four points is inconsistent, that is, it does not have a solutlon (Such a situation canroccui if three of the points lie on one line.) In t h ~ case s there n 11: be no sphere pass] thral:p5 these pnln+. You can try some exercises now. E4) Find the centre and radius of the sphere passing through (1,0, O), (0, 1. O), (0.0, 1) and

E5) Is there a sphere passing through (4,0, l), ( 10, -4,9), (-5,6, -1 1) and ( 1,2,3)? If so, find its equation.

A d i a ~ l e t e ro f a sphere. is a .line segment through its

centrc and with end points on tht sphere. .

Now, if instead of four points on the sphere, we only know the coordinates of the two ends of one of its diameters, we can still determine its equation. Let us see how. Let A (x,, y , , z,) and B (x', y2, zz) be the ends of a diameter of a sphere (see Fig. 2). Then, if P (x, y, z) is any point on the sphere, PA and PB will be perpendicular to each other. Thus. from (10) of Unit 4, we see that

This is satisfied by any point on the sphere, and hence is the equation of the sphere. I

For example, the equation of the sphere having the points (-3, 5. 1) and (3, 1,7) as the ends of l)+(z-l)(z-7)=0.

adiameteris(x+3)(~-3)+(y-5)(y-

thatis,~~+~~+z'=6~+8~-3. You can obtain the diameter form of a sphere's equation in the following exercise. Fig. 2

E6) Find the equation of the sphere described on the join of (3,4,5) and (l,2,3). By now you must be familiar with spheres. Let us now see when a line or a plane intersects a sphere.

2.3

TANGENT LINES AND PLANES

In this section we shall first see how many common points a line and a sphere can have. Then we shall do the same for a p!ane and a sphere.

2.3.1

Tangent Lines

Supnose you take a hollow ball and pierce it right through with a knitting needle. Then the ball think this is true of any line and the needle will have two points in common (see Fig. 3). that intersects a sphere? See what the following theorem has to say about this.

you

Fig. 3 : A line intersecting a sphere

Theorem 2: A line and a sphere can intersects in at most two points. y-b z-c t (say) be a given a=--P Y x-a

~ r o o f~: e t x ' +y ' + z 2 + 2 u x + 2 v y + 2 w z + d - ~ a n d

sphere and line, respectively: Then any po~nton the line 1s of the form ( a t + a, Pt + b, y t + c), where t E R. If this lies on the sphere, then ( a t + a)' + (Pt + b)' + (y t + c)' + 2u ( a t + a) +2v(Pt+b)+2w(yt+c)+d=O.

This is a quadratic in t. Thus. it gives two values oft. For each value oft, we will get a point of intersection. Thus, the line and sphere can intersect in at most two polnts.

Note that (4) can have real distinct roots, real coincident roots or distinct imaginary roots. Accordingly, the line will intersect the sphere in two points, in one point, or not at all. This leads us to the following definitions.

The Sphere

Definitions : If a line intersects a sphere in two distinct points, it is called a sacant line to the sphere. If a line intersects a sphere in one point P, it is called a tangent to the sphere at the point P; and P is called the point of contact of the tangent. For example, the line L in Fig. 3 is a secant line to the sphere; and the line L in Fig. 4 is a tangent to the sphere at the point P. Now, (4) will have coincident roots iff

Thus, (5) is the condition for

Fig. 4 : L intersects the sphere in only one point, P.

x - a - -y - b = -z-c

to be a tangent to

Let us consider an example.

Example 2: Find the intercept made by the sphere x2 + y2 + z2= 9 on the line x- 3 = y = z. Solution : Any point on the line is of the form (t + 3, t, t), where t E R. This lies on the sphere if (t+3)2+t2+tZ=9a3t2+6t=~=,t=0,-2,

Thus, the points of intersection are (3,0,0) and (1, -2, -2). Thus, the intercept is the distance between the two points, which is W

+ X = 2& .

You can try some exercises now. z x+3 E 7) Check if -- = ---- - - is a tangent to the sphere 4 3 5

E 8) If we extend the rule of thumb to find the tangent to a conic (see Unit 3) to a sphere, will we get the equation of a tangent line to a sphere? Why ? Let us now discuss the intersection of a plane and a sphere.

2.3.2

Tangent Planes

Consider a sphere and a plane that intersects it. What do you except the intersection to be? The following result will give you the answer.

Theorem 3: A planar section of a sphere is a circle. Proof: LetS be a sphere wit11 radius r and centre 0 (see Fig. S ) , and let the plane ll intersect it.

Fig. 5: A planar intersection of a sphere is a circle.

Drop a perpendicular ON from 0 onto 11, and let ON = a. Now let P be a point which belongs to Il as well as S. Then OP = r and OP' = ON' + NP'.

The Sphere. Cone and Cylinder

Thus, NP = d r 2 - a2 ,which is a constant. Thus, the intersection of S and n is the set of polnts in n which are at a fixed distance from a fixed point N. Thus, it is a circle m the plane n wlth centre N and radius -/,

. -

If a = 0 in the proof above, the plane passes through the centre of the sphere. In this case the circle of intersection is of radius r and is called a great circle (see Fig. 6 ) of the sphere.

Fig. 6: The intersection o f S ;od P i s a great circle o f the sphere S.

4 Z

Note that a sphere has infinitely many great circles. one for each plane that passes through the centre We have seen that the planar section of a sphere is a circle. Now let us find its equation. Let the equation of the sphere be x' + Y2 + z2 + 2ux + 2vy + 2wz + d = 0, and that of the plane intersecting it be A+; By + Cz + D = 0. Then the equation of the planar section can be written as

x2 + y2 + z 2 + 2 u x + 2 v y + 2 w z + d = O = A x + B y + C z + D , o r x 2 + y 2 + z 2+ 2 u x + 2 v y + 2 w z + d = 0 , A x + B y + C z + D = O y2

Fig. 7 : Planar sections of XI+

yz+ z'= I

...( 6 )

For example, the equation of the planar section of the sphere x' + y2 + z2 = 1 by the plane I 7 ? 3 I 2 2 ' z = - (see Fig. 7) is x + y + z- -- 1 = z- - = 0. This is the circle x- + y- = - ,in the plane 4 2 2 1 z= 2' Since the centre of the given sphere 1s (0, 0,O), we can get a great circle of the sphere by intersecting it with z = 0. Thus, one great circle is x2+ y- = 1 1n the plane z = 0. Let us consider an example of the use of Theorem 3. Example 3 :Find the centre and radius of the *:ircle. x Z + Y 2 + ~ 2 - 8 x + 4 y + 8 z - 4 5 = ~ , x - 2 y +2x=3.

* Solutiom: The centre of the sphere is C(4, -2, -4) and its radius is

The distance of the plane from the centre of the sphere is

Thus, the radius of the circle =

,/Gf = 4fi .

The centre of the circle is the foot of the perpendicular from C onto the plane. To find this, we first need to find the equations of the perpendicular. Its direction ratios are 1, -2,2. Thus, its equati~nsare

Therefore, any point on the perpendicular is given by (t + 4, -2t - 2,2t - 4), where t q R. This point will be the required centre of the circle if it lies on the plane, that is, if

Hence, the c e n a of the circle is

( y - 1."E).

T h e Sphere

You can do the following exercise on the same lines. E9) Find the centre and radius of the circle X ' + ~ ~ + Z 12x-12y-16z+ " 111=0=2x+2y+z-17. Now, if we take a = r in the proof of Theorem 3, then what happens to the circle of intersection? It reduces to a single point, that is, a point circle. And in this case the plane only touches the sphere (see Fig. 8). Definition : A plane is tangent to a sphere at a point P if it intersects the sphere in P only. In this case we also say that the plane touches the sphere at P. P is called the point of tangency, or the point of contact, ofthe tangent plane. Remark 1: If you go back to the proof of Theorem 3, you will see that the line joining the point of tangency to the centre of the sphere is perpendicular to the tangent plane. We will use t h ~ sfact to obtain the equation of q tangent plane. Let us find the equation of the tangent plane to the sphere x2+ y2 + z' + 2ux + 2vy + 2wz + d = 0 at the point P (a, b, c). As P lies on the sphere.

a'+b"+c2+2ua+2vb+2wc+d=0.

..(7)

Also, the centre of the sphere is C (-u, -v, -w). Thus, the direction ratios of CP are a + u, b + v, e c + w (see Equation (8) of Unit 4). Now, the tangent plane passes through P (a, b, c). Thus, its equation will be

Now CP is perpendicular to (8), and hence, is parallel to the normal to (8). Further, f, g, h are direction ratios of the normal to the plane. Therefore, a + u, b + v, c + w and f, g, h are proportional.

= L ==... t, say. a+u b+v c+w Then (8) gives us

, ~ x a + y b + z c + u x ' + v y + w z = a Z +2b+ C 2 + u a + v b + w c Using (7) and (9), we get x~+yb+~c+ux+vy+wz=-ua-vb-wc-d

...(9)

hug, the equation of the tangent plane at the point (a, b, c) to the sphere x ' + y 2 + ~ ' + 2 u x + 2 ~+y2 w z + d = 0 i s xa+yb+zc+u(x+a)+v(y+b)+w(z+c)+d=0.

11 i

i 1

In this the equation you got while doing E8? From the equation you may have realised that there is a similar thumb rule for the tangent plane (and not tangent line!) to a sphere. Rule of Thumb: To obtain the equation of the tangent plane io a sphere at the point (a, b, c), x+a

simply substitute ax for x', by for y2,cz for z'; and in the linear terms substitute -for x, 2

y+b for y, X + C for z in the equation of the sphere. 2

Fig. 8: The plane r is tangent to the sphere S at the point N.

The Sphere, Cone and Cylinder

For example, the equation of the tangent plane to x' + y2 + 2' = a2at (a, P, y ) is xa+y~+y*aZ. Why don't ygu try an exercise now ?

E 10)

Find the equations of the tangent planes a)mx2+y2+z2+2z=29at(2,3.+).

b ) t o ~ ~ + ~ ~ + z ~ - 4 ~ - 6 ~ + 4 = 01).a t ( 2 , 3 .

So, what we have seen so far is that if a plane is at a distance d from the centre of a sphere with radius r, then

if r ::d, the plane and sphere do not intersect;

ii)

if r = d, dhe plane is tangent to the sphere; and

I I

iiii) if r > d. the plane intersects the sphere is a circle of radius Now, if you qre given the equations of a sphere and a plane, can you tell if the plane is tangent to the sphere? An obvious way would be to check what the distance of the centre of the sphere from the plane is. Let us use this method to derive the condfiion for the plane is. Let us use this m e t h a to derive the condition for the plane Ax + By + Cz + D = 0 to be a tangent plane to the sphere x2 + y2+ z: + 2ux + 2vy + 2w2 + d = 0. Now, the radius of the sphere is Ju2 + v2 + w2 - d. The length of the perpendicular to the plane from the centre (-u, -v, -w) of the sphere is

This distance must equal the sphere's radius since the plane is tangent to&e sphere.

( A U + B V + C W + D=) ~( A ~ + B ~ + C ? ) ( ~ ~ + V ~ + W ~ - - ~ ) ,

...( 10)

which 1s the required condition.

Let us consider an example.

Example 4: Show that 2x - y - 22 = 16 touches the sphere of contact. x' + y2 + z2- 4x + 2y + 22- 3 = 0, and find the

Solution: The centre of the sphere is (2. -1. -1) and its radius is

= 3. .;

The length of the perpendicular from the centre to the plane 2x - y - 22 - 16 = 0 is 12.2+1+2+161 9 . . -. -. = - = 3, which is the same as the radlus ot the sphere. - 3 So the plane touches the sphere. Let (x,, y,, 2 , ) be the point of contact. Then the equation of the tangent plane is xx,+yy,+zz,-2(x+x1)+(y+y,)+(z+2,)-3=0

-(x,-2)x+[y,+l)y+(zI

+1) z - 2 x , + y , +z,-3=0.

But this should be the same as the given plane 2x - y - 22 - 16 = 0. So the coefficileint of x, y, z and the constant term ih both these equations must be proportional.

The Sphere

Uah$ d.$@@I

mahod, if we are given a point and a plane, we can find the sphere with the

pd~t gs w g @ g t ~and e tbe plane as a tangent plane. In the example below we illustrate this. 1

Fgorn~ld$1 md tho sphere with centre (-1,2,3), and which touches the plane 2x - y + 22 = 6. 84tiU~n:

distance ~ f , t hplane e from the point is

Th~rshadd $c the radius of the sphere. $0, sin;@ tfq cpptre of the sphere is (-1,2,3), its equation will be

my ~ Q P you Y do same exercises now? E 11)

&QW (hat 4b?plane x + y + z = f i touches the sphere x2 + y' + z2 = 1. Find the

of contact.

E 12)

P~QW

that the equation of the iphe;e which lies in the octant OXYZ and touches the @kidinateplanes is x2+ y2 + z2- 2k (x + y + z) + 2k2= 0. for some k E R. . I

E 131

tho equation of the sphere with centre (1,0, O), and which touches the plane q + y + 2 t -3f;o.

In thio prrptipn we have seen what sets can be got by intersecting a line or a plane with a 8pliot9. Mg1Y'Iet us discuss what fprm the intersection of two or more spheres can take.

III @? awtj@tyou will first see that the result of intersecting two spheres is the same as that d?@w bp igt,m~ciinga sphere and a plane, that is, a circle. And then you will see how to dbta19 fnf9qitsly many spheres whose intersection is a given circle.

Z4.1 Two fntemecting Spheres Let us o o ~ i & rtwo spheres given by t

f@ that sstisfies S, = 0 as well as S2= 0 will also satisfy the equation S, - S2= 0,

a(~,-~,~~+2(~,-~,)y+2(w,-w,)z+d,-d?.=0.

...(11)

T h e Sphere, Cone and Cylinder

But, from Theorem 1 of Unit 4 you know this is a plane. Thus, the points of iltersection of the spheres S, = 0 and S2 = 0 are the same as those of any one of the spheres a d the plane (1 1). Since the intersection of a plane and a sphere is a circle, the intersection ofS, = 0 and S, = 0 is a circle (see Fig. 9).

Fig. 9 : T w o spheres-intersect in a circle which can be (a) a real circle, (b) a point circle, or (c) and imaginary circle.

In Fig. 9 (a), the circle of intersection has positive radius, while in Fig. 9 (b) the two spheres intersect in only a point. In Fig. 9 (c), they don't intersect at all. While studying the motion of rigid bodies, you may come across a situation in which two spheres just touch each other, as in Fig. 9 (b). Remark 2 :If S, = 0 and S2 = 0 are any two spheres, which do not necessarily intersect, the. S, - S2= 0 is a plane, and is called the radical plane of the spheres. Now, when we discussed the intersection of lines or of planes, we spoke about the angle of intersection. Is it meaningful to talk abouAthe angle of intersection of two spheres? Definition: The angle of intersection of two spheres is defined to be the angle between the tangent planes to these spheres at a point of contact. You may wonder if the definition given above is 'proper' -the angle could vary from one point of contact to another. But, you will now see that the angle is irdependent of the point of contact. Let the spheres be S, = 0 and S2= 0, where 7 S i = x2 + y2 +z-+2uix+2viy+2wiz+d,=0,wherei=1,2.

Fig. 10: 0 is the angle hetween the two. spheres.

Then, their radiiare r, = ,/u:

+ v j +w: - d l and r, = ,/u: + v: + w: - d, ,respectively.

Let d be the distance between their centres C , and C, (see Fig. 10). Let P be a point of intersection. Thin the angle between the spheres is the angle between the tangent planes at P to each of the spheres. This, in turn, is the angle between the normals to these planes, which are PC, and PC,; Thus, if 8 is the required angle, then from elementary trigonometry you know that 2PC,.PC2cos 8 = P C , + ~ PC,' - c,c,~

:. cos 8 = Two spheres are called orthogonal i f angle of intersection is d2.

+ ri - d2

...( 12)

2r1r2

Thus, in particular, the spheres will be orthogonal iff r,' + rZ2= d2

+ W 1 2 - d1) +(u,'+

3( ~ 1 +v12

vZ2+ w,'-d2)

= (-U, t u,)'

+(-V, + v2)' + (-W. + w2)'

Thus, the spheres S, = 0 and S2= 0 intersect at 90" iff (13) is satisfied. Let us consider an example. 2

Example 6: Find the angle between x2 +,y2+ z2= 4 and x2 7 y2 + z - 2x = 0. Solution:Hereu, =O,v,=O,w, = O , d , = - 4 , u 2 = - 1 , v 2 = 0 , w 2 = 0 , d 2 = 0 . Thus, the centres of the two spheres are (0,0,0) and (1,0,O), their radii are 2 and 1, respectively, and the distance between their centres is 1. ,' ~ h e r f l r e ,by (12), the angle between the two spheres is /

You can see these spheres in Fig. 11. They intersect in only one point P, and the x-axis is the normal from the centres of the spheres to both tangent planes. You can try some exercises on intersecting spheres now. E 14)

Find the angle of intersection of the spheres x2 + y2 + z2- 2x + 2y - 42 + 2 = 0 and 2 2 2 x + y +z =4.

E 15)

Find the equation of the sphere touching the plane 3x + 2y - z + 2 = 0 at P (1, -2.1) and cutting the sphere x2 + .y2+ z2 - 4x + 6y + 4 = 0 orthogonally.

E 16)

a) Two spheres of radius r, and r2 and with centres C, and C,, respectively, will touch each other iff r, + r2 = CIC2.True or false? Why? b) Under what conditions on r,, r2 and C,C2 will. the spheres not intersect?

E17)

Showthatthespheresx + y +z, - 2 x - 4 y - 4 z = ~ a n d x 2 + y 2 + z 2 + l ~ x + 2 z10-0 + touch each other. What is the point of contact?

So far you have seen that two spheres intersect in a circle. Now let us see whether, given a circle, we can find two or more spheres passing through it.

2.4.2

Spheres Through a Given Circle

Suppose we are given a circle. Can we find two distinct spheres whose intersection the circle is? In fact, we can construct many spheres passing through a given circle (see Fig. 12).

Fig. 12: P a r t o f a family o f spheres through a circle.

In Fig. 12, the circle is a g;eat circle of the sphere S,, but not of S2, S,, etc. Let us see what the method of construction of this kind of family is. You know that a circle is the intersection of a sphere and a plane. So its equation if of the form S=O,n=O, where S ~ X ~ + ~ ~ + Z+2wz+d,andllm ~ + ~ U X +Ax+By+Cz+D. ~ ~ ~

T h e Sphere

T h e Sphere. Cone and Cylinder

Any sphere through this circle will be given by S+kIl=O,

...( 14)

where k is an arbitrary constant. Do you agree? Now. if you apply Theorem I , you can see that (14) represents a sphere. Further, every point that lies on the circle must satisfy (14). Thus, (14) represents a sphere through the given circle. So, for each value of k E R in (14) we get a distinct sphere passing through the given circle. Thus, we have infinitely many spheres that intersect in the given circle. Now, a circle can also be represented as the intersection of two spheres S , = 0 and S, = 0. In this case what will the equation of any sphere containing it be? It will be S, + kS, = 0, where k E R. Thus, the infinite set {S, + kS2 = 0 ( k E R ) gives us the family of spheres-passing through the given circle. Let us consider some examples of the use of (14).

Example 7: Show that the dircles x' + y' + z' - y + 22 = 0, x - y + z - 2 = 0 and

Solution: Thk equahon of any sphere through the first circle is

x '- + y2 + z2 - y + 2 z + k ( x - y + z - 2 ) - O , t h a t i s ,

~~+~~+z~+kx-(k+l)~+(k+2)z-tk=0. for some k E R Similarly, the equation of any spliere through the second circle is ~ * + ~ ~ + ~+~1)x-(k, + ( 2 +k3 )~y + ( 4 k l + 1 ) z - ( k , + 5 ) = O for some kl E R.

...(16)

To get a common sphere containing both circles, we must see if (1 5) and (16) coincide for spine k and k, in R. Comparing the coefficients of x, y and z, and the constant terms in ( 15) and ( 16), we get k = 2 k 1+ l , k + 1 = k , + 3 , k + 2 = 4 k , + 1 , 2 k = k I + 5 These equations are satisfied for k = 3 and kl

Thus. there is a sphere passing throi~gllboth the circles and its equation is xZ+y~zzl+3x-4y+5z-6=~.

Example 8: Find the equation of the sphere through the c~rclex' + y' + z' = 9 . 2 + ~ 3y + 42 = 5 and the orlgin. Solution: Let the equation of the sphere be x2 + y' + z' - 9 + k (2x + 3y + 42 4)= 0, where k € R.

9 Since, it passes through (O,O, 0), we get -9 -- 5k = 0. that is, k = - 5 Thus, the required equation is ~ ( X ' + ~ ' + * ~ ) " O ( 3y+4z). ~X+

Example 9: Find the path traced by the centre of a sphere which touches the lines y = x , z = 1 andy--x,z=-1. Solution:Let x' + y' + z' + 2ux + 2vy + 2wz + d = 0,be the equation of a sphere that touches the r e be only two lines. Since y = x, z = 1 touches it, the intersection of the line and t11f ~ ' ~ i zmust one point. Any polnt on the line is ( 1 . t, i), where t E R. I t l ~ e on s the sphere lf

T h ~ equation s has equal roots 11

Slmllarly. slnctl. y --x, z

- I touches the sphere. we get

Subtractinr! tlicse two conditions, we get 7 4uv. = 4at ( 1 + I ). that is. uv = 2w. Thus. the centre of the sphere, (-u, -v. -w), satisfiss t l x equation xy + 22 = 0. This is true 101 any sphere satisfy~ngthe given conditions. Thus, tlie required path is XY-22 50. Now, why don't you check if you've uiidcrstood what wg have done in this section so far'? Prove that the c ~ r c l e s

E 18)

lie o n tlie same sphsre. Find its equation also E 19)

Find tlie equations of the sp11el.e~that pass through x' + y2 + z' = 5 , 2 x + y + 32 = 3 and touch the plane 3x 4 y = 15.

E20)

F ~ n dthe equation ot'thc sphere f o ~\vhicli the circle 2x - 3y + 4z = 8. x' + ?' + z' t 7y - 22 + 2 = 0 is a great circle.

We will stop our discussion on spheres for now. though we shall refer to them off and on in tlte next block. Let us now do a quick review of what we l i a v e c t e r e d in this unit.

2.5

SUMMARY

In this unit we have covered the following polnts. 1)

The second degree equation x ' + y ' + ~ ~ 2i u x + 2 v y + 2 w z + d = O rcprcscnts a sphel-e with centre (-u, -v. -w) and radius Conversely. the equation of any sphere is of this form.

p+ v2 + w2 - d .

A line intersects a sphere in at most two points. It is a tangent to the sphere if itintersects the sphere in only thc polnt.

A plane ~ntersectsa sphere i11 a c ~ r c l eWhen . this ciiclc rcduces to a m ~ nct~ r c l eP, then the plane IS tangent to the sphcre at P.

The equation of the tangent plane to tlie sphere x- + y- + z2 + 2us + 2vy -+ 2wz +- d =; 0 atthepoint(xl,yl.zl)isxxl+yyl+zzl+~~(x+xl)+v(y+yl)+w(z+zl)+d=0. This is peipendicular to the line joining ( x , , y , , z , ) to the centre of'the sphere.

5) Two spheres intersects in a circle. 7

6) TI angle of intersrction of the two intersecting spheres x-' + y7

2 w l z + d l - 0 and .<-+ y- + z- + 2u,x + 2v,y + 2w,z + d, = 0 is cos

( r I 2+ r2? - d 2

2u!u + 2v,y 4

21.1r2

where r , and r, are their radii and d is the distance between their centres.

In particular, the two spheres are orthogonal is 2 ~ 1+,2vlv2 ~ ~ -t 2wIwI = d l + d y

Tltr S p l t e r ~Cone and Cylinder

T11el;e are infinitely many spheres that pass through a given circle.

You inay now like to go back to Sec. 5.1 and go through the list of unit objectives to see if you have achieved them. If you want to see what our solutions to the exercises in the unit are. we have given them in the follow~ngsection.

Its rttdius is ,/(-I)'

+ 2' + (-3)' - (-1 I) = 5. 9

9 E2) We can rewrite the equation as x2 + y' + z' + - = 0. 2 This represents an imaginary sphere with centre at the origin. E3) Its centre is (O,O, 2) and radius is 4. E4) Let the sphere be x2 + y2 + z' + 2ux + 2vy + 2wz + d = 0. Then, since the given points lie on it, we get 1+2u+d=O 17-2v+d=0 1+2w+d=0 2 (u+v+w)+d=O On solving these equations, we find that u = v = w = 0, d = -1.

T h s , the centre of the sphere is (O,0,0) and radius is 1. E5) Suppose such a sphere exists. Let its equatlon be ~~+~'+z~+2ux+2v~+2wz+d=0 Since the given points lie on it, we get the linear system 17+8u+2w+d=0 197 +20u-8v+ 18w+d=O 192- IOU+ 1 2 ~ - 2 2 w + d = O 14+2~+4v+6w+d=O. You can check that this systemis incons~stent.Thus. the point do not lie on a sphere E6) The required equation is (x-;)(x;l)+(y-4)(y-2)+(z-5)(~-3)=0. ~x~+y~+z'-4x-6y-8z+26=~. E7) Any point on the line 1s (4t - 3.3t - 4, 5t), where t e R. Tha will lie on the sphere d (4t-3)'+(3t-4)'+25t2+4(4t-3) 6(3t-4)+ 10(5t)=0. e50t2+36t-11 =O +

Since these are real distinct roots. the line will intersect the sphere in two distinct points. Hence,-it will not be a tangent to the sphere. E8) If7 we extend the rule of thumb to obtain the tangent at a point P (x,, y,, 2,)on the sphere 7 7 ~-+y-+z-+2~~+2vy+2wz+d=O.~eget xx, +yy, +zz, + u ( x + x , ) + v ( y + y , )t w ( z + z , ) + d = 0 . This is a linear equation, and hence represents a plane. and not a line. Thus. ~tcannot represent the tangent line.

T h e Sphere.

E9) The centre of the ~ p h e r eis C (-6,6,8). Its radius is r = 5. The distance of C from the plane is d = 3. = 4.

Thus, the radius of the circle =

x+6 -y-6 The equatioi~sof the perpendicular from C onto the plane are = -- -= z - 8. 2 2 ci

Thus, the centre of the circle is (-4,8,9).

E 11,) The radius of the sphere = 1. The dl tance of the plane from the centre (O,O, 0) of the sphere = 1. Thus, t e plane is tangent to the sphere. If the point of contact is (a, b, c), then the equation of the plane is ax + by + cz - I = 0,

a - -b - c - 1 . .. 1 1 1 - 3 . 1 iij 1 1 z ) .

Tbus, the point of contact is

E 12) Let the equation be x Z +y Z + z ' + 2 u x + 2 v y + 2 w z + d = 0 . Since the plane x = 0 is a tangent to it, the distance ( -u, -v, -w) from x = 0, must

(Note that I - u / = - u, since the centre lies in the octant in which the x, y and z coordinates are all positive). Similarly,-v = -w = r. 2 ~ h e n u ~ + v 'w+2 - d = r 2 = t , d = 2 r . Thus, the equation of the sphere is 7

X - + ~ - + Z - - ~ ~ ( X + ~ + Z ) + ~ ~ ' = O .

Thus, its equation is

E 14) Their centres are C, (1, -1,2) and C, (0.0, O), respectively. Both their radii are 2, and C,c:' = 6: Thus, the angle of intersection is

E 15) Let the sphere be given by

)

x 2 +y 2 + z 2 + 2 u x + 2 v y + 2 w z + d = 0 . Then the plane 3x + 2y - z + 2 = 0 is the same as ~ - 2 y + 7 ~ u ( ~ + l ) + v ( y - 2 ) +l )~+(d~=+O x-2y+z+u(x + l)+v(y-2)+w(z+ l)+d=O o x ( 1 +u)+y(v-2)+z(w+ l)+u-2v+w+d=O

T h e Sphere, Cone and Cylinder

Further, this sphere cuts x2 + y' + z2- 4x + 6y + 4 = 0 orthogonally. Thus, using ( 13) we see that -4~+6v=d+4. 7 5 Substituting the values of v and d, we get u = - . And then v = 5, w = - - , d = 12. 2 2 Thus, the required sphere is x2+ y"

z2+ ./x + 1Oy - 5z + 12 = 0.

E 16)

a) This is true only if one sphere doesn't lie inside the other. Otherwise, as in Fig. 11, ClC2# r, + r2. b) If one lies outside the other and r, + r, > C,C2,then they won'dntersect. If one lies inside the other and I r, - r;l> C,C2,they won't inters&

E17)

TheircentresareC,(l,2,2)andC2(-5,0,-1).

C,C2= 7 =sum of their radii. Thus, they touch each other. The plane S, - S2= 0 is the common tangent plane, where S, = 0 and S2= 0 are the two sphere. ThiswilIbe6~+2y+3~+5=0. The point of contact will be the intersection of the line C,C2with the plane. Now, x+5 y z+l C,C, is given by -= - = -. Any point on this is (6t - 5,2t, 3t - 1). This lies 6 2 3

-11 8 5 Thus, the point of contact is (T7 7 i )

E 18)

Solving this on the lines of Example 7, you can check that they lie on the sphere x2 + y2 + Z2 - 2 ~ - 2 ~ - 2 ~ - 6 = 0 .

Its centre is (-k7 - 77 - 3k) . Its distance from 3x + 4y = 15 is the same as the radius of the sphere.

Thus, the two spheres that satisfy the given conditi0.q~are ~ ~ + y ~ + z ~ + 4 ~ + 2 11 y+ = 06 az n- d 5 ( ~ ~ + ~ ' + ~ ~ ~ - 8 ~ - 4 ~ 13=0. -12zE20)

Any such sphere will be given by x 2 + y 2 + ~ 2 + 7 Y - 2 z + 2 + k ( 2 x - 3 Y + 4 z - 8 ) = ~ , w h e r eR.k ~ Since the given circle is a great circle of the sphere, the centre of the sphere must lie an the plane 2x - 3y + 42 = 8.

Thus, the equation af the $here IS

UNIT 3

CONES AND CYLINDERS

Structure 3.1

Introduction Objectives

3.2

Cones

3.3

Tangent Planes

3.4

Cylinders

3.5

3.1

summary

INTRODUCTION

In the previous unit we discussed a very commonly found three-dimensional object. In this unit we look at two more commonly found three-dimensional objects, namely, a cone and a cylinder. But, what you will see in this unit may surprise you -what people usually call a cone or a cylinder are only portions of very particular cases of what mathematicians refer to a_s a cone or a cylinder. We shall start our discussion on cones by defining them, and deriving their equations. Then we shall concentrate on cones whose vertices are the origin. In particular, we will obtain the tangent planes of such canes. The other surface that we will discuss in this unit is a cylinder. We shall define a general cylinder, and then focus on a right circular cylinder. The contents in this unit are of mathematical interest, of course. But, they are also of interest to astronomers, physicists, engineers and architects, among others. This is because of the many applications that cones and cylinders have in various fields of science and engineering. The surfaces that you will study in this unit are particular cases of conicoids, which you will study in the next block. So if you go through this unit carefully and ensure that you achieve the following objectives, you will find the next block easier to understand.

Objectives After studying this unit you should be able to : obtain the equation of a cone if you know its vertex and base curve; prove and use the fact that a second degree equation in 3 variables represents a cone with vertex at the origin if it is homogenous; obtain the tangent planes to a cone; obtain thc equation of a right circular cylinder if you know its axis and base curve. Let us now see what is cone is.

T h e Sphere, Cone a n d Cylinder

3.2

CONES

When you see an ice-cream cone, do you ever think that it is a set of lines? That is exactly what it is, as you will see in this section. Definition: A cone is a set of lines that intersect a given curve and pass through a fixed point which is nolt in the plane of the curve. The fixed point is called the vertex of the cone, and the curve is called the base curve (or directrix) of the cone. Each line that makes up a cone is called a generator of the cone. Thus, we can also define a cone in the following way. Definition : A cone is a surface generated by a line that intersects a given curve and passes through a fixed point which doesn't lie in the plane of the curve. For example, in Fig. 1 (a), we give the cone generated by a line passing through the point P, and intersetting the circle C. The base curves of the cones in Fig. 1 (b) and Fig. 1 (c) are an ellipse and a parabola, respectively.

Fig. I : (a) A circular cone, (b) an elliptic cone, (c) a parabolic cone.

At this point we would like to make an important remark. >,*. * Remark 1 : A cone is a set of lines passing through its vertex and;,base curve. Thus, it

extends beyond the vertex and the base curve. So the cones in our diagrams are only a portion of the actual cones.

Id I

Fig. 2: A circular cone which is not right circular.

Now let us introduce some new terms. Definitions :A cone whose base cuive is a circle is called a circular cone. The line joining the vertex of a circular cone to the centre of its-base curve is called the axis of the cone. If the axis of a circular cone is perpendicular to the plane of the base curve, then the cone is called a right circular cone. nus, the cone in Fig. 1 (a) is a right circular cone, while the one in Fig. 2 is not. Cones were given great importance by the ancient Greeks who were studying the problem of dqubling the cube. A teacher of Alexander the Great, Manaechmus, is supposed to have geometrically proved the following result. This result is the reason for the continuing imporfance of cones.

Theorem 1 : Every planar section of a cone is a conic.

This theorem is the reason for an ellipse, parabola or hyperbola to be called a conic section (see Fig. 1 of Unit 3). It was proved by the Greek astronomer Appolonius (approximately 200 B.C:). We will not give the proof here.

Cones, and Cylinders

Now, according to Theorem 1 would you call a pair of intersecting lines a conic? If you cut a right circular cone by a plane that contains its axis, what will be the resulting curve be? See Fig. 3. Let us now see how we can represent a cone algebraically. We shall first talk about a right circular cone, which we shall refer to as an r.c. cone. So, let us take an r.c. cone. Let us assume that its vertex is at the origin, and its axis is the z-axis (see Fig. 4). Then the base curve, which is a circle of radius r (say), lies in a plane that is parallel to the XY-plane. Let this plane be z = k, where k is a constant. Then, any generator will intersect this curve in a point (a, b, k), for some a, b E R. So the angle between the generator and the axis of the cone will 9 = tan-'

(i)

, which is a constant. This is true for any generator

Fig. 3 : A pair of intersecting lines is a conic section.

of the cone. Thus, every line that makes up the cone makes a fixed angle 9 with the axis of the cone. This angle is called the semi-vertical angle (or generating angle) of the cone. We can now define an r.c. cone in the following way. Definition : A right circular cone is a surface generated by a line which passes through a fixed point (its vertex), and makes a constant angle with a fixed line through the fixed point. Let us obtain the equation of an r.c. cone in terms of its semi-vertical angle. Let us assume that the vertex of the cone is 0 (0,0,0) and axis is the z-axis. (We can always choose our coordinate system in this manner). Now take any point P (x, y, z) on the cone (see Fig. 5). Then, the direction ratios of OP are x, y, z, and of the cone's axis are 0,O, 1. Thus, from Equation (9) of Unit 4, we get

cos e = Fig. 4 : A right circular cone with vertex at the origin and base curve x2 + y' = IJ, z = k.

Thus, x2 + y2 + z2 = z2 sec29 x2-+ y2 = z2 tan2 9. (1) is called the standard form of the equation of a right circular cone. Now, why don't you try the following exercises?

Show that the equation -

x-a y-b axis ---

the r.c. cone with vertex

2-C

and semi-vertical angle 9 is

[a(x a)

+ P(Y b) y (z c)12(a2 P2 J ) { (X a)2 (y b12+ (Z c)' ) cos2e ...(2) -

E2) Can you deduce (1) from (2)? F.3) Find the equation of the r.c.,cone whose axis is the x-axis, vertex is the origin and X

semi-vertical angle is - . 3 Fig. 5: x2 + y2 = z2 tan1 0

Let us now look at a cone whose vertex is the origin. In this situation we have the following result.

The Sphere. Cone and Cylinder

Theorem 2 :The equation of a cone whose base curve is a conic and whose verteh is (0,0,O) is a homogeneo.us equation of degree 2 in 3 variables.

An equation is homogeneous of degree 2 if each of its terms IS of degree 2.

Proof: Let us assume that the base curve is the conic.

cl~~+2hx~+b~~+2~~+2f~+c=0,z=k. Any generator of the cone passses through (0,O. 0). Thus, it is of the form

Ihis line intersects the plane z = k at the point

[-,Y PkY ak

-9

k, '

This point should lie on the conic. Thus,

Eliminating a,P, y from this equation and (3), we get

This is the equation of the cone. As you can see, it is homogeneous of degree 2 in the 3 variables x, y, and z. x2 Y 2 For example, the equation of the cone whose base curve is the ellipsc - + - = 1 in the 4

plane z = 5, and whose vertex is the origin, is

Do you see a pattern in the way we obtain the equation of the cone from the equation o the base curve? The following remark is about this. Rerpark 2: To find the equation of the cone with vertex at (O,O, 0) and base curve in the plane Ax =By + Cz = D, D # 0, we simply homogenise the equation of the curve, as Ax+By+Cz follows. We multiply the linear terms by ,and the constant ten? by D +

Bd,

* ;and we leave the quadratic terms as they are. The equation that we get by

this procesq is a homogeneous equation of degree 2, and is the equation of the cone. Let us look at a few examples of cones with their vertices at the origin. Example 1: Show that the equation of the cone with the axe= fyz + gzx + hxy = 0, where f, g, h E R

generators is

Solution. By Theorem 2, the equation of the cone is ax2+ by2+ cz2+ 2fky + 2gzx + 2hxy = 0, for some a, b, c, f, g, h E R.

Since the x-axis is a generator, (1,0,O) lies on it. Therefore, a = 0. Similarly, as it passes through (0, 1,O) and (0,0,1), b = c = 0. So the equation becomes fyz + gzx + hxy = 0. Example 2: Find the equation of the cone with vertex at the origin, and whose base curve is z~= the ~ i r c l e x ' + ~ ~ +16,x+2y+2z=9.

Cones and Cylinders

The Sphere. Cone and Cylinder

E6).If

, X

= - = - is a generator of the cone given by the homogeneous equ,ation (4). then

P Y

show that (a,p, y) lies on (4). E7) Whit,. of the following equations represents a cone? 3x+4y+5z=o;x2+y2+z'=9;3(x2+y'+z')=xy;xyz=~zx+xy.

u2 v2' w2 I E8) If -+ - + - = d, show that ax2+ by2+ cz2+ 2ux + 2vy + 2wz + d = 0 represents a b c a cone.

Let us now go back to Example 1. This is an example of a cone with three mutually perpendicular generators. Its equation has no term containing x2, y2 or z2.Does this mean that whenever a cone has three mutually perpendicular generators, its equation,must have no term with x2, y2 or z2? The following theorem helps us to answer this question.

Theorem 4: If the cone ax2 + by2+ cz2+ 2fyx + 2gzx + 2hxy = 0 Ms 3 mutually perpendicular generators, then a + b + c = 0. Proof: Let the direction cosines of the three mutually perpendicular generators be li, mi, n;, where i = 1,2,3. Since they are mutually perpendicular, we can rotate o w coordinate system so that these lines become the new coordinate axes. Then the direction cosines of the previou~coordinate axes with respect to the new axes are 11,12,13;m,, m2, m3 and n,, n,, n,, respecti+ely. So ~ n G (Equations 4 (3) and (10)) tell us that

Further, since the perpendicular lines are generators of the cone, using E6 we get

Adding these equations, and using (6), we get a + b + c = 0. Actually, the converse of this result is also true. The proof uses a fact that you have already seen in Fig. 3 in the case of an r.c. cone, namely, If the lines of intersection of a cone and a plane through the cone's vertex ate imaginary, the intersection reduces to a single point, namely, the cone's vertex (as in Fig. I(e) of Unit 3).

any plane through the vertex of a cone intersects the cone in two lines, which may or may not be distinct. C

The following result, which we shall not prove, tells us about the angle between the lines of intersection.

Theorem 5: The angle between the two lines in which the plane ux + vy + wz = 0 intersects the cone C(x, y, z) E ax2+ by2+ cz2+ 2fyz + 2gzx + 2hxy = 0 is

Cones and Cylinders

h. b f v

wherep2 = g u

c w

u . v w

Looking at (7), can you give the condition under which the angle will be n/2? The lines of intersection of the plane and the cone will be perpendicular iff ~(u,v,w~)=(a+b+c)(u~+v~+w~)

....(8)

Let us use (8) to solve the following example, which includes the converse of Theorem 4.

Example~3:Show that if a + b + c = 0, then the cone C(x,y,z)

P$+by2+~~2+2fyl+2gzx+2h~y=~

has infinitely many sets of three mutually perpendicular generators. X Y Z Solution : Let ==be any generator of the cone. Then, by E6, we know that

C (a, B, y ) = 0. Therefore, using the fact that a + b + c = 0 and (S), we see that the plane a x + p y + y z = 0 intersects the cone in two mutually perpendicular g'enerators, say L and X Y Z Now ==is noimal to the plane a x + P y + yz = 0. Thus, it is perpendicular to both L

and L'. Thus, these three lines f o m a set of three mutually perpendicular cone.

of the

X Z Note that we choose = Y =arbitrarily. Thus, for each generator chosen we get a set of a P Y three mutually perpendicular generators. Hence, the cone has infinitely many such sets of generators.

Why don't you try some exercises,now? -

Find the anglk between the lines of intersection of 3x+y+5z=Oand6y~-2~+5xy=O.

-E10)

rrove f i t a3 +by + cz = 0, where abc # 0, cuts the cone yz + zx + xy = 0 in 1 1 1 perpendicular lines iff - + - + - = 0. a b c

E 11)

Prove that if a right circular cone has three mutually perpendicular generators, its semi-vertical angle is tan-' JZ . .

Let us now discuss the intersection of a line and a cone.

3.3

TArGGZNT PLANES

In the previous unit you saw that a line can intersect a sphere in at most two points. What do you expect in the case ofthe intersection of a line and a cone? Let's see. -Letthe equation of the cone be (4), that is, ax2+ bya + cz2+ 2fyz + 2 p + 2bxy = 0.

The Sphere, Cone and Cylinder

Note that, by shifting the origin if necessary, we can always assume this equation as the cone's eqyation. For convenience, we will write C (x, y, z) = ax2+ by2+ cz2+ 2fyz + 2gzx + 2hxy.

- 1' - yl Now consider the line -a P

. Any point on this line is given by

(x, + ra, y, + rp, z, + ry), for some x E R. Thus, the line will intersect the cone, if this point lies on the cone for some r E R. This happens if

a ( ~ ~ + r a ) ~ + b ( ~ ~ ~ + r ~ ) ~ + ~ ( ~ +~ r+y )r+~2 g) (~i l + r2y )f ( (x l~+~r a+) +r ~ ) ( ~ ~ 2h (x, + ra) (y, + rp) = 0.

o r 2 c(a, P, y ) = 2r {a(=, +by, + gf,) + P (hxl +by1+fill + Y (gx,+ fy1+ C Z I ) ~ +C(x13Y ~ ~ Z ~ ) = O. .

...(9)

Now, if (x,, y,, 2,) doesn't lie on the cone, then (9) is a quadratic in r, and hence has two roots. Each root corresponds to a point of intersection of the line and the cone. Thus, we have just proved the following result.

Theorem 6 :A straight line, passing through a point not on cone, meets the cone in at most two points. I

X-x, - y - y , 2-2 Now suppose @at the line --- -= ---I is a tangent to the cone (4) at (x,, y,, 2,). a P Y Then, since (x,, y,, z,) lies on the cone, C (x,, y,, z,) = 0. So (9) b c ~ ~ m e s r2c(a,p,y)+2r {a(ax,+hy,+gz,)+P(hx, + b ~ , + f ~ , ) + ~ ( g x , + f y ~)=O +cz~) This equation must have coincident roots, since the line is a tangent to the cone at (x,, y,, 2,). The condition for this is

- XI

2-2

-1 = 2to be tangent to the cone So, (10) is the condition for -a 3I Y ax2+ by2+ cz2+ 262 + 2gzx + 2hxy = 0. Note that (10) is satisfied by infinitely many values of a , P, y,Ths,

at each point of a cone we can draw infinitely many tangents to the cone. Now, from See. 4.3.3, you know that (10) tells us that each of these lines is perpendicular to the line with direction ratios. ax, +hy,+gz,,hx,+by, +fz,,gx,+fy,+cz,. Thus, the set of all the tangent lines at (x,, y ,, z,) is the plane (x-x,)(ax, +hy, +gzI)+(y-yl)(hxl +by, +fz,)+(z-z,) (gx, +fy, +cz,)=O \

~ x ( a x , ~ h y , + g z , ) + y ( h x , + b y , + f z , ) + z ( g x , ++cz,)=O. fy, since C (xl, y,, 2,) = 0. This plane is defined to be the tangent plane to the cone at (x,, y,, z,). Thus, (1 1) is tbe equation of the tangent plane at (x,, y,, z,) to the cone ax2+ by2+ cz2+ 2fyz + 2gzx + 2hxy = 0. There is a very simple working mle for writing (1 1).

Rule of thumb: To write the equation of the tangent plane at any point (a, B, y) on the cone (4),

Cones and Cylinders

replace x2by ax, y2 py by, z' by yz,2yz by yy + pz, 2zx by az + y x and 2xy by px + ay. I

For example, the tangent plane tothe cone 2x2+ y2 -2xz = 0 at (1, 0 , l ) is 2x (1) + y(0) -(x+z)=O,tbatis,x=z. So far we have only found the tangent plane to a cone whose vertex is at the origin. What about a general cone? The following remark is about this. Remark 4: We can find the tangent plane to a cone with vertex at (a, b, c) in the same manner as above. All we need to do is to shift the origin to (a, b, c) and find the tangent plane in the new coordinate system. And then we can shift back to the old coordinate system, making the required tra~sformationsin the equation of the plane. This will give us the required equation. Now, if you look closely at (1 l), you will notice that (0, 0, 0) satisfies it. Thus, the tangent planeto a cone passes through the vertex of the cone. Therefore, the tangent plane at P (x,, y,, z,) contains P as well as the vertex 0of the cone. Hence, it contains the generator OP of the cone. Thus, the tangent plane to a cone touches the cone along the generator passing through the point of contact. This generator is called the generator of contact of the plane. In Fig. 6 OP is the generator of contacf of the tangent plane T.

Fig. 6 : T is the tangent plane to the cone at P.

You can try some exercises now.

1 1 Find the equation of the~angentplane at the point (Tr q r 1) to the cone

E 12)

E 13)

Use Theorem 5 to obtain the condition under which a given plane is tangent to a cone.

If you've solved El 3, you would have seen that the condition for ux + vy + wz = 0 to be tangent to the cone (4), that is, ax2+ by2+ cz' + 2hxy + 2fyz + 2gzx = 0 is a h u

h b f v

g f c w

u v = 0, that is, w o

Thus, the line - = - = - ,which is the normal at (0, 0, 0) to the tangent plane, is a generator of

...(12)

the cone AX' + ByZ+ cz2+ 2Hxy + 2Fyz + 2Gzx = 0.

n i u s , (12) is the cone generated by the normals to the tangent planes at the vertex (0, 0,0) of the \ cone (4). Since it is homogeneous, it vertex is also (0, 0,O). Note \ that (I?) is r othing but the determinant equation.

The Sphere, Cone and Cylinder

a h g x

h g b f f c y ' z

x y z=O 0

Now, if we bonsider the surface generated by the normals at (O,O, 0) to the tangent planes to (12), what do we get? On calculatin ,you will get a surprise! The cone is (4), because Bc-~=*.cA-G2=bA,AB-3=c~GH-AF=fA7HFBG=gA,FG-cH=M, where, a A= h g

h b

g f

Because of this relationship between (4) and (12) we call them reciprocal cones. Actually, the following example shows why the name is appropriate. x2 Y 2 z2 , Example 41 Show that the cones ax2+ by2+ cz2= 0 and - + -+ - = 0 are reciprocal. a b c (Here abc # 0). Solution: The reciprocai cone of ax2+ by2+ cz2= 0 is given by the determinant equation

b e a O

O c

O Y z - X O

b 0

O c = o

I $ I *I $1' 1 I

= 0, dividing throughout by abc. dkA

This is the nequired equation. Now you can do the following exercises. This will help you to understand reciprocal cones better.

E 14)

Find the cone on which the perpendiculars drawn from the origin to tangent planes to thecone 19x2+11y2+3z2+6yz-1Ozx-26xy=Olie.

E 15)

Prove that the cone 4 has three mutually perpendicular tangent planes iff bc+ca+ab=~=d!h~.

And now ldt us shift our attention to another surface that is generated by lines.

3.4

CYLINDERS

You must have come across several examples of the surface thai we are going to discuss in this section. For instance, a drain pipe is cylindrical in shape, and so is a pencil. But for us, the pencil will pot be a cylinder, only its surfac~will, according to the following definition.

Cones and Cylinders

Definiton :A. cylinder is the set of all lines which intersect a given curve and which are parallel to a fvred line which does not lie in the p h e of thecurve. The fixed line is called the axis of the cylinder and the curve is called the base curve (or directrix) of the cylinder. All the figures in Fig. 7 represent portions of cylinders.

Flg. 7: (a) A circular cyilnder, b) an elliptic cyiinder, c) a parabolic cyiinder.

In Fig. 7 (b) the cylinder's base curve is an elipse, while it is a parabola in Fig. 7 (c). Fig. 7 (a) is an example of a right circular cylinder according to the following definition. Definition: A cylinder whose base curve is a circle, and whose axis passes through the centre of the base curve and the perpendicular to the plane of the base curve, is called a right circular cylinder.

As you can see, in common parlance when people talk of a cylinder, they mean a portion of a right circular cylinder. Henceforth, in this section, by a cylinder we shall mean a right circular cylinder.

Let us now find the equation of a cylinder. We shall first assume that its axis is the z-axis, and its base curve is the circle x2+ y2 r2,z = 0 (see Fig. 8). Let P (x,, y,, 2,) be any point on the cylinder. Let the generator through P intersect the plane of t+e base curve (that is, the XY-plane) in M. Then the perpendicular distance of P from the axis

Jm.

i s 0 ~ =

Fig. 8: The qllnder xz +

But this is also r. Thus,

This equation is true for every point P (x,, y,. cylinder is

) r * 2 .' .

.' nder. Thus, the equation of the

;' r'.

You may wonder why z is not figuring in the e uation. This is because whatever value of z you take, the equation of the cylinder remains x9 + y2 = r2

What does this mean geometrically? It says that whatever plane parallel to the XY-piane you take, say z = t, and take its intersection with the cylinder, you will always get the circle 2 2 x +y2=r. Thus, in a sense, the cylinder is made up of infinitely many circles, each piled up on the other!

The radius of a cylinder is the radius of its bw curve.

Note that the plane z = t is perpendicular to the axis of the cylinder x2 + y2 = r2

The Sphere, Cone and Cylinder

Also note that the length of the perpendicular from any point on a cylinder to its-axisis equal to its radius. Using this fact, let us find the equation of a cylinder of radius r and whose axis is x - a= = - b =

z - c (see Fig. 9). Y

Fig. 9 : A right circular with axis AM.

Let P (x,, y,, z,) be any point op-the cylinder. Let A be the point (a, b, c), which lies on the axis, andM be the foot of the perpendicular from P onto the axis. Then PM = r. Also, AM = AP cos 9, where 9 is the angle between the lines AM and AP.

:. A M =

(x, - a)a + (Y, - b)P + (2, - C)Y

,/+pl+u'

,using Equation (9) of Unit 4.

Since Al&' is jght-angled triangle, we get A P ~ = AM^ + M P ~Thus, .

This equation holds for any point (x,, y,, z,) on the cylinder. x-a y-b z-c -=Thus, the equation of the right circular cylinder with radius - r and axis a P Y

Let us logk at an example. Example 5: Find the equation of the cylinder having for its base the circle x2 + y2 + z2= 9, x-y+zr3. Solution :The centre of the sphere is (0, 0, O), and radius 3. The distance between (0, 0,0) and

. So the radius of the base circle is

the plane x ~y + z = 3 is Fig. 10

= & (see Fig. 10).

The axis of the cylinder is perpendicular to the plane x - y + :.:= 3 and passes through (O,0,0). X

-. Z

SOits equations are - = - = 1 -1 1

Cones and Cylinders

Thu's, using (14), we get the required equation as ~ ( X ~ + ~ ~ + Z ~ - ~ ) 2= ( X - ~ + Z ) 3

~+ z +xy+yz-zx-9=0. ~ ? ~

Why don't you try an exercise now? El@

Findlthe equation of the cylinder a) whose axis is x = 2y = - z and radius is 4. x-1 - y 2-3 b) whose axis is -- - = -and radius is 2. 2 3 1 our discussion on cylinders here. Let us now briefly review what we have

In this unit we have discussed the following points : 1) A cone is a surface generated by a line passing through a fixed point (its vertex) and intersecting a given curve (its base curve), such that the vertex does not lie in the plane oft& base curve. 2)

A cqne whose base curve is a circle, and for which the line joining its vertex to the centre of the base curve is perpendicular to the plane of the base curve, iscalled a right circular cone.

3) A planar section of a cone is a conic. 4)

The equation of a right circular Cone with semi-vertical angle 8 is x2+ y2 = z2tan28.

A second degree equation in x, y, z represents a conk with vertex at the origin if it is homogeneous.

6) The cone ax2+ by2+ cz2+ 2fyz + 2gzx + 2hxy = 0 has 3 mutually ~erpendiculargenerators ifa+b+c=O.

7) Any plane through the vertex of a cone intersects the cone in two distinct or coincident lines. The angle between the lines obtained by intersecting ux + vy + wz = 0 with C (x, y, a) = ax2+ by2+ cz2.+2fyz + 2gzx + 2hxy = 0 is

tan-'

2 ~ ~ 2 ~ 7 (a + b + c)(u2 + v2 + w2)-C(U,V, W)

Thus, the plane is tangent to the cone iff p2 = 0.

8) The equation of the tangent plane to the cone ax2+ by2 + cz2+ 2fyz + 2gzx + 2hxy = 0 atthepointP(x,, y,,z,)is (ax, +hy, +gz,)x+(hx, +by, +fz,)y+(gx, +fy, +cz,)z=O.

This contains the line OP, where 0 is the vertex of'the cone.

The Sphere, Cone and Cyllnder

aa' + bpZ+ cy + 2@y+ 2gya + 2hap = 0.

Cones and Cylinders

(a, P, y) lies on (4).

E7) only ~ ( X ~ + ~ ~ + Z ~ ) = X ~ . E 8) Substituting the value of d in the equation, we can write it as

which is a homogeneous equation of degree 2 in x +

Thus, it is a cone with vertex at

.. =, Y + -L, z + -. a b c

(- a - jv; f). U

-9

r -

E9) The required angle is

a = tan-'

0 - 6(1) (5) + 2 (5) (3) - 3 3 ) (1) '

where P'

-1

-3

1-1

:. a = tan-' J35 . E 10)

In this situation (8) tells us that the lines will be perpendicular iff bc+ca+ab=O:

-a

E 11)

Let its semi-vertical angle be 8. Then the equation of the cone is (I), that is, x2 + y2 = z2 tan' 8. Since this has three mutually perpendicular generators, Theorem 4 tells us that 1+ I - t a n 2 8 = 0 j 8 = t a n - ' A .

E 12)

The required equation is

E 13)

A tangent plane must touch the cone along a generator. Thus, the two linas of intersection of the plane and the cone must coincide. Thus, the angle between these lines must be 0.

Thus, from Theorem 5, we see that ux + vy + wz = 0 is a tangent to the cone ~ ( x , ~ , z ) = ~ ~ ~ = ~ ( s i n c e ~ ~ + ~ ~ + w ~ ~ ~ . )

lil

b f v

c w

v =o w o

g u

T h e S p l ~ e r e ,Cone and Cylinder

The required cone is the reciprocal of the given cone. Thus its equation is

The cone will have three mutually perpendicular tangent planes iff the reciprocal cone has three mutually perpendicular generators. Using Theorem 4 and its converse, we see that this happens iff In ( 12), A + B + C = 0, that is, iff (bc - t') + (ca -,g2) + (ab - h') = 0, that a iff bc + ca + ab = t- + g' + h'.

b) The required e uation is 14 {(x-1) 2 +y-+(z-3)'-4)={2(x-1)+3y+(z-3)}2 e 10x"5y2+ 13z'-12xy-6yz-4xz-8x+3~y-74z+59=~.

MISCELLANEOUS EXERCISES (This siction is optional) In this section we have gathered some problems related to the contents of this block. YOUmay like to do them to get a better understanding of these contents. Our solutions to the questions follow the list of problems, in case you'd like to counter-check your answers. AZ 1)

x.-2 y - 3 2 - 4 Find the equations to the planes thrqugh the line -= -- -,which are 2 4 5 parallel to the.coordinateaxes.

Y b

x + l _--y - 3 --z + 2 and the point 2)- Find the equation of the plane passing thrdugh --3 2 1 I 2-y z+2 (0,7, -7). Also check if x = -= - lies in the plane. 3 2

)

X Fig. 1

+ - + - = 1 meets the axes in the points A, B and C. Find the equations 3) The plane a b c

determining the circumcircle of the triangle ABC (see Fig. 1). 4)

Prove that if every planar section of a surface given by a quadratic equation is a circle, the surface must be a sphere.

5) Find an equation of the set of points which are twice as far from the origin as from (-191, 1). 6) If the sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 cuts x2+ y2 + z2+ 2u8x+ 2v'y + 2w'2 + d' = 0 in a great circle then show that 2 (uu' + vv' + ww') - (d + d') = 2f2,where r' is the radius of the second sphere.

7) Find the equation of the sphere inscribed in the tetrahedron whose faces are

Fig. 2

8) Show that the sum of the squares of the intercepts made by a given sphere on any three mutually perpendicular lines through a fixed point is constant. (Hint: Take the fixed point to be the origin.)

9) Find the equations to the lines m which the plane 2x + y - z = 0 cuts the cone 4x2- Y2+3z2=0.

Y 10) If x = - = z represents one out of a set of three mutually perpendicular generators of 2

the cone 11yz + 6zx - 14 xy = 0, find the equations of the other two. (Hint: Take a plane through the given line, and apply the condition for the lines of intersection of this plane and the cone to be perpendicular.)

,11) Find the equation of the right circular cone with vertex (1,1,3), axis parallel to the line -=-=-

and with one of its generators having direction ratios 2, I, -1.

12) Find the eqnation of the cone which passes through the common generators of the cones x2+ 2y2+ 3z2= 0 and 5xy - pz + 5xz = 0 cuid the line with direction ratios 1,O.l. (Hint: The cone passing through the common generatorsof the cone C, = 0 and C, = 0 isC1 +kC2=0,whereks R) 55

T h e Sphere, Cone and* Cylinder

13) Find the equation of the right circular cylinder that is generated by lines which are X

parallel to - = - = - ,and which are tangent to the sphere x2 + y2 4- z' = r2 (see Fig. 3). a b c (Note: Such a cylinder is called the enveloping cylinder of the sphere.) x

14) The axis of a cylinder of radius 3 has equations 5 =

y + l z-1 = . Find the equation 2 1

of the cylinder. 15) Prove that for a cylinder the tangent plane at any point to its axis (see Fig.-4).

SOLUTIONS %"

The equation to any plane through the given line is a(x-2)+b(y-3)+c(z-4)=O,where2a+4b+5~=0 If this is parallel to the x-axis, we must have a(l)+b(O)+c(O)=O-a=O. Thus, the equatipn of such a plane is 5y - 42 + 1 = 0. Similarly, you can check that the planes parallel to the y and z axes are 5x - 22 - 2 = 0 and2x - y - 1 = 0, respectively.

Theplanewillbea(x-l)+b(y-3)+c(z-+2)=0, where -3a + 2b + c = 0. Since (0,7, -1) lies on it, we have a + 4 b -5c=0. Eliminating a, band c from (2) and (3), we get

Fig. 3: The cylinder envelops the sphere.

....(1) ....(2)

....( 3)

-- a

b ------c a -.= b- c =-= -10-4 1-15 -12-2 1 1 I :. (1)becomes I(x+ I ) + 1 ( ~ - 3 ) +l ( z + 2)-0 *x+y+z=o. D

The line

y-2-z+2

-7 will lie on this plane, if it is parallel to the plane and any

point on it lies on the plane. Since 1 (1) + (-3) ( 1 ) + 2 (1) = 0, the line is parallel to the plane. Also, (0,2, -2) is a common point. Thus, the line lies in the plane. Fig. 4 : The plane i s tangent to the cylinder.

The circurncircle will be the intersection of the given plane with any sphere passing through A, B and C. So, let us take the sphere through 0, A, B and C. The coordinates of these points are (0,0,0), (a, 0, O), (0, b, 0) and (0,0, c). You can check that the equation . ., i s x2 + y2 + z2 -ax-by-cz=O. Thus, the equations that give the circurncircle are x2 + y7- + z 2 -ax-by-cz=O,

-x+ - y+ - z a

Let the :quation of the surface be ax2+ by- + cz2+ 2hxy + 2gzx + 2fyz t 2ux + 2vy + 2wz + d = 0. It intersects z = 0 in the conic a~~+b~~+2h~~+2ux+2v~+d=0. This will be a circle iff a = b and h = 0. Similarly, on intersecting with x = 0 and y = 0 we will get a = b = c and f = 0 = g = h. Thus, the e uation of the surface reduces to ' a ( ~ ' + ~ ~) + 2zu x + 2 v y + 2 w z + d = O , which represents a sphere.

Let (x, y, z) be any point in the set. Then

p + y 2 + z ' = 2 \ j ( x + 1 ) ~+ ( y - I ) +~( * - 112 56

3 3 (x2 + y2

+ z2)+ 8x - 8y - 82 + 12 = 0, which represents a sphere.

If the two spheres are S = 0 and S, = 0, then (-u', -v', -w') lies on S - S, = 0, that is, on2(u-u')x+2(v-v')y+2(w-w')z+d-d'=0. :. 2(u - u') u' + 2(v - v') v' + 2(w - w') w' = d - d'. + 2(uut + vv' + WW')- d + d' = 2(ur2+ vt2+ wt2)= 2rt2+ 2d' + 2 (uu' + vv' + ww') - (d + d') = 2rt2. Let the equation be x'+ y'+z'+2ux t2vy 4 2 w z + d = 0 . Since the givc I-11.e~ are tangent to it, the distance of (-u, -v, -w) from these planes --

A*+ v2 + 2 d . Thus, we see that -

IS r =

Solving these equations, we get r =

3+& 6

Thus, the equation of the sphere is, 7 2 7 3+& X - + y + ~ - - 2 r ( x + ~ + z ) + 2 r ' = 0 , w h e r e r =6 '

Let us assume that the fixed oint is (O,O, 0) and the three lines are the axes. Then let 1: the sphere be given by x2 .+ y- + z' + 2ux + 2vy + 2wz + d = 0. Its intercept on the x-axis, that is, y = 0 = Z. is 2 , / r d Similarly. the other intercepts are 2 NOW,( 2 ~ d ) ' + ( 2

Jz and 2 , / / .

~~2-d)' +(2

which IS a constant, since the sphere is a given one. x y z 2 7 2 Let a line of intersection be - = - = - . Then 21 + m - n = 0 and ul - m- + 3n = 0. 1 m n Solving these equations. we get nz=-21.n=00rm--41,11=-21. Thus, the two lines are

2x --y + k (y - 22) = 0, k E R, gives any plane through the given line. This will cut the given cone in perpendicular lines if 11 (k- I)(-2k)+6(-2k)(2)- 14(2)(k-l)=O-k=-2,7111. Thus, the planes are 2x - 3y + 42 = 0 and 1l x .- 2y - 72 = 0. Now, 2x - 3y + 42 = 0 intersects the cone in two perpendicular lines of which one is the given one which lies in the plane. Therefore, the other one has to be the normal to the plane at (0, 0,O). This is

y 1 3 ix = II 2 / = a. So this will be another of the required set of

mutually perpendicular generators. Similar1y;the third generators will be the normal to 11x - 2y - 7a = 0 at (O,O, O), that is,

If 0 is its semi-vertical angle, then 2+2-2 2

Miscellaneous Exercises

The Sphere, Cone and Cyllnder

A - 1

y - 1

Also, the axis.is given by -=-1 2

--A - J 2

n u s , the equation of the r.c. cone is

@x2+ 10y2+1 0 z 2 + 1 2 x Y + 2 4 ~12xz-5oX-104y-96~+221 + =O. Let the cone be (x2+ 2y2+ 3z2)+ k (5xy - yz + 5xz) = 0, where k E R. Since the line with 4 direction ratios 1,0, 1 lies on it, (1,0, 1) must satisfy i t This gives us k = - - . 5 Thus, the required cone is 5 ( ~ ~ + 2 ~ + 3 ~ ~ ) - 4 ( 55xz)=O. x~-~z+ Let (a, p, y) be any point on the cylinder. A generator through this will be given by x - a +---. y-p-z-y -a b c This line intersects the sphere in (ak + a , bk + B, ck + y), where k is given bv

(ak + a? + (bk + P ) +~(ck + Y)2= r2. This quadratic equation in k gives two values of k, which correspond to two points of intersection. Thus, the generator will be a tangent to the sphere if these points coincide, that is, iff . (aa + bp + cy12= (a2+ b2 + c2)(a2+ p2 + $ - r2). Thus, the locus of (a, P, y ), which is the equationof the enveloping cylinder, is (ax +by + C Z )=~(a2+ bZ+ c2)(x2 + y2+ z2- r2).

We can always assume the equation of the cylinder to be x2 + y2 = r'. Its axis is the z-axis, that is, x = 0,y = 0. /

As in the case of a cone, you can show that its tangent plane at a point (x,, y,, z,) is xx, +yy, = r2. This is parallel to the z-axis. Hence, the result.

GENERAL THEORY OF CONICOIDS Structure 1.1

Introduction Objectives

1.2

What is a Conicoid?

1.3

Change of Axes 1.3.1 Translat~on o f Axes 1.3.2 Projection 1.3.3 Rotation of Axes

1.4

Reduction to Standard Form

1.5

Summary

1.6

Solutions and Answers

1.1

INTRODUCTION

You have seen in Block 7 that the general equation of second degree in two variables x and y represents a conic. In analogy with this we can ask: what will a general second degree equation in three variables represent? In Block 8 you have studied some particular forms of second degree equations in three variables, namely, those representing spheres, cones and cylinders. In this unit we study the most general form of a second degree equation in three variables. The surface generated by these equations are called quadrics or conicoids. This name is apt because, as you will see in Unit 8, they can be formed by revolving a conic about a line called an axis. Alexis Clairaut (17 13- 1765), a French mathematician, was one of the pioneers to study quadric surfaces. He specified that a surface, in .general, can be represented by an equation in three variables. He presented his ideas in his book 'Recherche Sur Les Courbes a Double Courbure' in which he gave the equations of several conicoids like the sphere, cylinder, hyperboloid and ellipsoid. We start this unit with a small section in which we define a conicoid: In the next section we discuss rigid body motions in a three-dimensional system. We shall consider two types of transformations: translation of axes and rotation of axes. You can see that a conicoid remains unchanged in shape and size under these transformations. Lastly, we shall discuss how to reduce the equation of conicoid into a more simple form.

Objectives After studying this unit you should be able to define a general conicoid; obtain the new coordinates when a given coordinate system is translated or rotated; use the fact that translation and rotation of axes are rigid body motions; I

check whether a given conicoid has a centre or not; prove and apply the fact that if a conicoid has a centre, then it can be reduced to standard form.

Conicoids

1.2

WHAT IS A CONICOID?

In his section we shall define surfaces in a three-dimensional coordinate system which are analogous to conic sections in a two-dimensional system. Let us start with a definition. Definition: A general second degree equation in three variables is an equation of the form ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0, ... (1) where a, b, c, d, f, g, h, u, v, w are real numbers and at least one of a, b, c, d, f, g, h is non-zero.

Note that if we put either z k, a constant, x = k or y = k, in (I), then the equation reduces to a general second degree equation in two variables, and therefore, represents a conic. Now, we shall see what a general second degree equation in three variables represents. Let us first consider some particular cases of (1). Case 1 : Suppose we put a = b = c = 1 and g = h = f = 0 in (1). Then we get the equation

Does this equation seem familiar to you'? In Unit 4 of Block 8 you saw that if u' + v2 + w2 - d > 0, then (2) represents a sphere with centre (-f, -g, -h) and radius Ju2 + v2 + w2 - d . Case 2 : Suppose we put u = v = w = d = 0 in (I), then we get ax2 + by' + cz2 + 2fyz + 2gzx + 2hxy = 0. What does this equation represent? You know from Unit 4 of Block 8 that this equation represents a cone. Case 3 : If we put a = b = 1, h = 0 and z = k in (I), then it reduces to x2 + y2 + 2ux + 2vy + d = 0, z = k

...(3)

This represents a right circular cylinder (see Unit 4 of Block 8, Sec. 6.4). Similarly, you can see that if we put x again (3) represents a cylinder.

k or y = k and a = b = 1, h = 0, then

We will discuss the surfaces represented by (1) in detail in the next unit. The particular cases 1, 2. and 3 suggest that the points whose coordinates satisfy (1) lie on a surface in the three-dimensional system. Such a surface is called a conicoid or a quadric. Algebraically, we define a conicoid as follows : Definition : A conicoid (or quadric surface) in the XYZ-coordinate system is the set S of points (x, y, z) E R3 that satisfy a general second degree eqution in three variables. So, for example, if F(x, y, z) = ax2 + by2 + cz2 + 2fyz 4- 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 is the second degree equation, then S = {(x, y, Z) E R3 I F (x, y, z) = 0: Note that S can be empty. For example, if F (x, y, z) = x2 + y' + z2 + 1 = 0, then S = {(x, y, z) I F (x, y,z) = 0) = 0.tlie empty set. In such cases we call the conicoid an iniaginary conicoid.

Since the above expression is very lengthy, for convenience we often denote this coniocid by F (x, y, z) = 0.

Note : In future, whenever we use the expression F (x, y, z) = 0, we will mean the equation (1). In this unit you will see that we can always reduce F(s, y, z) to a much smaller quadratic polynomial. To do this we need to tradsform the axes suitably. Lct us see what this means.

1.3 I

CHANGE OF AXES

In Block 7, you saw that a general second degree equation can be transformed into the standard form using a suitable change of axes. You also saw that these standard forms are very useful for studying the geometrical properties of the conic concerned. Here we shall show that in the case of the three-dimensional system also we can transform an equation F (x, y, z ) = 0 into the corresponding standard form by means of an appropriate change of coordinate axes. As .in the case of the two-dimensional system, the transformations that we apply are of two types: change of origin and change of direction of axes. Let us consider these one by one in the following sub-sections.

1.3.1 Translation of Axes Here we shall discuss how the coordinates of a point in the three-dimensional system get affected by shifting the origin from one point to another point, without changing the direction of the axes. The procedure is the same as in the two-dimensional case. Let OX, OY. OZ be the coordinate axes of a three-dimensional system. What happens when we shift the origin from 0 to another point 0' (see Fig. .I)?

Fig. 1: Translation of axes through 0'.

Let the coordinates of 0' in the XYZ-system be (a, b, c). Let O'X', O'Y' and O'Z' be the new axeswhich are parallel to the OX, OY and OZ axes. Suppose P is a point as given in Fig. 1. Let the coordinates of P in the XYZ-system be (x, y, z) and in the X'Y'Zf-system be (x', y', z'). We first find a relationship between z and 2'. For this we draw a line through P, which is perpendicular to the XOY plane. L.et it cut the planes XOY and X'O'Y' in L and L', respectively. Then we have PL = z and PL' = z'.

Gerteral Theory 0:' (:otticoids

From Fig. 1 we have PL = PL'+ L'L Now L'L is also the length of the perpendicular from the point 0' to the KOY plane. Therefore L'L = c Thus, we have z = z' + c. Similarly, we get x = x' + a and y = y' + b. Hence if we shift the origin from 0 (0, 0, 0) to another point 0' (a, b, c) without changing the direction of the axes, then the new coordinates of any point P (x, y, z ) with respect to the origin 0' will be x'=x-a, y'=y-bandz'=z--c ...(4) So, for example, what will the new coordinates (x', y'. z') of a point P (x, y, z) be when we shift the origin to (2, -1, I ) ? They will be x' = x - 2, y' = y + 1 and z ' = z - 1. When we transform the axes in such a way, we say we have shifted the origin to (2, -1, l), or translated the axes through (2, -1, 1). Now, what will the effect of such a transformation be on any equation in x, y, z? If, in an equation in the XYZ-system, we replace x, y, z by x' + a, y' + b, z' + c, then we get the new equation in X'Y'Zf-system. For example, when we shift the origin to (2. -1, l), then the equation of the plane 3x + 2y - z = 5 will be transformed into 3x' + 2y'- z' = 2. Note that the respective coefficients of x. y. z and of x', y', z' remain unchanged under a shift in origin. Thus, the direction ratios of the normal to a plane do not change when we shift the origin (Recall the definition of direction ratios from Unit 4). Can you guess why this happens'? This is obviously because we are not shifting the direction of the coordinate planes, we are only shifting the origin. Now, let us consider the effect of translation of axes on a general second degree equation. Theorem 1 : Let the coordinates of a given surface S in a given coordinate system XYZ satisfy a second degree equation in three variables. Let us shift the origin from 0 to another ponit 0' giving rise to a new coordinate system X'Y'Z'. Then S is still represented by a general second degree equation in three variables in the new coordinate system. having degree 2, namely ax' + by' + cz' + 2fyz + 2 g x + 2hxy i s called the second dcgree part of the equat!on and the group of the telm 2ux + 2vy + 2wz is called the li~iearpart

Proof : Let the given surface S satisfy the equation F(x, y, z) = ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 For convenience we write the equatioh in the form F(x, y, z) = Zax2 + Z2fyz + 2Zux + d = 0

... ( 5 )

Let (p, q, r) denote the coordinates of 0' in the XYZ-system. Consider now the new system of coordinate axes O'X', O'Y'. O'Z' parallel to the given system with origin 0'. You know that the relation between the coordinates in the original and new system are given by

Substituting these expressions for x, y, z in (5), we get Ca(x + p)' + Z2f(y + q) (z + r) + X2u(x + p) + d = 0. Now, we expand the above expression and simplify by collecting like terms. We get ... (7) + cz'? + 2fy'z' + 2gz'x' + 2hx'y' + 2u'x' + 2v'y' + 2w'z' + d = 0, axtZ +

where u' = (ap + hq + gr) + u v' = (hp + bq + fr) + v w' = (gp + fq + er) + w and d' = ap2 + bq2 + cr2 + 2fqr + 2grp + 2hpq + 2up + 2vq + 2wr + d

General Theory o f Conicoids

... (8)

Let G (x', y', z') denote the expression in the left hand side of (7). Then we see that any point (k', y', zf).belonging to S satisfies the equation G (x', y', z') = 0, which is again a general second degree equation. Hence we have proved the result. If you compare (5) and (7), then you will see that the second degree part of the equation remain unchanged whereas the linear part changes. Hence, we can conclude that under the transformation of shifting the origin of a coordinate system, the second degree part of a general second degree equation does not change. Why don't you try some exercises now?

El) a) What will tha new equation of a right circular cone, with vertex 0 , axis OZ and semi-vertical angle a be, when we shift the origin to (-1, 1, O)? b) What does the new equation represent? Sketch the surface. E2) Obtain the transformed equation of the following equations when the origin is shifted to (1, -3, 2). a) x2 + y2 + z2 - 4x + 6y - 22 + 5 = 0. b) x2 - 2y2 - 32 = 0. Let us now consider the transformation in which the direction of the axes is changed. For this we need to understand the concept of a projection. So, let us first see what a projection is.

1.3.2 Projection '

In this section we shall talk about an important concept in geometry. This concept has even been used by artists through centuries for giving depth to their works of art. Let us define it.

Definition : Let A be a point in'the XYZ coordinate system. The projection of A on a line segment PQ is the foot of the perpendicular drawn from the point A to the line. From Fig. 2 you can see that the projection of A on PQ will be the point 0 where the plane through A and perpendicular to PQ meets the line PQ.

Definition : The projection of a line s ~ g ~ n e AB n t on a line PQ is the segment A'B' of the line PQ, where A' and B' are respectively the projections of the points A and B on the line PQ. (see Fig. 3).

I Fig. 3 : T h e Projection o f the line segment A B on the line P Q is the line segment A'B'.

Remark : From Fig. 4 you can see that the length of A'B' = I AB / cos 0, where 0 is the angle between AB and A'B'. We also call this number the projection of AB on PQ.

So what will the projection of BA be? It will be I BA 1 cos (n + 0) that is - AB cos 0. This shows that the projection can be positive or negative depending upon B the direction of the line segment.

I I I

I I

Whenever you come across the term projection. from the context it will be clear whether we are referring to a line segment or a number. We shall now state a simple result which we shall need while discussing rotation of

I I axes. A' Fig. 4

Theorem 2 : Suppose A,, A,, ..., A,, are n points in space. Then the algebraic sum of the projections of A,A,, A,A,, .........., All-.,AII on a line is equal to the projection of A , All on that line. We will not prove this result in general here; but shall give you a proof in a particular case only. The proof, in any situation, is on the same lines. Proof : Consider the situation in Fig. 5 concerning 4 points A , , A,, A,, A, and their projections A',, A',, AP3,A', on a given line L.

Then

G e n e r a l T h e o r y o f Conicoids

This shows that the sum of the projections of the lien segments A,A,, A2A3, A3A4, is equal to the projection of AlA4. So, we have proved the result for the situations in Fig. 5. ;\low, consider another useful result involving projections. Theorem 3 : Let P (x,, y,, z,) and Q (x2, y2, z2) be two points in the XYZ coordinate system. Then the projection of the segment PQ on a line with direction cosines 1, m, n is given by (x2 - x,) 1 + (y2 - Y,) m + (z2 - z,) n. Proof : In Unit 4, sec. 4.3.2, you have seen that the direction ratios of the line joining P and Q are (x2 - x,), (y2 - y,), (z2 - z,).

Let I PQ I denote the distance between P and Q, i.e.,

Then the direction cosines of the line segment PQ are

Let 8 be the angle between the lines PQ and L. Then the projection of the line segment PQ on the line L is I PQ ( cos 9. But, from Unit 4, Sec. 4.3.3, we have

Therefore, the required projection = (x2 - x,)l + (y2 - yl) m + (2, - 2,) n For example, what will the projection of the line segment joining 0 (0, 0, 0,) to the point P (5, 2, 4) on the line having 2, -3, 6 as its direction ratios be'? We know that the direction cosines of the line with direction ratios 2, -3, 6 is. 2 -3 J(212 + (-3)2

;(612 J(212 + (-3)2 + (6)2

6 J ( z ) ~+ (-3)'

+ (6)2

2 - 3 6 i.e., -, -, -. 7 7 7 .

2 6 Thus, the projection of OP = 5 x - + 2 x ( - 3 + 4 x i = 4 . 7 Now here is an exercise for you.

E3) Let P(6, 3, 2), Q (5, 1, 4), R (3, -4, 7) and S (0, 2, 5) be four points in space. Find the projection of the line segment PQ on RS.

Now we are in a position to discuss how the coordinates in space are affected by changing the direction of the axes without changing the origin.

1.3.3 Rotation of Axes Let us now consider the transformation of coordinates when the coordinate system is -,t-t-A

o h n ~ ~ thn t

nr;n;n

thrnllnh ~n "nnln Ll

T nt

tho

nr;n;nol

r.rrtnm

hn nYV7

Conicoids

Suppose we rotate the coordinate axes through an angle 0 in the anti-clockwise direction. Let OX', OY', OZ' denote the new coordinate axes (see Fig. 6). Suppose the direction cosines of OX', OY' and OZ' be l,, m,, n,; lZ, m2, n,;. l,, m, n,, respectively.

Fig 6: The axes OX', OY' and OZ' are obtained by rotating the axes

OX, OY and OZ through an angle 0.

Let, P be any point in space having coordinates (x, y, z) and (x', y', z') with respect to the old and new coordinate systems. Let PN be the perpendicular from P on OY'. Then OY = y' The line segment ON is also the projection of OP on the line OY' with direction cosines I,, m,, n,. Therefore, by Theorem 2, we have ON = (x - O)l, + (y - O)m, + (z - O)n,. Hence we get y' = xl, + ym2 + zn,

... (9)

Similarly, we can show that X' = x1, + ym, + zn,

and z' = xl, + ym, + zn, Therefore, given (x, y, z) and the direction cosines of the new coordinate axes, we can get the new coordinates (x', y', 2') using equations (9), (10) and (1 1). Now how can we find x, y, z in terms of x', y', z'? For this we draw PM perpendicular to OY (see Fig. 6 ) . Then,

OM is also equal to the projection of OP on OY. Now let us sek what the direction cosines of OY with respect to the new coordinate axes OX', OY', OZ' are. We know that the direction cosines of OX',.OY', OZ' with respect to OY are m,, m2 and m,. Do you agree that the direction cosines of OY with respect to OX', OY', OZ' are also my m,, m,? (We leave this as an exercise for you to verify.) Therefore, by Theorem 2, we get y = OM = (x' - 0) m, + (y' - O)m, + (z - O)m,, ...( 12) i.e., y = m,x' + m,y' + m,z' Similarly, we get + 1,z' and z = n,x' + n,y' + n,z' X = 1 , ~+ ' 1,y'

Hence, (12), (13) and (14) give the coordinates of x, y, z in terms of x', y' and z'. You may find that these equations are not easy to remember. For easy reference, we arrange the equations in a table, as shown in Table 1.

Note that for finding x, y z we make use of the elements in the respective columns, apd to find x', y' and z' we make use of the elements ini the respective rows. Let us consider an example. Example 1 : Find the new equation of the conicoid 3x2 + 5y2 + 3z2 + 2yz + 2zx + 2xy = 1 when the coordinate system is transformed into a new system with the same origin and with the coordinate axes having direction ratios -1, 0, 1; 1, -1, 1 ; 1, 2, 1 with respect to the old system. Solution : The given surface is 3x2 + SY2+ 3z2 + 2yz + 2zx + 2xy = 1

... (15)

Let OX' Y' Z' be the new coordinate system. Then the direction cosines of OX', OY' and OZ' with respect to the original axes are 1 -

1 1 - 1 1 -, --, -,

1 2 1 --, -, -, respectively.

0, -,

A' AhJ3hJzJzJz

We form the transformation table :

From the table we have

and y

= O x x ' + ( ~ ) x X Y 1 + ( ~ ) x z '

and z

A+JS+Z

To find the new equation, we substitute the expressions of x, y, z in (15). Tl~enwe get

General Theory o f Conicoids

Con .aids

Simplifying each term of the above expression and collecting the coefficiients of like terms, we get 2 ~ +' 3y12 ~ + 6 ~ =' 1.~ This is the new equation of the conicoid. YOUcan do the following excercises on the same lines.

E4) Find the new equation of the following conicoids when the coordinates system is changed into a new system with the same origin and direction ratios 1, 2, 3; 1, -2, 1; 4, 1, -2; with respect to the old system. a) x2 - 5y2 + z2 = 1. b) x2 + y2 + z2 - 2xy - 2yz + 2zx + x - y + z = 0

E5) For the conicoid in Example 1 and E4, calculate the sum of coefficients of the square terms in the original equation and in the new equation. Can you infer anything from the outcome?

E6) What will the new equation of the plane x + y + z = 0 be when the coordinate system XYZ istransformed into another coordinate system X'Y'Z' following equations?

by the

E7) Does a cone remain a cone under rotation of axes? Give reasons for your answer. Now let us consider the effect of rotation of axes on F(x, y, z) = 0.

Theorem 4 : Let S be a conicoid satisfying a second degree equation in a coordinate system XYZ is transformed into another coordinate system X'Y'Z' by the direction of axes, without changing the origin, S will still be represented by a second degree equation. Proof : Suppose S is represented by the second degree equation h x 2 + 2Xfyz + X2ux + d = 0 in the XYZ-systems. Let I,, m,, n,; 12, m2: n2; 13, in3, n3 be the direction cosines of the new coordinate axes OX', OY', OZ' respectively. Then you know that the coordinates (x, y, z), (x', y', z') of a point in the old and new system respectively, satisfy the following relationship.

Let us substitute these expressions in the given equation. We consider the second degree parts and linear parts of the equation separately. The second degree part is Xax2 + 2Xfyz. When we substitute the expressions for x, y, z in this part, we get

{Ca (l,xl + 12y' + 13z')l + 2f (mix' + m2y' + m3z') (n,x' + n2yf + n3z)).

General T h t o r y of Conicoids

The coefficient of xf2 in the above expression is ~ + 2gn,11 + 2hl,m,), (all2 + bmI2 + ~ n + , 2fm2n2 Similarly the coefficient of yt2 in the above expression is (alZ2+ bmZ2+ cn22 + 2fm2n2 + 2gn212 + 2hl,m2), and that of zf2 is + bm,2 + ~ n +, 2fm,n3 ~ + 2gh313+ 2hl3m3). Similarly we collect the coefficients of y'z', z'x' and x'y'. Then we get an expression of the form a'~+ ' ~vyl2+ ~ ' 2 '+~ 2fy'z' + 2g'z'x' + 2h'x'y' ... (16) You can also see that the linear part becomes U'X + v'y + w'z L

Where u' = ul, + vm, + wn, V' = u12 + vm2 + wn2, and W' = u13 + vm3 + wn, From (16) and (17) we see that the transformed equation is a general second degree equation. Now looking at expression (17), can you say anything about the change in the constant term? It remains unchanged under rotation of axes. Another interesting fact that you may have observed in the proof above is given in the following exercise.

E8) Suppose that the second degree part Cax2 + 2Cfyz of a general second degree equation transforms into Ca'xf2 + 2Cfy'z' under rotation of axes. Show that a+b+c=a'+b'+c'. Well, let us see what we can gather from Theorems 1 and 4. They say that if a conicoid S is represented by a second degree equation F(x,y, z) = 0 in one coordinate system, then it is still represented by a second degree equation in any other coordinate system obtained by a translation or rotation of axes. Thus b -

a conicoid remains a conicoid under translation or rotation of axes. b

I I

In fact, every geometrical figure remains unchanged in shape and size under translation or rotation of axes. Therefore these transfom~ationsare called rigid body motions. In this section we have discussed two important transformations of a threedimensional coordinate system. We also said that the importance of these transformations lies in the fact that we can use them to reduce any general second degree equation in variables into a simple form. Let us see how this happens.

1.4

REDUCTION TO STANDARD FORM

In this section we shall show that by suitably applying the transformations that we have discussed in the previous section, we can write the general equation of a conicoid in a simpler foml.

Ii 1 I

Let us consider a conicoid given by the.equation F(x, y, z) = Cax2 + 2Cfyz + C2ux + d = 0

Let us assume that there exists a new Cartesian coordinate system, obtained by translating the origin, in which the linear part of F(x, y, z) = 0 vanishes. You will see that this is possible only for a particular type of conicoids. Let the coordinates of 0' be (x,,y,, z,) in the new system. Then we know that in the transformed equation, the second degree part is unchanged and the linear part becomes.

+ v'y' + w'z' where u', v' and w' are as in (8). We have assumed that the linear part vanishes.

U'X'

Therefore u' = v' = w' = 0. This means that we should have ax, + hy, + gz, + u = 0 hx, + by, + fz, + v = 0 gx, + fy, + CZ, + w = 0 In other words, (x,, yo, z,) is a solution of the system of equations

If the sysem of equations (18) has a solution for (x,, yo, z,) E R ~ then , the point, (x,, yo, zo), is called a centre of the given conicoid and we say that the conicoid has a centre at (x,, yo, z,). You will understand why this is called a centre later in Unit 8. Now let us assume that the given conicoid S has a centre. We transfer the origin to the centre (xo, yo, z,). Then the transformed equation becomes Xax2 + 2Xfyz + 2 u k + 2v'y + 2w'z + d' = 0. (Recall that the second degree part does not change by shifting the origin.) Since (x,, yo, z,) is a solution to (18), we see that u' = v' = w' = 0. Therefore, the above equation reduces to (Xax2 + 2Xfyz) + il' = 0. This equation does not have any linear part. We have just proved the following result.

Theorem 5 : Suppose that S is a conicoid which is represented by a general second degree equation F(x, y, z) = 0 in a coordinate system XYZ. Suppose that S has a ' centre 0' (i.e., the system of equations (18) has a solution (x,, yo, z,) Then by shifting the origin to the centre O', the equation assumes the form ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy + d' = 0 in the new coordinate sysem X'Y'Z'. Let us consider an example.

Example 2 : Consider the conicoid given by the equation 2x2 + 3y2 + 4z2 - 4x - 12y - 242 + 49 = 0. Does it have a centre? If so, find it.

Solution : The given equation is F(x, y, z) = 2x2 + 3y2 + 4z2 - 4x - 12y - 242 + 49 = 0 ,

We shall first check whether the given equation has a centre or not, that is, if the system of equations (18) has a solution or not. Here a = 2, b = 3, c = 4, u = -2, v = - 6, w = -1 2. Then we have 2x-2=0 3y-6=0 42- 1 2 = O This set of equations has a solution, namely, (1, 2, 3). Hence, (1, 2, 3) is a centre of S.

Note : Let us go back to (18) for a moment. There we saw that lf the system of equatlons has a solution (x,, yo, z,), then (x,, yo, z,) is a centre of the conlc. In U n ~ t 5 of the course 'Elementary Algebra', you have seen that the system of equatlons has a solutlon if

In fact ~f A # 0, then there exists a unique solution. Why don't you try some exercise now?

f!:b j

!: f

!i 1.'.: ; ,.

E9) Check whether the following conicoids have a centre or not a) 3x2 + 7y2 + 3z2 + lOyz - 2zx + lOxy + 4x - 12y + 42 + 1 = 0 b) x2 + y2 + z2 - 2yz + 2zx - 2xy -'2x + 2y - 22 - 3 = 0. c) 5x2 + 6y2 - 2x = 0

E10) Find a centre of the conicoid 14x2 + 14y2 + 8z2 - 4yz - 4zx - 8xy + 18x - 18y + 5 = 0. What will its new equation be if the origin is shifted to this centre?

81 @i

: ::

+.. 5.:

8bii i a9::.;

gi ) <.

.#.. kii .? c t? " .~ . r !\

i *> 9 : i.

3; .F: ..; . ?

:;t

We shall now state a theorem without proving it. To prove this we need some advanced techniques in calculus which are beyond the scope of this course. Theorem 6 : Suppose S is a given conicoid whose equation is given by F (x, y, z) = ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0. with respect to a given system of coordinates XYZ. Then there exists a new Cartesian coordinate system X'Y'Z' obtained by rotating the axes of the given system XYZ, without shifting the origin, such that the representation of S in the new system takes the form G(x', y', z') = a'xI2 + b'y'2 + C ' Z ' ~ + 2u'x' + 2v'yf + 2w'z' + & = 0.

.; ,: " ?':

:;<

?$!

That is, the new equation does not contain the product terms, yz, ex and xy.

&!;

IS $

!j :16I I!

$it

4$ ;f

combining Theorems 5 and 6 we have the following result. Corollary 1 : Let 'S be a conicoid given by the equation F(x, y, z) = 0, which has a centre 0' in a coordinate system XYZ. There exists a new coordinate system .obtained by shifting the origin from 0 to 0' and then rotating the system about 0', in which the equation takes the simpler form + bYy'2 + C'Z'2 + d' = 0 ... (19) If d' # O, then we can divide throughout by d', and we get the form ax2 + by2 + cz2 = 1, wherea=-:,b=--andc=--. a' b' d d'

c' d'

;i 4I

(19) is called the standard equation of a conicoid.

Recall that in the case of the two-dimensional system also we have seen that we can reduce any second degree equation into a simple form.

iI !

i i

Let us now consider some examples. Example 3 : Show that the conicoid given by x2 + 2yz - 4x + 6y + 22 = 0 has a centre. Reduce it to standard form by shifting the origin to the centre, and then rotating the axes to get a new system in which the direction ratios of the new axes are given by 1 -.I1 0 --1 1 - - 1 -- 1 '

JZ'JZ",h',h"

J Z JZ

G e n e r a l Tl~eoryo r ( o n ~ c o i d c

Conicoidr

with respect to the original coordinate system.

We first check whether the conicoid has a centre or not. Using (18) we see that (2, 0, 0) is a centre of the given conicoid. Shifting the origin from (0, 0, 0) to (2, 0, 0), we get the new equation as x2 + 2yz - 4 = 0 Now, we apply a rotation of axes to the new equation. We note that the direction cosines of the new axes are

0 -- 1 ' "

1 1 -. 1 -- 1 -

- -1

JZ' J;T

Form Sec. 7,3,3 we have

Substituting these equations in the given equation of the conicoid, we see that

which is in the standard form. Now here is an exercise for you. E l l ) Find the standard equation of the following conicoids. a) x2 + y2 + z2 - 2x - 2y - 22 - 1 = 0 , by shifting the origin to the centre. b) 3x2 + 5y2 + 3z2 - 2yz + 2zx - 2xy + 2x + 12y + 102 + 20 = 0, by shifting the origin to the centre and then rotating the system so that the direction ratios of the new, axes are -1, 0, 1; 1, 1, I; 1, -2, 1. We will stop our discussion on general theory of conicoids for now, though we shall refer to them off and on the following units. In the next unit we will discuss the surfaces formed by (19). Let us now do a quick review of what we have co,vered in this unit.

1.5

SUMMARY

In this unit we have covered the following points : 1)

A general second degree equation in three variables ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 represents a conicoid.

Translation of axes: the transformation of a coordinate system in which the

origin is shifted to another point without changing the direction of the axes. The equation of transformations are given by x=x'+a y=y'+b z=z'+c 3)

Rotation of axes: the transformation of a coordinate system in which the direction of axes is changed without shifting the origin. The equations of transformation are given by the following table:

where Ii, mi, ni, i = 1, 2, 3 are the direction cosines of the axes. 4)

A conicoid remains a conicoid under a translation or rotation of axes.

There is a Cartesian coordinate system in which the equation of a conicoid with a centre takes the standard form a'x'2 + b'y'2 ,+ C'Z'~ + d' = 0.

And now you may like to check whether you have achieved the objectives of this unit (see Sec. 7.1). If you would like to see our solutions to the exercises in this unit, we have given them in the following section.

El) a) In unit 6 of Block 2 you have seen that the equation of a right circular cone with vertex 0, axis OZ and semi-vertical angle a is x2 + y2 = z2 tan2 a. When we shift the origin to (-1, 1, 0), then the coordinates in the new system are given by x ' = x + I, y ' = y - 1, 2' = 2, i.e., x = x' - 1 z = z' y = y ' + 1, Substituting for x, y, z in the given equation of the cone, we get = z" tan2 a. (x' - 1)' + (y' + b) This equation represents a right circular cone with vertex at the point (-1, 1, 0) axis along the line parallel to the z-axis through the vertex and semi-vertical angle a (see E l and Sec 6.2 of Unit 6, Block 8).

E2) The equations of transformation are given by x = x ' + 1, y = y ' - 3 , z = z ' + 2 a) the given equation is x2+y2+z2-4x+6~-2z+5=0 Substituting for x, y, z, in the above equation, we get (x' + + (y' - 3)2 + (z' + 2)2 - 4(xf + 1) + 6(y' - 3) - 2(z' + 2) + 5 = 0 Simplifying, we get X'2 + y'2 + z'2 - 2x' + 22' - 7 = 0 which represents a sphere b) The transformed equation is x" - ZY" + 2x' - 4y' - 3z' - 7 = 0.

E3) The direction ratios of RS are -3, 6, -2. 3 6 2 Therefore, the direction cosines of RS are - - -, - -. 7'7 7

(;I

IIPI

JI Theory of Conlcoids

Hence the projection of PQ on RS is

W) a) The given equation is

x* - 5y2 + z2 = 1 The direction ratios of the axes are given by 1 , 2 , 3; 1,-2, 1; 4, 1, -2. Therefore, the direction cosines of the axes are given by 1

The transformation Table is

Then we have

Substituting this in the given equation, we get

Simplifying we get 10 xt2 - 3y1= + 15 z ' 2 + 48 24 x'y' + ---

fi&

&fi y'z' - ,/Ti24f i xlz' = 1

8 The new equation is xQ + y" + 2'' + ~ 6 -

Y'Z'

- f i ?j x'zt = 1.

5) From E4 (a) we get a+b+c=-5+2=-3

This shows that under rotation the sum of coefficients of the square terms remains unaltered in value, i.e., a + b + c = a' + b' + c'. Similarly, you can observe the same for E4 (b) and Example 1 also.

E6) From the given equations of transformations we see that the coordinate system is changed into another coordinate system with the same origin and the direction cosines of the new axes, with respect to ~e old system, are given by

To get tbe new equation, we substitute the values of x, y, z in the equation x + y + z = 0. Then we get

Therefore, under the transformation the plane x + y + z = 0 becomes the plane z' = 0.

E7) Yes. This is because with vertex at the origin is represented by a homogeneous second degree equation. ax2 + by2 + cz2 + 2hxy + 2fyz + 2gzx = 0. Now, from the proof of Theorem 2, we see that under rotation of axes, the above equation becomes a'xt2 + bfyf2+ c'zf2 + 2h'x'yf + 2fyfz' + 2gfz'x' = 0, which is again a homogeneous second degree equation. Therefore, it repreents a cone. .*r

E8) From the proof of theorem 2, we have

a ' = all2 + bmI2 + cn12+ 2fmlnl + 2gn,11 x + 2hl,m1 b' = a122 + bm22 + cn + 2fm2n2 + 2gn212 x + 2h12m2 c' = a132 + bm,' + cn? + 2fm3n3 + 2gn313 x + 2h13m3 then a' + b' + c' = a (lI2 + lZ2+ )1: + b (mI2 + m22 + mj2) + c (n,2 + n22 + n32) + 2f (m, nl + m2 n2 + m, n,)f2g(nll, + n212 + n313) + 2h (llml + 12m2+ 13m3). We know that 3

z c = ~ = C m ' = C 2n ~ i =1

i= 1

i=l

(see Unit 4, Block 8). Also, since the axes are mutually perpendicular, by condition for perpendicularity given in Unit 4, Block, 8, we get

C mini = o = C ni( =

-.3

i=l

i= 1

i=l

hmi

Therefore, we get a' + b' + c' = a + b + c

E9) a) We have to see whether t h e following system of linear equations is consistent or not. ax+hy+gz+u=O hx+by+fz+v=O gx+fy+cz+w=O Herea=-3,b=7,~=3,f=5,g=-l,h=5,~=2,~=-6, w = -2. So we have 3x+5y-z+2=0 5x+7y+5z-6=0 . -x+5y+3z-2=0 On solving this system of equations, we find that it has a unique solution

1 1 4 given by x = 3 , y = - - and z = - . Therefore, the given conicoid has a

unique centre at

(: - 31 4 . 9

General 'Tlicory o f C,.~niccritls

b) This conicoid has infinitely inany centres.

c) This conicoid has no centre since the system of equations (18) is inconsistent in this case.

EIO) The conicoid has a centre at

(0, 0, 0) to

1 1 (- 2. ?, 0) . Now we shift the origin from

I (- ?1 5.0) . The equations of transformation are 7

1 1 y=yl+-,z=zf 2' 2 Substituting for x, y, z in the given equation, we get the new equation as 14x2 + 14y2 + 8z2 - 4yz - 4zx - 8xy = 4. x=x'---

Ell) a) We know that the given equation represents a sphere with centre at (1, 1, 1) (see Unit 5). Then by shifting the origin to the centre we get the standard equation as x'2 + y'2 + 2'2 = 4. b) The given equation is 3x2 + 5y2 + 3z2 - 2yz + 2zx - 2xy + 2x + 12y + 1oz + 20 = 0. We first cheek whether the conicoid represented by this equation has a centre.

Herea=3,b=5,c=3,h=-l,g=1,f=-1,u=1,~=O,w=5 and d = 20. The system of equations for transformations are given by 3x-.y+z+ 1 = o -x+Sy-z+6=0 x- y+3z+5=0 1 5 13 Solving these equations, we get x = - -, Y = - - and = - - Hence a 6 3 6 .

Now we shift the origiin to the centre. Then we get the new equation as 3x2 + 5y2 + 3z2 - 2yz + 2zx - 2xy + d' = 0, Where

Now we apply the rotation of axes. The equations of transformation are x = - x ' + y'+z' y = y' - 22' z = x' + y' + z' Substituting for x, y, z in the given equation of the conicoid, we get 3(-x' + y' + z')' + 5(y' - 2 ~ ' )+~3(x' + y' + z')' - 2(yf - 22') (x' + y' + 2') + qx'+ y' + 2') (--x' + y' + 2') - 2 (-x' + y' + 2') (y' - 22') + d' = 0 i.e., 4 ~ +' 9y" ~ + 36zf2+ d' = 0 This is the standard form of the given conicoid.

UNIT 2 CENTRAL CONICOIDS Structure 2.1

Introduction Objectives

2.2

A Conicoid's Centre

2.3

Classification of Central Conicoids

2.4

Ellipsoid

2.5

Hyperboloid of One Sheet

2.6

Hyperboloid of Two Sheets

2.7

Intersection With a Line or a Plane 2.7.1 Line Intersections 2 . 7 . 2 Planar Intersections

2.8

Summary

2.9

Solutions and Answers

2.1

INTRODUCTION

In the previous unit you were introduced to certain surfaces in a three-dimensional system, which are called conicoids or quadric surfaces. There we discussed some general theory of conicoids and showed that a conicoid remains a conicoid under translation and rotation of axes. You also saw that some conicoids possess a centre and some don't. Based on this the conicoids are classified into two types - central and non-central conicoids. In this unit we shall concentrate only on central conicoids. We first observe that a central conicoid can be reduced to a simpler form by an appropriate change of axes. Then we use these simpler form to discuss the geometrical properties of different types of central conicoids. You will see that there are four types of central conicoids - cone, imaginary ellipsoid, conicoid, hyperboloid of one sheet and hyperboloid of two sheets. You have already studies cones in detail in Unit 6. The remaining three real conicoids are the three-dimensional versions of the central conics that you studied in Block 1, namely, ellipses and hyperbolas. Ancient mathematicians like Euclid, Archimedes and Appolonius were familiar with the above mentioned geometrical objects, though they did not study them analytically. The analytical study of these objects started much later, with the application of the three-dimensional coordinate system to geometry. The first mathematician to suggest the extension of the two-dimensional coordinate system to three dimensions was the Swiss mathematician John Bernoulli (1667-1748). But the actual application of space coordinates to geometry was done by another Swiss mathematician Jacob Hermann (1678-1733). He applied them to obtain the equations of several types of quadric surfaces. Even though we are only concerned with central conicoids in this unit, in the first section we shall consider some necessary general theory of conicoids which we have not covered in the previous unit. We shall define a centre of a conicoid and obtain a characterisation for central conicoids. Then we shall discuss the four different types of central conicoids in separate sections. We will end this unit with a discussion on sections obtained by intersecting a central conicoid by a line or a plane. In this connection we also discuss tangents to a central conicoid. In the next unit we will tackle non-central conicoids. That will be easier for you to grasp if you ensure that you have achieved the objectives given below.

Conicoids

Objectives After studying this unit'you should be able to : @

check whether a conicoid is central or not if you are given its equation;

obtain standard f m s of an ellipsoid, hyperboloid of one sheet and hyperboloid of two sheets; trace the standard forms of the above mentioned three conicoids; obtain tangent lines and tangent planes at a point to a central conicoid; check whether a plane is a tangent plane to a conicoid or not; use the fact that a planar section of a central conicoid is a conic.

2.2

A CONICOID'S CENTRE

In the last unit you saw that a point P is called a centre of a conicoid F (x, y, z) = 0 if its coordinates satisfy a system of linear equations (see Equations (18) of Unit 7). In this unit we define a centre geometrically and then see the relationship between the geometrically and analytical defections. Let us consider the conicoid S given by ax2 + by2 + cz3 + d = 0, abc # 0 Let P(x,, y,, z,) be a point on the conicoid. Then you can see that P (-x,, -y,, -z,) also lies on the conicoid. This means that S is symmetric about the origin 0. Because of this property 0 is called the centre of the conicoid (see Fig. 1).

Fig. 1 : T h e pair o f points P and P' a r e symmetric about t h e centre 0 of t h e conicoid.

Definition : A conicoid S is called symmetric with respect to a point P if, when the origin is shifted to P, the transfom~edco~?icoidS is symmetric with respect to the origin. Let us formally define a centre. To do so let us first see what we mean by symmetry.

Definition : A point P is called a centre of a conicoid S if S is symmetric with respect to P. Using the definition above we can easily see that the origin O(0, 0, 0) is a centre of the spehre. Let us consider an example now.

Example 1 : Show that the origin is the only centre of the cone ax2 + by2 + cz2 = 0, abc # 9 Solution : From the definition you can see that the origin is a centre of the sphere. NOW let us take another point (x!,, y,, z), which is non-zero. Sh~ftingthe origin to

(x,, yo, z,), we get the relationship.

where (x', y', z') denote the coordinates in the new system. Substituting the equations above in the equation of the cone, we get

This is the transformed equation of the cone. Because of the non-zero linear summand (inside brackets) of this equation, we see that (0, 0, 0) is the only point about which the transformed cone is symmetric. Hence the origin (0, 0, 0) is the only centre of the cone. ax2 + by2 + cz2 = 0, abc ;m 0. We will discuss conditions under which you can easily judge if a point is centre of conicoid or not. But first try these exercises.

El) Show that (- u, - v, - w) is the only centre of the sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 E2) Show that every point on the Z-axis is a centre of the cylinder x2 + y2 - z2 = 0 (see Fig. 2).

Fig. 2 : A cylinder with z-axis as axis.

Next, we shall prove a result which tells us something about the equation of a conicoid with the origin as a centre. Theorem 1 : The origin 0 is a centre of the conicoid ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 If and only if u = v = w = 0.

... (1)

Proof : Suppose that u = v = w = 0. Then the given equation of the conicoid takes the form ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy + d = 0. Let P(x,, y,, z,) be any point on the conicoid. Then we have

We can rewrite this equation is a(-x, j2 + b(-y1)* + c(-z,)~+ 2f(-Y,) (-2,) + 2g(-zl) (-XI) + 2h(-x,) (-YJ +d=0. This shows that (-x,, -y,, -z,) lies on the given conicoid. Hence, 0 is the centre of the conicoid. Conversely, suppose that 0 is a centre of the conicoid given by (1). Suppose P is and point (x,, y,, z,) lying on the conicoid. Then the point P (-x,, -y,, -z,) also lies on the conicoid, since 0 is the centre. Then we have ax12+ by,? + cz12+ 2fyIzI + 2gzIxI + 2hxIyI + 2u1xI + 2vIyI + 2w1z, + d = 0 ...( 2) and

Subtracting (3) from (2), we get ux, + vy, + wz, = 0 This shows that (x,, y,, z,) lies on the plane ux + vy + wz = 0. This is true for any point (x,, y,, z,) on the conicoid. But how can every point on a conicoid lie on the plane ux + vy + wz = O? This can happen only if u = v = w = 0. Hence. the result. Now, here is an exercise for you.

E3) Which of the following conicoids has a centre at the origin? a) x2 + y2 + z2 - 23x = 0. b) 2x2 + 3y2 - z2 = 1. c) 14x2 + 26y2 + 2' ,& z2 = 1. d) 41x2 - 28y2 = 0. In Unit 7 you saw that the existence of a centre is connected with the solvability of a system of equations. In the next theorem we establish the fact.

Theorem 2 : A conicoid S, given by the equation F(x, y, z) = ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0, has the point P (xo, yo, 2,) as a centre if and only if ax, + hy, + gz, + u = 0 hx, + by, + fz, + v = 0 gx, + fy, + CZ, + w = 0

Proof : Let us first assume that P(x,, yo, z,) is a centre of the given conicoid in the coordinate system XYZ. Now let us translate the origin from 0 to the centre P. Then from Unit 7 you know that the equation of the conicoid in the new coordinate system is given by

where u' = ax, + hy, + gz, + u V' = hx, + by, + fz, + v w' = gx, + fy, + cz, + W and d' = ax: + byo2 + czo2 + 2fyozo + 2gzoxo + 2hxoyo + 2ux0 + 2vy0 + 2wz, + d Now, this conicoid has a centre at the origin. Therefore, by theorem 1, we have u' = v' = w' = 0. This means that

ax, + hy, + gz, + u = 0 hx, + by, + fz, + v = 0 gx, 4- fy, + cz, + w = 0 Cunversely, let us assume that (4) holds for some point P(x,, yo, z,). We shift the drigin from 0 to P. Then we get an equation of the form (5). But since (4) holds for P, we see that uf = = Wf = 0. Therefore the equation reduce to a'x'2 + b'y'' + c'zf2 t 2fy'z' + 2gzfx' + 2hxfy' + d' = 0. This equation does not have any first degree terms. Therefore, by Theorem 2, we see that the origin P is a centre. Do you find any connection between Theorem 2 and Theorem 3? Yousmight have noticed that Theorem 2 is a special case of Theorem 3 in the case when the point P is the origin. Now let us go back to E l and E2. From there you know that a conicoid may have a unique centre of infinitely many centres. In Unit 7 you have also seen examples of conicoids which have no centre. Mathematicians have divided all conicoids up into two types depending on whether they have a unique centre or not. We define these conicoids as follows. Definition : A conicoid is called a central conicoid if it has a unique centre. A ~onicoidis called non-central, if it either has no centre or it has infinitely many centres. Thus, a sphere is an example of a central conicoid and a cylinder is an example of a non-central conicoid. Here is an exercise for you.

Ecl) Examine whether the conicoides given in E3 have a unique centre or not. We will look at non-central conicoids in the next unit. In this unit we will only discuss central conicoids. Let us now start this discussion.

2.3

CLASSIFICATION OF CENTRAL CONICOIDS

In this section we shall obtain the different forms of central conicoids. We shall also study the shape of these conicoids. Let us consider a central conicoid. Then, from Corollary 1 in Sec. 7.4 of Unit 7 you know that by first shifting the origin to the cenre and then rotating the axes suitably about the centre, we can reduce the equation to its standard form

Now, let's go back for a moment to Theorem 2. Over there you saw that a conicoid, F(x, y, z) = 0 has a unique centre iff the system of equations.

has a unique solution. This means that

Central Conieoids

a h

h b

g . f#O.

Now, if the conicoid is in the form (6), then the condition reduces to

10 0

abc # 0. This means that a # 0, b # 0, c # 0. Then, depending upon the signs of a, b, c and d, we have the following five possibilities: Case 1 (d = 0): In this case the equation reduces to ax2 + by2 + cz2 = 0. You know from Unit 6 that this represents a cone, irrespective of the signs of a, b and c. Case 2 (d # 0 and a, b, c, dare of the same sign) : In this case there are no real values of (x, y, z) which satisfy (6). This is because for any (x, y, z) €R3, the left hand side is either positive or negative, never zero. We call such a conicoid an imaginary conicoid. Infact it represents an imaginaq ellipsoid. Case 3 (d # 0 and the sign of the coefficients a, b, c are different from d): In this case we write (6) in the form ax2 + by2 + cz2 = -d,

d d d Note that the numbers - -, - - and - - are positive. a b c

This equation is the three-dimensional analogue of the equation of an ellipse. We call the conicoid represented by this- equation an ellipsoid. Case 4 (d # 0 and two of the four coefficient a, b, c and d are of the same sign). Let us assume that a > 0, b > 0, c < 0 and d < 0. d d d Then - -, - - and - are positive. We put a, = a b c b2 =

E1 {. and C2 =

Then (6) gives us

The conicoid generated by this equation is called a hyperboloid of one sheet. (You will see why in Sec. 8.5).

Similarly, we can obtain the equations of hyperboloids of one sheet in casesa < 0, b < 0, c > 0, d > 0, and so on. Case 5 (d ;t: 0 and two of a, b and c have the same sign as d) : As in the other d d d case we assume that a > 0, b < 0, c < 0 and d < 0. Then - - > 0,- > 0,- > 0. a b c

Put a3

i- .

b, =

\/$ and c, J:- .Then, we have =

The conicoid is called a hyperboloid of two sheets. (You will see why in Sec. 8.6.) The other forms of hyperboloids of two sheets can be similarly obtained. Thus, we saw that central conicoids can be classified into 5 types namely : cone, imaginary conicoid, ellipsoid, hyperboloid of one sheet and hyperboloid of two sheets. We have tabulated this fact in Table 1. Table 1: Standard Forms of Central Conicoids

'Ispe

Standard

Cone

ax2 + by2'+ cz2 = 0

Imaginary ellipsoid

x2 y2 z2 --+-+-=-I a2 b2 c2

Ellipsoid

x 2 y2 -+-+--=I a2 b2

z2 c2

Hypel-boloid of one sheet

-x2 + - - - =y2I a2 b2

z2 c2

b2 x2

z2 + 7 = l

. Y2

+-+-=I a b2 c2

Hyperboloid of two sheets

x2 a2

y2 b2

x 2 y2 --+---=1 a2 b2 x2 a2

z2 =1 c z2 c2

y2 z2 +-=I b2 c2

Why don't you do an exercise now?

E5) Identify the type of the conicoid from the following equations a) x2 + 4y2 - z2 = 1 d) z2 = 3x2 + 3y2 b) 16z2= 4x2 + y2,+ 16 e) x2 - y2 - z2 = 9 2

c) -x+ y 4

2 + - z= I 4

Conicoids

If you look closely at the cases where d # 0, you w ~ l lsee that the equations in these four cases can be given by a single equation ax2 + by2 + cz2 # 1 This will represent an ellipsoid if a, b, c are all positive;

ii) a hyperboloid of one sheet if two of a, b, c are positive and the third is negative; iii) a hyperboloid of two sheets if two out of a, b, c are negative and the third is positive;

iv) an imaginary conicoid if all the a, b, c are nepativc, We shall study the shapes of the real coni'coids listed above one by one. Let us start with ellipsoids.

2.4

ELLIPSOID

Let us consider the ellipsoid given by the equation

x 2 y 2 z2 ~ + ~ + 7 = where 1 , a, b. c > 0 a b c

...

Let S denote the surface generated by this equation. Form your experience of Unit 2, can you note some geometr~calproperties of S from the above equation? Of course, if a = b = c, then the equation represents a sphere. And you have studied the geometry of a sphere in deta~lIn Block 2. So, let us look at a more general case. From the following exercises you can get some idea of the geometrical aspects of the ellipsoid.

Show that the surface represented by (8) is symmetric about the YZ-plane, ZX-plane and XY-plane.

E7) Do all the coordinate planes, intersect the surface (8)? If so, find the sections obtained by the intersections.

ES) Check whether the coordinates axes intersect, the surface (8).

We say that a surface given by an equation F(x, y, z) = 0 is symmetric with respect to the XY-plane, if, when we replace z by z in F(x, y, z), we get the same equation. Symmetry with respect to the YZ-Plane and XZ-plane is similarly defined.

If you have done the exercises, you will have noticed that the surface (8) Intersects the coordinate axes in A (a, 0, 0) and A (-a, 0, O), 13 (0, b. 0) and R' (0, -b, 0), C(0, 0, c) and C' (0, 0, -c). Now let us consider the intersections of the surface by planes parallel to the coordinate planes. Here we assume that a # b. Let us first consider a plane parallel to the XY-plane, say z k, a constant.

Putting z = k in (a), we get

k If 7 < 1, i.e., - c < k < c, the equation represents an ellipse. C

This is true for all values of k such that 1 k I 5 c. If we put k = 0, we get an ellipse whose semi-axes are a and b. Thus, the surface can be considered as a family of ellipses placed one on top of another lying between the planes z = c and z = -c. Now what if k > c in (9)? The equation only has imaginary roots. Therefore no portion of the surface lies beyond the plane ] z j = c. Similarly, we can show that no part of the surface lies below z = -c. You can also check that the surface lies between the planes y = -b and y = b, as well as between the planes x = - a and x = a. Consequently, the surface is a bounded surface formed by ellipses.

Collecting all this information about (8), we get a figure as shown in Fig. 3.

Fig. 3 : T h e ellipsoid

z2 +c+- 1. Its intersection with the planes b2 c2 -

z = L a n d z = - 5 are the ellipses El and 2 2

E2.

We shall make some remarks here. Remark : Suppose we have a > b = c in (8). Then (8) can be rewritten as

The intersection of this ellipsoid with the plane z = 0 is the ellipse

If we revolve this ellipse about its major axis, i.e., the x-axis, then you can see that the surface formed is nothing but the given ellipsoid. This is the reason why the surface is known as an ellipsoid. Now, what can you say about the ellipsoid in the case when a = b > c in (8)? If a = b > c, the ellipsoid can be obtained by revolving the ellipse x2 z2 - + - = 1, y = 0 about its minor axis (i.e., the z-axis).

Let us consider an example now. x2 y2 2 Example 3 : Trace the conicoid -+ - + z = 1 9 16

...(10)

Solution : You know that the equation represents an ellipsoid. Let us try to trace it.

We first consider the intersection of the surface with the coordinate axes. From (10) we see that the x-axis intersects the surface in points (3, 0, 0) and (-3, 0, Oj, the y-axis intersects the surface in (0, 4, 0) and (0, -4, 0) and the z-axis intersects the surface in two points (0, 0, 1) and (0, 0, -1). Now let us consider the intersection of the ellipsoid with planes parallel to coordinate planes. Consider the plane z = h, a constant. Putting z = h in (10) we get

You know that this represents an ellipse for all ellipse for all h such that I h 1 < 1. If h = 1, we find that the intersection of surface wirh the plane z 1 is the point (0, 0, 1). Similarly for h = -1, we get the point (0, 0, -1).

If I h 1 > 1, then there is 110 (x, y) satisfying (1 1). This shows that no portion of t!le surface lies above the plane z = 1 and below the plane z = - 1. Now we draw the ellipse corresponding to z = 0 (E, in Fig. 4). Note that the major axis of the ellipse is 4 and the minor axis is 3. The figure shows some more ellipses corresponding to a few plancs parallel to z 0. Note that as h increases, the ellipses become smaller and smaller. ;

Likewise, if x = h, a constant, then again we get ellipses for I h I < 3. Since there is no (x, y) satisfying (10) for ( h ( > 3, we see that no portion of the surface lies to the right of the plane x = 3 and left of the plane x -3. For x = 0, we get the ellipse, as shown in Fig. 4.

Similarly, you can see that the intersection with the planes y = h are also ellipscs for ( h ( I 4. and no portion of the surface lies to the right of the plane y 4 and left of the plane y = - 4. The ellipse corresponding to y = 0 is shown in Fig. 4.

Having got the intersections with the coordinate planes and the coordinate axes, we obtain the ellipse -is shown in Fig. 4.

Fig. 4: T h e ellipses, El. E, and E, a r e t l ~ e iotersectio~~sol' ellipsoid x2 +K+z2 = 1 with the coordinate planes. z = 0. x = 0 and y = 0 respectively.

Why don't you trace an ellipsoid now?

E9) a) Trace the ellipsoid x'

+I 51

4 1 + r' = I

b) Check whether the ellipsbid iL (a) can be obtained by revolving an ellipse about any one of its axes.

So you have seen how to trace an ellipsoid in standard form. Actually, now you are in a position to trace any ellipsoid. How? Simply apply the transformations given in Sec. 1.3 of Unit 1, and reduce the given equation of ellipsoid to standard form! But we shall not go into such detail in this course. Let us now consider an application of ellipsoids. In Unit 2, you came across the reflecting property of an ellipse. This property is made use of in constiucting whispering galleries. Whispering gaIleries are galleries with a rectangular base and ceiling in the form of an ellipsoidal surface. Because any vertical cross section of the ceiling is elliptical, the sound produced at one focus will be reflected at the other focus with little loss of intensity. This is called the reflecting property of an ellipsoid. This property is used by architects. We shall now stop our discussion on ellipsoids and shift our attention to another central conicoid.

In this section we shall study the shape of a hyperboloid of one sheet in detail and study some of its geometrical properties. As in the case of ellipsoid, we shall restrict our attention to the standard forms. Let us consider the hyperboloid of one sheet given by

Do you agree that this equation is represented by the surface in Fig. 5?

x2 z2 Fig. 5 : T h e hyperboloid o f one sheet Y- - ,2- I ,2 +b2

You can obtain some geometrical properties of this surface by doing the following exercises. E10) Show that the surface formed by (12) is synunetric about the coordinate planes. Ell) Check whether the coordinate axes intersect the surface formed by (12). if so, what are the intersections'?

C e n t r a l Conicoids

Conieolds

The next point to check is the intersection of (1 1) with the coordinate planes. Its intersection with the plane z = 0 is

Which is an ellipse. In fact, its intersection with any plane z = h will be an ellipse (see Fig. 6), and the size of the ellipseincrease as h increases in both positive1 and negative directions. You may wonder why we don't call the surface an ellipsoid too.

Fig. 6 : Sections of

c2 - 1 obtained o f planes parallel to the XY-plane. + -a2 b2 a

But, now look at what happens when we intersect the surface with the YZ-plane. The intersection is given by

which represents a hyperbola (see Fig. 7)

x2 y2 z2 Fig. 7 : Intersection o f - + - - - = I . a2 b2 c2

b y the XY-plane.

Similarly, you can see tkpt the intersection of the surface with the plane y = 0 is the hyperbola

zL -= 1 (see Fig. 8).

~ h n t r aConieoidr ~

xZ Fig. 8 : Intersection of -+ a2

- -- 1. by the ZX-plane. c2 -

From the above properties you surely agree with us that Fig. 5 represents (12). Sometimes we say that the surface (12) is generated by the variable ellipse

which is parallel to the XY-plane and whose centre (0, 0, h) moves along the z-axis. This is because, as you can see, it is made up of these ellipses piled one top of the other. Now, why do you think (12) is called a hyperboloid of one sheet? Firstly, it is of one sheet because it i s a connected surface. This means that it is possible to travel from one point on it to any other point on it without leaving the surface. In the next section you will see that we also come across hyperboloids of two sheets. To see why it is called a hyperboloid see what happens if, for example, a = b in (12). Then the equation reduces to the form

If we put x = 0 in this, then we get the hyperbola

whose conjugate axis is the z-axis. If we revolve this hyperbola about its conjugate axis, then we get the hyperboloid (13). Similarly, we can obtain certain such hyperboloids by revolving a hyperbola about its transverse axes. So far we have been discussing only one standard form of a hyperboloid of one sheet. You know, from Table 1, that there are two more types of hyperboloid of one sheet. The following exercises are about them and the standard form (12). E12) Consider the hyperboloid of one sheet given by x2 + y2 - z2 = 1. a) What are its horizontal cross-sections for z = -+ 3 , f 6? b) What are the vertical cross-sections for x = 0 or y = O? Describe the sections in (a) and (b) geometrically. E13) a) Sketch the surface defined by the equation

Conicoids

b) Sketch the curves of intersection of the surface given in (a) by the plane z = k, when I k I = 3, when 1 k I < 3 and when I k I > 3.

E14) a) Obtain the surface defined by the equation

b) What curve is formed by intersecting it with the plane x = l ? Another interesting property of a hyperboloid of one sheet which makes it very useful in architecture is that it is a ruled surface. This means that the. surface is composed of straight lines, and is therefore easy to make with a string (see Fig. 9).

Fig. 9 : A model of a hyperboloid of one sheet.

Let us now discuss another type of hyperboloid.

2.6

HYPERBOLOID OF TWO SHEETS

In this section we shall concenrate on the geometrical features of a hyperboloid of two sheets. Its analytical properties are very similar to those of a hyperboloid of one sheet or an ellipsoid. So, it will be easy for you to bring out these properties yourself. Let us start with a hyperboloid of two sheets given by the equation

Note that this equation has two negative coefficients while the equation of a hyperboloid of one sheet has only 1 negative coefficient. So let us see what (14) looks like. To start with, why don't you try the following exercises concerning (14)?

E15) Discuss the symmetry of the surface obtained by (14) with respect to the coordinate planes. E16) Do all the coordinate axes and coordinate planes intersect the surface? Give reasons for your answer. In E l 6 you must have observed that the XZ-plane and XY-plane intersect the surface in hyperbolas. What about the YZ-plane? You must have observed that the YZ plane does not intersect the surface.

So, now you know why this surface is called a hyperboloid. But, you may wonder why this surface is called a hyperboloid of two sheets. This is because of the following property. Let us consider the intersection of the surface and the plane x = h, a constant. You can see that the curve of intersection is the ellipse

in the plane x = h.

This ellipse is real o111y if

h2 > 1, i.e., h > a or h < - a. a

Therefore, it follows that those planes which are parallel to x = 0 and lie between the planes x = -a and x = a do not cut the surface. This means that no portion of the surface lies between the planes x = - a and x = a. We note from (15) that the semi-axes of the ellipses are given by

and the semi-axes increase as h increases. Hence, we see that the surface has two branches: one on the left of the plane x = -a and one on the right of the plane x = a. Both these are generated by a variable ellipse. In fact, the shape of the surface is as shown in Fig. 10.

Fig. 10: The hyperboloid of two sheets

xz a2

z2 e2

In the following exercises we ask you to trace the other two forms of hyperboloid of two sheets:

E17) Check whether the coordinate planes intersect the surface

If so, what are the curves of intersection?

E18) Sketch the surfaces given in E17.

Central Conieoids

Conicoids

So, you have seen what ihe standard forms of the various central conicoids look like. Now let us see what their intersections with various lines and planes are.

2.7

INTERSECTION WITH A LINE OR A PLANE

In this section we shall first discuss the intersection of a line and a central conicoid. This will help us to derive conditions under which a line is a tangent to a central conicoid, and to obtain the tangent planes. Then we shall discuss the intersection of a plane and a central conicoid.

2.3.1

Line Intersection

Consider a central conicoid given by ax2 + by2 + cz2 = 1 Let us prove the following result about the intersection of given line with this conicoid.

Theorem 3 : Any line intersects a conicoid in two points, which tray be distinct feal points, coincident real points or imaginary points. Proof : Let L be a given line with direction ratios u,p, y that passes through a given point (x,, yo, 2,). Then the equation of the line L is

If B(x, y, z) is another point on the line L, which is at a distance r from A, then the coordinates of B are given by x = xo + a r ; y = y, -t pr, z = zo + y r. If B is a point of intersection of L with the conicoid (16) then its coordinates must satisfy (1 7). This means that (aa2 + bP2 + c?)

+ 2r(axoa + byoP + CZ,~)+ ax,'

+ by:

+ czO2- 1 = 0

-.(18)

(18) is a quadratic in r. So it will give us two values of r each of these values will give us a point of intersection of L with the conicoid (16). Thus, L, will meet (14) in two points, which may be real and distinct, coincident or imaginary. This is true for any line. Hence, Theorem 1 is hue for a central conicoid. Now let us suppose that the point A (x,, yo, 2,) lies on the conicoid (16) itself. Then ( 18) becomes.

The line L will be tangent to the conicoid at A if the points of intersection coincide with the point A (x,, yo, zo). i.e., if (19) has coincident roots. As you can see, the condition for this is

Thus, (20) gives the condition that the line

is a tangent to the conicoid ax2 + by2 + cz'

1 at A (x0, yo, z,),

Central Conicoids

x- yi For example, the line x = z, y = 4 is a tangent to the ellipsoid - + - + z' = 1 at 9 16 (0, 4, 0) because a'= 1 = y, p = 0. You can also check that the line through the point (0, 4, 0) and parallel to the x-axis is also tangent to this ellipsoid. In fact, there are infinitely many lines that are tangent to this ellipsoid at the point (0, 4, 0). This means that at each point of the ellipsoid we can draw infinitely many tangents to the conicoid. This is not only true for this ellipsoid. It is true for any conicoid. Let us see what the set of all tangents at a point of a conicoid look like. Let us eliminate a, p, y from (17) and (20). Then we get ax, (x - x,) + by, (y - y,) + cz, (z - z,) = 0. ax,x + by,y

+ cz,z - ax:

- by:

- cz:

= 0

e axx,+ b y y o + c z z o = 1, since (x,, y,, z,) lies on (17). (20) is the equation of a plane. Thus, the set of all tangent lines to (16) is the plane (21). Definition : The set of all tangent lines to a conicoid at a point on the conicoid is called the tangent plane. So, let us assume that (17) is a tangent of (16) at (x,, yo, z,). x2 Y 2 For example, if the conicoid is an ellipsoid - + -+ z 2 = 1, the equation of the 9 16 tangent plane at any point (0, 4, 0) on it is y = 4. Similarly, the equation of the tangent plane at the point (0, 4, 0) on the hyperboloid x2 + y2 - z 2 = 1 is y = 4. of one sheet 9 16 In Fig. 11, we have shown both these tangent planes.

Fig. 11: x is a tangent to (a) an ellipsoid (b) a hyperboloid.

Note that the tangent plane y = 4 intersects the given ellipsoid in only one point on the other hand, this plane intersects the given hyperboloid along the two tangent lines x = +3z, y = 4 at (0, 4, 0). The should not be surprising, since the plane is built up of tangent lines. Now, suppose we are given a plane and a conicoid. Can we say when the plane will be tangent to the conicoid? Let us see. Let us consider the plane given by

ux + vy + wz = p and the conicoid given by ax' + by2 + cz2 = I. You ki~owfrom (20) that the plane ux + vy + wz = p will be a tangent plane to the given conicoid at some point (x,, yo, 2,) if and only if its equation is of the form axx, + byyo + czz, = 1 ...(23) Therefore, if (22) represents a tangent plane, the coefficients of (22) and (23) must be proportional, i.e.,

(Remember that a # 0, b # 0 and c # 0.) Now, the point (x,, yo, z,) lies on the given conicoid. Therefore,

which is the required condition that a plane ux + vy + wz = p touches the conicoid ax2 + by2 + cz2 = 1. Also note that the point of contact of the plane and the conicoid is given by (24). Let us consider some examples.

Example 4 : Show that the plane 3x + 12y - 62 = 17 touches the hyperboloid 3x2 - 6y' + 9z2 + 17 = 0, and find the point of contact. Solution : We first rewrite the equation of the hyperboloid in the standard form as

The condition that 3x i- 12y - 62 = 17 touches the conicoid is given by (25).

Here a = (-$.b=-

6 17,c=

(-;I,

u = 3, v = 12, w = -6, p = 17. Then

w 2 9 x(-17) 144 x 17 3 6 x (-17) + -------+ c 3 6 9 = 17[-3 + 24 - 4 1 = 172 = p2.

u2 v2 -+-+-= a b

Thus the condition (25) is satisfied. Hence the plane touches the conicoid. The point of contact is given by

Example 5 : Find equations of the tangent planes to the conicoid 7x2 + 5y2 + 3z2 = 60 which pass through the line 7x + 10y - 30 = 0, 5y - 32 = 0.

Central Cot~icoids

Solution : From Block 1 you know that any plane through the given line is of the form 7x + 1Oy- 30 + A ( 5 y - 32) = 0, where h is a real number. Since this plane is a tangent plane to the given conicoid, we see that the equation of the plane must be of the form 3YYo "R + +- = I, for some (xo, yo, 1,). 60 60 60 Comparing the coefficients, we get

Since (x,, yo, z,) lies on the given conicoid, we get 7 x 4 + 5 ( 2 h + 4)2 + 12h2 = 60 a 28 + 20h2 + 80h-7 80 + 12h2 = 60 a 32h2 + 80h + 48 = 0 2h2 + 5h + 3 = 0. 3

This is a quadratic equation in 1. Its roots are -1 and - -. For each of these 2 values of h we get a tangent plane. Therefore, there are two tangent planes passing through the given line. 'The required equations of the planes are 7x + 3y + 32 = 30 and 14x + 5y + 9z =60. You can now try some exercises. E19) Find the equation of the tangent plane to the hyperboloid x2 + 3y2 - 3z2 = 1 at the point (1; -1, 1) E20) Find the equations of the tangent planes to the conicoid 2x2 - 6y2 + 3z2 = 5 which pass through the line x + 9y - 32 = 0, 3x - 3y + 62 - 5 = 0. So far we have been discussing the tangent planes to a central conicoid. You have found that sometimes such planes intersect the conicoid in a point, and sometimes in a pair of lines. Does this give you a clue to what 7~ and S will be where ~c is a plane and S is a central conicoid? We discuss this now.

8.7.2 Planar Intersections In this unit you have seen that the section of a standard ellipsoid or hyperboloid by a plane parallel to the coordinate planes is either an ellipse, a hyperbola or their degenerate cases, i.e., the section is a conic. What do you expect in the case of the section by any plane which is not parallel to the coordinate planes? Will it still be a conic? Let's see. Let us consider a central conicoid given by ax2 + by2 + cz2 = 1, abc # 0. We want to find the section of this conicoid by a plane ux + vy + wz = p. The following result tells us about this. (We shall not prove it here. If you are interested in knowing the proof, refer to the iniscellaneous exercises at the end of the block.) Theorem 4 : The section of a central conicoid by a given plane is a conic section.

Further, if the conicoid is given by ax2 + by2 + cz2 = 1 and the plane is given by ux + vy + wz = p, then the section will be a hyperbola, parabola or an ellipse according as

Conicoids

In case abc > 0, the condition reduces to u2 v2 w Z u2 v2 w 2 -+-+-<o,-+-+-=o,-+-+->o. a

u2 a

v2 b

w2 c

This theorem is not difficult to prove. We can obtain the condition by eliminating either x, y or z from the equations ax2 + by2 + cz' = 1 and ux + vy + wz = p. Since it is lengthy, we have not included it here. From Theorem 4, you know that a planar section of central conicoid need not be a central conic. In Fig. 12 we illustrate this.

Fig. 12: A Planar section of the hyperboloid

Y + E- -- = 1.

can be

a) ellipse b) a parabola.

Let us consider an example.

Example 6 : Show that the section of the hyperboloid 9x' + 6y' - 14z2 = 3 by the plane x + y + z = 1 is a hyperbola. 14 Solution: Here a = 3, b = 2, c = - - and u = v = w 3 Therefore, bcu2 + cav2 + abw2 = 2 x

(

14 + 6 < O .

Hence by Theorem 4, the section is a hyperbola. You can try some exercises now.

E21) Find the sections of the following conicoids by the plane given alongside. a) .2x2 + y2 - z2 = 1; 3x + 4y + 5z = 0. b) 3x2 + 3y2 + 6z2 = 10; x + y + z = 1. Let us now end our discussion on central conicoids by summarizing what we have covered in this unit.

2.8

SUMMARY

In this unit we have covered the following points : 1)

The definition of a centre of a conicoid.

The necessaj and sufficient condition for the conicoid to ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz +

d = 0 to have a centre is

la h 16

h b

bl f + 0. cl

Conicoids are divided into two groups - those with a unique centre (that is, central conicoids) and those which have either to centre or infinitely many centres (that is, non-central conicoids).

The standard form of a central conicoid is ax2 + by' + cz2 + d = 0, abc + 0. If d # 0, then there are four categories, as given in the table below:

Table 1: Standard Form of central conicoids Standard form Cone

ax2 + by2 + CZ* = 0

Imaginary ellipsoid

x2 ,J2 z2 -+-+-=-I. a 2 b2 c 2

Ellipsoid

x 2 y2 z2 -+-+--=-I. a 2 b2 c 2

Hyperboloid of one sheet

x2 yZ z2 -+---=I a 2 b2 c 2 x2 a2

z2 +-=I b2 c 2 Y2

- -.+x- -2+ - - =YI2 a 2 b2

Hyperboloid of two sheets

z2 c2

- - +x2 - - - = Iy2 a 2 b2 xz

z' c2 z2 +-=I c2

n d = 0, then the equation represents a cone.

How to trace the standard forms of an ellipsoid, hyperboloid of one sheet and hyperboloid of two sheets.

The condition for a line to be a tangent to the central conicoid ax" + d = 0 at (x,, yo, z,) is ax,a + byup + czoy 0, where a,p, y are the direction ratios of the line.

The equation of the tangent plane to a central conicoid ax2 + by2 + cz' = 1 at a point (x,, yo, z,) is axx, + byy,, czz, = 1.

The condition that the plane ux + vy + wz = p is a tangent to the central conicoid ax' + by' + cz2 = 1 is

A planar section of a central conicoid is a conic section

by' + cz2

Central Conicoids

Conicoids

Now you may like to go back to Sec. 3.1 to see if you have achieved the objeectives listed there. You must have solved the exercises in the unit as you came to them. In the next section we have given our answers to the exercises. You may like to have to look at them.

El) We first show that (-u, v, -w) is a centre of the sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 You can check that by shifting the origin to (-u, -v, -w), we get the transformed equation as xf2 + Y'2 + 2'2 = u2 + v2 + W* - d where x', y', z' denote the coordinates in the new system. Certainly (0, 0, 0) to (x,, yo, 2,). Then, we get the transformed equation as xf2 + Y'2 + 2'2 + 2xf (u + xn) + 2y' (v + yo) + 22' (w + 2,) + [2ux, + 2vy0 + 2wz0 + dl = 0 This is the transformed equation of the sphere. Since at least one of the u, v, w are different from (x,, yo, z), respectively, the linear summand is non-zero, therefore the equation is not symmetric with respect to' the origin (0, 0, 0). Hence, (-u, -v, -w) is the only centre of the sphere.

E2) Let us take any point (0, 0, z) on the z-axis. If we transform the origin to this point, we get the same equation is symmetric with respect to the origin. This shows that any point on the z-axis a centre of the sphere. E3)

a) Origin is not a centre. b) Origin is a centre. c) Origin is a centre. d) Origin is not a centre.

E4) a) The conicoid is x2 + y2 + z2 - 23x = 0 x, - 23 = 0, yo = 0 and z, = 0 a = 1, b = 1, c = 1, u = -23. So we get The system has a unique solution. Thus the conicoid has a unique centre. b) Unique.centre at (0, 0, 0) c) Unique Centre at (0, 0, 0) d) Infinitely many centres-every point on the z-axis.

E5) a) Hyperboloid of one sheet. b) Hyperboloid of one sheet. c) Ellipsoid. d j' Cone. e) Hyperboloid of two sheets.

E6) If we change x to -x, there is no change in the equation. This implies that the surface (8) is symmetric about the YZ-plane. Similarly, the surface is symmetric about the XZ-plane and XY-planes.

E7) Yes. All the coordinate planes intersect the surface. Let the equation of the ellipsoid be

The equation of the YZ-plane is x = 0. To find the intersection, we put x = 0 in the equation of the ellipsoid. Then we get

which represents an ellipse (or a circle).

Similarly, substituting y = 0 or z = 0, we get an ellipse (or a circle) in the XZplane or the XY-plane.

EB) The equation of the ellipsoid is (8). We first check whether the x-axis intersects the surface. Any point on the x-axis is of the form (r, 0, 0). So, to find the intersection of the x-axis, we substitute (r, 0 0) in (8). We get

r2 a. a2 Thus, the x-axis intersects the surface in the two points (a, 0, 0) and (-a, 0, 0). Similarly, the y-axis intersects it in two points (0, b, 0) and (0, --b, 0) and the z-axis intersects it in the points (0, 0, c) and (0, 0, -c). - = 1, i.e., r =

y2 E9) a) The given ellipsoid is x2 + - + z2 = 1. 4 We first find the intersection of the ellipsoid with the coordinate plane. Let z = h, a constant. Then, we et

If / h / < 1, this is an ellipse centred at the origin. Likewise if x = h, a constant, we get

This again represents an ellipse if / h I < 1. Putting y = h, we get circles given by

Let us now draw the ellipses and the circle for z = 0, x = 0, y = 0 respectively (see Fig. 13).

Fig. 13: The ellipsoid x2 +

+ z2 = 1.

The shaded portion in Fig. 13 shows the ellipsoid.

b) Suppose we revolve the circle x2 4. z' = 4 around the y-axis. We get the ellipsoid. xL yL zL ElOj Let the equation of the surface be - + - - - = 1 a 2 b2 c2 Changing x to -x, y to -y and z to -2, there is no changing in the equation. Therefore, the surface is symmetric about the XY, YZ and ZX planes.

Central Conicoids

Conicoids

El 1) Let the equation of the surface be

x2 y2 z2 - + -- - = 1 a 2 b' c2

Then, we see that the x-axis intersects the surface at the points (a, 0, 0) and (--a, 0, 0). Similarly, the y-axis intersects the surface at the points (0, b, 0) and (0, - b, 0) Next we put z = 0 in the given equation of the conicoid, to get z2 = - c'. This shows that the points of intersection are imaginary. That is, the z-axis does not intersect the surface. E12) a) The given equation is x2 + y2 - z2 = 1. For z = 3, k 6, the horizontal cross-sections are circles with centres on the z-axis and radius 1 and 2 respectively (see Fig. 14).

Fig. 14: Circles obtained b y intersecting the planes s = f 1, f 2 with t h e hyperboloid o f one sheet xz + y2 + z2 = 1.

b) When x = 0, the equation is y2 - z2 = 1, Which represents a hyperbola (similar to Fig. 7) in the plane x = 0. When y = 0, the equation is x' - z2 = 1 or x = f z, which again represents a hyperbola. E13) a) By renaming the axes, we can obtain the surface as shown in Fig. 15.

Central Conicoids

Fig. 15: The hyperboloid of one sheet --+ 16

zZ L+-= 1.

b) When ( k 1 = 3, we get

which represents a pair of lines. When I k I < 3, we get

which represents a hyperbola with transverse axis parallel to the y-axis. When ( k I > 3, we get k2 -y2 - - = 1 x2 -9 16 9 .1.e.---=--I x2 Y 2 k 2 16 9 9 which represents a hyperbola with transverse axis parallel to the x-axis. The cross-sections are shown in Fig. 15

Fig. 16: The I~yperboloid of one sheet

+-=I. 9

Conieoids

z2 Y2 b) The hyperbola - - - = 1 in the YZ-plane. 8 3219 E15) The surface is symmetric about the coordinate planes.

E l 9 The coordinate planes y = 0 and z = 0 intersects the surface in hyperbolas Y2 x2 - z2 = 1 and x2 - = 1 respectively.

The plane x = 0 does not intersect the surface. The x-axis meets the surface in points (a, 0, 0) and (-a, 0, 0). The y-axis and the z-axis do not meet the surface.

Eli') a) The XY-plane does not intersect the surface. The YZ-plane and the XZ-plane intersect the surface in hyperbolas

22 - y2 = 1 and z2 - - = 1.

b) The XZ-plane does not intersect the surface. The XY-plane and the YZ-plane intersect the surface in hyperbolas

and

x2 Fig. 17: The hyperboloid o f two sheets a) z2 - - - y2 4

b) q - T - - z l = 1.

E19) The equation of the tangent plane at (xo, yo, z,) is axxo + byy, + czzo = 1. The given equation is x2 + 3y2 - 3z2 = 1. So, here a = 1, b = 3, c = -3, x, = 1, yo =-1 and z, = 1. Thus the required plane is x - 3y - 32 = 1. EO) Any plane through the line x + 9y - 32 = 0, 3x x+9y-3z+h(3~-3y+6z-5)=0 where h is a real number.

3y + 62 - 5 = 0 is

Now, we know that this plane touches the given conicoid. Therefore, we must have 2u0

322 6YYo + A = 1 for some (x,, yo, zo). Then we get

Substituting this in the equation 2x2 - 6y2 + 3z2 = 5, we see that h = 1 and h = -1. Hence, there are two tangent planes given by 4x + 6y + 32 = 5 and 2x - 12y + 9z = 5. a) Here a = 2, b = 1, c = - I , u = 3, v = 4, w = 5. Then bcu' + cav2 + abw2 = - 9 - 32 + 50 > 0. Therefore the section is an ellipse. b) The section is an ellipse.

C e n t r a l Conicuids

UNIT 3 PARABOLOIDS Structure 3.1

Introduction Objectives

3.2

Standard Equation of a Paraboloid

3.3

Tracing Paraboloids

3.4

Intersection with a Line or a Plane

3.5

Sunnnary

3.6

Solutions/Answers

INTRODUCTION

3.1

In the previous unit we discussed central conicoids. In this unit, which is the last unit of this course, we look at non-central conicoids. You are already familiar with one type of non-central conicoid, namely, a cylinder. Here, we discuss another such surface, called a paraboloid.

We shall start this unit with a discussion on the standard forms of a paraboloid. You will see that paraboloids can be divided into two types : elliptic and hyperbolic paraboloids. In Sec. 3.3, we discuss the shapes of the two types of paraboloids. The last section contains a brief discussion on the intersection of a paraboloid with a line and intersection with plane. . Like central conicoids, paraboloid s are also used in various fields. The most commonly found paraboloidal surfaces are dish antennas, which most of us are familiar with. You can see some more applications in the unit. The way this unit unfolds is the same as the previous one, except for one difference. In this unit we assume that you have enough experiencr by now to bring out many properties of the surface by yourself. Accordingly, you will find that we have left most results to you to prove. Now, please go through the following list of objectives. If you achieve these, then you can be sure that you have grasped the contents of this unit.

Objectives After studying this unit you should be able to check whether a given equation of a conicoid represents an elliptical paraboloid or a hyperbolic paraboloid; trace the standard elliptic or hyperbolic paraboloid; obtain the tangent lines and tangent planes to a standard paraboloid.

3.2

STANDARD EQUATIONS OF A PARABOLOID

In this section we shall obtain the standard equations of a non-central conicoid. Then we shall define a paraboloid and discuss its standard equations.

To begin with, let us go back to Theorem 4 in Unit 7 for a moment: According to this theorem any second degree equation can be reduced to an equation of the form ...(1) ax2 + by2 + cz2 + 2ux + 2vy + 2wz + d = 0 Now let us assume that (1) represents a non-central conicoid. Since the conicoid has no centre, by Theorem 2 i n Unit 8 we find that either i) ii)

exactlytwoofthea, bVandcarezero,or only one of the a, b and c is zero.

Let us look at these cases separately. We first consider the case (i). Let us assume that a = 0, b = 0 and c # 0. (We can deal with the cases a, c = 0, b # 0; b, c = 0, a # 0 similarly.) In this case (i) becomes cz2 + 2ux + 2vy + 2wz + d = 0

): , we see that the equation takes the form

By shifting the origin to (0.0, -

cZ2 + 2uX + 2vY + do = 0, where X, Y, Z, are the coordinates in the new system. What does this equation represent? Let's see. If both u and v are zero, then the surface represents a pair of lines. If one of the coefficients u and v is non-zero, say v # 0 and u = 0, then you can see that the surface is built up of a series of parabolas along a line parallel to the x-axis. Thus, it is a parabolic cylinder. In fact, even if both u, v are non-zero, the surface is a parabolic cylinder. Let's now go to case (ii) Here we assume that a = 0 and b, c # 0. (We can deal with the other two cases b = 0, c, a # 0; and c = 0, a, b # 0 similarly.) In this case (1) reduces to the form by2 + cz2 + 2ux + 2vy + 2wz + d = 0,

v2 b

= - 2ux + d,, where dl = -+--d

By shifting the origin to

w2 c

(0 i- ): , we see that the, above equation takes the form 9

- -9

by2 + cZ2 + 2ux + d, = 0 where X, Y, Z are coordinates in the new system. For example, y2 + 10z2 = 2 and 2y2 + z2 = 12x represent non-central conicoids. But is there a difference in the type of conicoid represented by them? Let's see. cz2 + d = 0. Do you recognize Suppose that u = 0 in (2). In this case we get by" the surface given by this, equation? It represents a cylinder or a pair of planes. We have already discussed these surfaces in detail in Block 2. Now, let us assume that u # 0. Then we rewrite (2) in the form by2 + cz2 = -2ux - d i.e., by2 + cz' = 2u'

( 7 29 where u' X -

= - U.

Conieoids

Now, by translating the origin to the point

- 0,0 , the 'equation reduces to (,",I.

Do you agree that this equation is a three-dimensional version of the standard equation of a parabola? We call this surface a paraboloid. So, for example, the equation 2y2 + z2 = 12x represents a paraboloid. What are the other forms of an equation of a paraboloid? We leave thls as an exercise for you (see El).

El) Discuss what happens to (1) in the following cases: a) b = 0, a, c # 0. b) c = 0, a, b # 0. If you've done E l , you must have found that there are two more types of equations which represent a paraboloid, namely ...(4) ax2 + by2 = 2wz, and ax2 + cz2 = 2vy Now let us look at the coefficients of these equations more closely, as we did in the case of central conicoids. Let us consider (4). We have the following two cases: 't

Case 1 (a and b are of the same sign): Suppose a and b are positive. Let a, = & , and b, = h then (4) becomes

Similarly, if a and b are negative, we can write (4) in the above form. Thus, when a and b are of the same sign, (4) reduces to the form

The paraboloid represented by this equation is called an elliptic paraboloid. Case 2 (a and b are of opposite signs): In this case y w can see that (4) reduces to the form

The surface represented by this equation is called hyperbolic paraboloid. If you do E2, you will see why the adjectives 'elliptic' and 'hyperbolic' are appropriate.

E2) Show that the intersection with any plane parallel to the XY-plane of the paraboloid i) x2 + 2y2 = 32 is an ellipse. ii) 3x2 - y2 = 4x is a hyperbola.

E3) Check whether the following equations represent a paraboloid or not? for those that do, classify the paraboloids as elliptic or hyperbolic. a) 4y2 - 4z2 - 2x - 14y - 222 + 33 = 0 b) x2 + y2 + z2 - 2x + 4y = 1 c). 4x2 - y2 - z2 - 8x - 4y + 82 - 2 = 0 ' d) 9x2 + 4z2 - 36 = 0 e) 2x2 + 20y2 + 22x + 6y - 22 - 2 = 0

. I

From E2 you may have already realised that the two types of paraboloids are geometrically different. Let us now see whether there are more differences.

- -

3.3

TRACING PARABOLOIDS

In this section we shall discuss the geometry of the two types of paraboloids, and see how to trace their standard forms. We shall trace an elliptic paraboloid here and leave the tracing of a hyperbolic paraboloid as an exercise for you.

I 1

So, let us consider (S), the standard equation of an elliptic paraboloid. We can observe some geometrical poperties, similar to the properties you have seen in Unit 3 for an ellipsoid or hyperboloid. In E4 we have asked you to obtain them, using the knowledge you have gained in previous units.

E4) Check whether the surface (5) is symmetrical about the coordinate planes.

E5) Do all the coordinates axes intersect the surface (5)? If so, what are their intersections?

E6) Obtain the intersections of the surface (5) with the XZ-and YZ-planes. If you've done E6, then you must have realised why this surface is called a paraboloid You already know why the surface is called elliptic from E2. You may be more convinced about this fact if you look at the following property. Let us look at sections of the surface by the plane z = k, where k is a constant. It is given by

The left hand side of (7) is positive for all values of x and y. Therefore, w and k must be of the same sign. So, if w > 0, then k > 0. In this case (7) represents an ellipse (or circle if a = b) with centre at (0, 0, k) on the positive direction of the z-axis. Note that the size of the ellipse increases as k increases. If k = 0, the plane z = 0 just touches the surface at the point (0, 0, 0). If k < 0, the plane z = k does not intersect the surface. Therefore, no portion of the surface (5) lies below the plane z = 0. We have drawn the surface in Fig. l(a).

Fig. 1 : The elliptic Paraboloid

7 + - = 2wk in case (a) w > 0. (b) w < 0

Conicoids

Now what if w < 0 in (7)? Then k has to be negative. In this case we get an ellipse whose centre lie on the negative direction of the z-axis. As above, you can see that no portion of ( 5 ) lies above the plane z = 0. We have drawn the surface in Fig. l(b). Now you know geometrically why this surface is called an elliptic paraboloid. Why don't you try to trace some paraboloids on your own now?

Fig. 2: Antenna in the shape

of a circular paraboloid.

E7) Trace the paraboloid given by a) x2 + y2 = Z; and describe its sections obtained by the planes x = 0 and y = 0. b) y2 + 4z2 = x and describe its sections by the planes y = 0 and z = 0. The paraboloid you got in E7 (a) is called a circular paraboloid. There you can see that the plane section of the surface by the plane x = 0 is a parabola with focus at the point (0, 0, 114). When we revolve this parabola about the z-axis, we get the surface you have traced in E7 (a). Therefore, we also call this surface a paraboloid of revolution. Paraboloids of revolution have many applications. Circular paraboloids are used for dish antennas and antennas in radio telescopes (see Fig. 2). This is because of tlie property that of all the paraboloids having the same area, a circular paraboloid has he largest reflecting surface. Circular paraboloids are also used for satellite trackers and microwave radio links. Now let us consider the hyperbolic paraboloid given by (6), that is, x2

= 2wz.

As in the case of the elliptic paraboloid, we have two cases; w < 0 and w > 0. We shall restrict our discussion to w < 0. (Exactly similar properties hold for the case w > 0.) In this case we ask you to find out the properties of the following exercises.

E8) a) What are the properties of a hyperbolic paraboloid which are analogous to those obtained by you in E4 for an elliptic paraboloid? b) What are the sections of the paraboloid (6) with z = k, k < 0, and k > 0. In E8 (b) you must have obseived that the section of a hyperbolic paraboloid by the x2 y2 = 2wk. This hyperbola is real for all plane z = k (k # 0) is the hyperbola 7a b non-zero values of k (# 0), positive or negative. If k > 0, it will have its transverse axis parallel to the x-axis, and if k < 0, its transverse axis will be parallel to the y-axis. You can see one branch of the hyperbola in Fig. 3(a). In Fig. 3 (b) you can see the parabolas which are sections of the paraboloid by the planes x = 0 and y = 0. You can also observe that for k > 0, the length of the semi-transverse axis of the hyperbola is

J2ka,which increases as k increases. Similarly, for k < 0, the length

increases as 1 k / increases. You can now try these exercises.

x2 Fig. 3: T h e planar section o f the hyperbolic paraboloid - - - = y2 ~ W Zby the planes a2 b2 (a) z = 1 and z = -1, are the hyperbolas H, and H2 (b) x = 0 and y = 0, are the parabolas

P, and P2.

E9) Consider the hyperbolic paraboloid given by xZ - 2y2 = z a) What are its sections by the planes x = 0 and y = Or! b) What are its sections by the planes z = 0, _+ l ? c) Sketch the surface described by the given equation. E10) Sketch the surface given by the equation 2z2 - y2 = x.

So far we have discussed how to trace the standard paraboloids. For this purpose we considered their intersection with planes parallel to the coordinate planes. Now let us discuss the intersection of the paraboloid with a general plane, as well as with a line.

3.4

INTERSECTION WITH A LINE OR A PLANE

In this section we shall discuss some results for paraboloid similar to those obtained for central conicoids in Sec. 2.7 of Unit 2. We shall present some of these result in the form of exercises for you to do. Let us start with an exercise on the analogue of Theorem 1 in Unit 2 for the paraboloid given by ax'

+ by' -- 22,

Conicoids

Ell) Prove that a line intersects a paraboloid at"two ". points which may be real or imaginary. v

What can you say about the intersection of a line parallel to the z-axis with the elliptic paraboloid (8) (when a and b are of the same sign)? Let us go back for a moment to Fig. 1. There you can see that there is only one real point of intersection on the paraboloid. Let us see what happens to the intersections with the x- and yaxes. Again from Fig. 1 and E7, we observe that the lines just touch the paraboloid, that is, the points of intersection are coincident. You know that such lines are called tangent lines. As in the case of central conicoids, the set of all tangent plane lines at a point of the surface is a plane, called the tangent plane. You should be able to write the equation of the tangent plane from the corresponding results on central conicoids. In fact, this is what the following exercises are about. E12) Prove that the condition for a line with direction ratios a , P, y through the ljoint (x,, y,, z,) to be a tangent to the paraboloid (8) is ax,a + byoP - y = 0. E13) Prove that the equation of the tangent plane at a point (x,, yo, z,) on the paraboloid (8) is axx, + byy, = (z + 2,). E14) a) Prove that the plane ux + vy + wz = p will be a tangent plane to the paraboloid (8) iff

b) Obtain the point of contact.

If you have done E14, you know that the point of contact is that situation will be

Let us consider cn example Example 1: Show that the plane 8x - 6y - z - 5 = 0 touches the paraboloid x 2

y2 = 2 , and find the point of contact.

Solution: Let us check the condition given in (9). Here a = 1, b = --, u = 8, 3 v=-6, w=-1,p=5. Substituting in (9), we get 64 - 54 - 10 = 0, which is true. Therefore, the plane touches the paraboloid. The point of contact is (8, 9, 5). In the following exercises we ask you to apply what you have proved in E l 3 and E14. E15) Find. the,equation of tangent plane to the given conicoid at the indicated point a) x2 + y2 = 42, (2, -4, 5) b) x2 - 3y2 = z, (3, 2, -3). E16) Show that the plane 2x - 4y - z + 3 = 0 touches the paraboloid x2 - 2 ~ =* 32, and find the point o f , contact.

Let us now see what the intersection of a paraboloid with a general plane is. Consider the following theorem, which is analogous to Theorem 4 in Unit 2. We will not be doing its proof here, but leave it as an exercise for you to do if you are interested in proving it. (See Miscellaneous Exercises.)

Theorem 4 a) The section of a paraboloid ax2 + by2 = 22 by a plane ux + vy + wz = p is a conic. b)

If w = 0, the section is always a parabola.

If w # 0, then the section is i) a hyperbola if a and b are of opposite signs, ii) a parabola if at least one of a and b is zero, iii) an ellipse if a and b are of the same sign.

In Fig. 4 we have diagrammatically illustrated some particular cases.

Fig. 4: Planar section o f ax2 + by2 = 22 by the plane us + vy + wz = p when u, v. w, p # 0.

In Fig. 4 you can see the elliptical section. Why don't you try an exercise now?

E17) Sketch the sections of the following conicoids by a plane perpendicular to the XY-plane. a) 3x2 - y2 = ' z . b) 2x2 + y2 = Z. We shall end this unit now with a brief review of what we have covered in it.

9.5

SUMMARY

In this unit we have discussed the following points. 1)

The standard form of a non-central conicoid is ax2 + by2 + 2wz + d = 0. If w = 0, the equation represents a cylinder or a pair of straight lines. If w # 0, the surface represented by the equation is called a paraboloid.

The standard equation of a paraboloid is ax2 + by2 = 2wz, w # 0. There are two types of paraboloids. When a and b are of the same signs, we get an elliptic paraboloid. W h e n a anrl h are n f nnnnaite cionc

w e opt a hvnerhnlir narahnlnirl

ionicoids

How to trace an elliptic paraboloid and a hyperbolic paraboloid.

The condition for a line with direction ratios a, P, y to be a tangent to the central conicoid ax2 + by2 = 22 at (x,, yo, z,) is ax,a + by,P = y.

The equation of the tangent plane to the paraboloid ax' + by2 = 22 at a point (xi, yo, z,) is axx, + byy, = (z + 2,).

The condition that the plane ux + vy + wz = 0 is a tangent plane to the paraboloid ax2 + by2 = 22 is

The planar section of a paraboloid is a conic section.

Now you may like to go back to Sec. 3.1 to see if you've achieved the objectives listed there. You must have solved the exercises as you came to them in the unit. In the next section we have given our answers to the exercises. You may like to have a look at them.

3.6

SOLUTIONSIANSWERS

El) ,-a) Putting b = 0, a, c # 0 in (1) we get ax2 + cz2 + 2ux + 2vy + 2w2 + d 4 0.

= - 2vy + d,,

. ( -: -, 09 -

By shifting the origin to

u2 W 2 where d2 = - + - - d a c

, the above equation reduces' to the

f m

a x 2 + cZ2 + 2vY + d = 0. where X, Y, Z denote the coordinates in the new system. b) Similarly, putting c = 0, a, b # 0, in (1) we gct that the equation reduces to the form a x 2 + b y 2 + 2wZ + d = 0, where, X, Y, Z denote the coordinate in the system obtained by shifting the origin to

(- u u 0) -

ET) Any plane parallel to the XY-plane is of the form z = k, where k is a constant, k # 0. i) The given ellipsoid is x2 + 2y2 = 32. Putting z = k in this equation, we get

which represents an ellipse.

ii) Similarly, putting z = k in the equation of the hyperboloid, we get

E3) (a) and (e) represent a paraboloid. (b), (c) and (d) do not represent a parabolo~d.(a) represents a hyperbolic paraboloid, whereas (e) represents an elliptic paraboloid. E!2) Surface (5) is symmetric about YZ and ZX-planes. It is not symmetric about the XY-plane. ~ 5 ) Yes. ' When we put y = 0 and z = 0 in (5), we see that x = 0. Thus, (0, 0, 0) is the only point of intersection of the x-axis with (5). Similarly, we can show, that (0, 0, 0) is the only point of intersection with the y and z-axes.

E6)

a) Putting z = 0 in (5), we get

The only point which satisfies the above equation is (0, 0, 0). Therefore the XY-plane intersects the surface in the point (0, 0,'O). b) Putting y = 0 in (5) we get I;

i.e., x2 = 2a2wz, which represents a parabola. c) Similarly, the YZ-plane intersects the surface in a parabola.

Fig. 5: T h e parabolas E , and E, a r e the sections obtained by intersecting the elliptic paraboloid w i t h the planes x

0 and y

0, respcctivcly.

Fig. 6: F , a n d FI a r c thc sections obtained by intcrsccting the elliptic paraboloid w i t h thc plancs y = 0 and z = 0, respcctivcly.

Conicoids

E8) The surface has the following properties.

0 It is symmetrical about the XZ-plane and the YZ-plane. ii) The coordinate axes intersects the surface in the point (0, 0, 0).

iii) By putting z = 0 in (6), we see that the XY-plane intersects the surface in two lines.

b Y=fax. Similarly, by putting x = 0 in (6), we get y2 = - 2b2wz, which represents a parabola, i.e., the intersection with the YZ-plane is a parabola. The intersection of (6) with the ZX-plane also is a parabola given by the equation x2 = 2a2wz.

~-.

b) Putting z = k in (6), we get

When k < 0 and k > 0, this represents a hyperbola. b

When k = 0, this represents a pair of lines y = f - x . a

E9) a) The section by the plane x = 0 is the parabola I 2 The section by the plane y = 0 is the parabola x2 = z (see Fig. 7) yZ = - - z (see Fig. 7)

b) The section by the plane z = 0 is the pair of lines

The section by the plane z = 1 is the hyperbola x2 - 2y2 = 1. and the section by the plane z = -1 is the kL~erSola 2y2 - x2 = 1. This shows that the set of all tangent lines i.e., tangent plane is given by the equation. axx, + byy, = z + z,.

E10) The figure is similar to the figure in E9 with a change of the coordinate axes. Ell) The equation of a line through (xo, yo, z,) with direction cosines a, p, y is

x-xo --=y-yo --

z-z,

Any point on this line is of the form (ar + x,, pr + yo, yr + z,) for some r. When this line meets the paraboloid ax2 + by2 = 22, we have a(ar + x,)~+ b(pr + yo)' = 2 a (yr + z,), ...(10) i.e., (aa2 + bp2)s + 2r(aaxo + bpy, - y) + ax: + by: - 22, = 0

I i

This is a quadratic equation in r which gives two values of r, which may be real or imaginary. Hence the result. E12) The equation of the line L passing through (x,, yo, zo), having direction ratios a , P, Y is x-x, -. y-Yo z--0 ---a P Y In E13 we saw that this line L meets the conicoid in two points, which may be real and distinct, real and coincident or imaginary. If the line is a tangent to the conicoid at the point (x,, yo, z,) then the points of intersection conicide. That is, (0), in (El 1) has real coincident roots. The condition for this is a m , + bay, - y = 0. Note that since (x,, yo, z,) lies on the conicoid, we have ax:

E13) The conicoid is ax2 + by2 = 2z

+ by:

...( 11) 22, = 0.

...(12)

We know that the tangent plane at (x,, yo, 2,) is the set of all tangent lines at (x,, yo, zo). Let us assume that the line

where a,p, y are the direction ratios of the line, is a tangent to the conicoid (11). Eliminating a,8, y between (1 1) and ( l2), we get axo (X - x,) + byo (Y - yo) - (z - z,) = 0 axxO+ byyo - (axo2 + byo2) = z - 2, hxxo + byy, - 22, = z - z,, since ax: + byo2 = 22, axx, + byyo = z + 2,. E14) By E13, the plane ux + vy + wz = p will be a tangent plane to the paraboloid ax2 + by2 = 22 if it is.of the form axx, + byyo = z + z, for some point (x,, yo, z,) on the paraboloid. That means, we have ax0 by0 -=-=-=u

'-1 w

Since (x,, yo, z,,) lies on the conicoid ax2 + by2 = 22, we get

The point of contact is given by

(z. - w -P) -v,

E15) Equation of the tangent plane is axx, + byyo = z + 2, a) The given equation can be written as

1 2 Then, we have from (14)

So, in this case a = - = b and xo = 2, yo = - 4 and z , = 5.

E16) T'he given conicoid can be rewritten as

The given plane is 2x - 4y - z = -3. The condition that this plane touches the paraboloid is given by (13) in E14. In this case u = 2, v = -4, w = -1, 2 4 p = -3, a = - b = - c. Then, we have 3' 3

This shows that the plane touches the paraboloid. The point of Gontact is given by (3, 3, -3).

Fig. 8: F is the section of the paraboloid 3x2 - y' = z by a plane perpendicular to the XY-plane.

Pi#. 9: F is the section o f the paraboloid 2xz + yz = z by a plane perpendicular to the XY-plane.

MISCELLANEOUS EXERCISES (This section is optional)

In this section we have gathered some problems related to the contents of this block. You may like to do them to get a better understanding of these contents. Our solutions to the questions follow the list of problems, in case you'd like to countercheck your answers. 1)

Which type of conicoids do the following equations represent? i) x2-16z2=4y2 ii) 5x2 + 2y2 - 6z2 = 10 iii) 4y2 = x iv) x2 + 4 y 2 + 16z2 = 12 V) 2y2 + x2 = 42 vi) 4x2 - 3y2 - 6z2 = 10 vii) x2 + Y2 + z2 = 4 viii) 222 + x = y2 K) 4y2 + 9x2 - 36z2 + 36 = 0 25x2 - 9y2 = 225 x) z2 y2 a) The hyperbola - - - = 1 (a > b) is rotated about the z-axis. What is a2 b2 the surface formed in this situation? Obtain the equation of the surface. b) Find the value of x, such that the plane x = xo intersects the surface

obtained in (a) in a pair of straight dines.

a) A normal at a point P of a conicoid is the line through P which is perpendicular to the tangent plane at that point. Find the equation of a normal to a central conicoid ax2 + by2 + cz2 = 1 at a point (x,, yo, 2,).

b) Using (a) obtain a normal at the point

( 2 1 -

to the ellipsoid

Find the equation of a normal to a paraboloid ax2 + by2 = 22 at any point (x,, Yo, 2,) on it.

Suppose that the XYZ-coordinate system is transformed into another coordinate system with the same origin and with the coordinate axes having direction cosines

(J Z ' JZ (L JZ'Ji

0) and (0, 0, 1) with respect to the old ,0) system. What does the equation xy = z represents in the new system?

Find the equations to the tangent plane of the given surfaces at the indicated points

Prove that the section of a central conicoid by a given plane is a conic section. Further, if the conicoid is ax2 + by2 + cz2 = 1 and the plane is ux + vy + wz = p, then prove that the section will be

u2 v2 w2 i) an ellipse if - + - + -- > 0 . a b c u2 v2 w2 ii) a hyperbola if - + - + -< 0 a b c u2 v2 w2 iii) a parabola if - + - + -= 0 . a b c

a) Prove that the section of a paraboloid ax2 + by2 = 22 by a plane ux + vy + wz = 0 is a conic section. b) If w = 0, show that the section is a always a parabola. c) E w # 0, then show that the section is i) a hyperbola if a and b are of opposite signs ii) a parabola if at least one of a and b is zero iii) an ellipse if a and b are of the same sign.

a) Show that the intersection of the plane y = 2 and the ellipsoid x2 9

z2 16

- + - + - = 1 is an ellipse.

b) Find the lengths of the semi-major axis and semi-minor axis, the coordinates of the centre, and the coordinates of the focii of the ellipse obtained in (a). x2 y2 z2 10) A tangent plane to the ellipsoid - + - + - = 1 meets the coordinate axes a 2 b2 c2 in the points A, B, C. Prove that the centroid of the triangle ABC lies on the surface z2 x2 = 9 . (A centroid of a triangle is the point of intersection of +a' b2 c2 the medians of the triangle.) -+

Answers 1)

i) Cone ii) hyperboloid of one sheet iii) cylinder iv) ellipsoid v) elliptic paraboloid vi) hyperboloid of two sheets vii) sphere viii) hyperbolic paraboloid ix) hyperboloid of two sheets x) cylinder.

a) The surface formed is a hyperboloid of one sheet. It's equation is given by

b) Putting z = z, in the above equation, we get

The equation represents a pair of straight lines only if z, f a. 3)

a) The equation of the tangent plane at (x,, yo, z,) is axx, + byy, + czz, = 1 where ax,, by, and xz, are the direction ratios of the normal to the plane. This means that the direction ratios of any line perpendicular to the plane are ax,, by, and cz,. Since a normal to a conicoid at (x,, yo, z,) is a line perpendicular to the tangent plane and passes through (x,, yo, z,), its equation is given by

Conicoids

b) The equation of the normal to the ellipsoid

Then, we have x - 1 - y-1-- z - l l f i 114 114 1

JZ This is the equation of a normal at the point

X--Xo --Y - Y o - z-zo by, -1

The equations of the transformations are

Substituting for x, y and z in the equation xy = z, we get

xv2- -+-=zI yt2 i.e., 2 2 which represents a hyperbolic paraboloid. 6)

i) The equation represents an ellipse. Therefore the tangent plane is x + 2y + 2z = 5. ii) The equation represents an elliptic paraboloid. The equation of the tangent plane is 3~ - 2y - 32 = 4. iii) The equation represents an hyperboloid of one sheet.

The equation of the tangent plane is x + 2y - 22 = 0. 7)

Let the equation of the central conicoid be ax2 + by2 + cz2 = 1, abc + 0.

--..

Suppose that the plane ux + vy + wz - p = 0 (u # 0) intersects the conicoid..

Substituting for x in the given equation of the conicoid, we get

This is a general second degree equatien and therefore represents a conic section. Hence the result. Let us now find the nature of the conic section, i) You know from Block 1 Unit 3 that the above equation represents an ellipse if

since abc # 0, we can divide throughout the left hand side by abc. Then we get u2 a

v2 b

-+-+->o.

w2 c .

ii) Similarly, we can show that the section will be an hyperbola if u2 a

v2 b

-+-+-<o.

w2 c

iii) The section will be a parabola if

-u2+ - +v2 - = a w2 a

a) The given paraboloid is ax2 + by2 = 22. The given plane is ux + vy + wz = p. Now ux + vy + wz = p wz = - 4x - vy + p. Substituting for wz in the given equation of the conicoid, we get ax2 + by2 + 2(ux + vy - p) = 0 i.e., wax2 + wby2 + 2ux + 2vy - 2p = 0. This is a general second degree equation and therefore represents a conic section. The section will be a hypcrbo!a, a parabola or an ellipse according as w2ab $ 0, w'ab = 0, w2ab > 0 respectively. b) If w = 0, we get w2ab = 0 and @erefore in the case the section is always a parabola. c) If w # 0, then the condition in (a) reduces to ab < 0, ab = 0 or ab > 0

Conicoids

Thus, the section in this case will be i) a hyperbola if ab < 0 i.e., a and b are of opposite signs. ii) a parabola if ab = 0 i.e, at least one of a and b is zero. iii) an ellipse if ab > 0, i.e., a and b are of the same sign. 9)

a) Substituting for y = 2 in the given equation of the ellipsoid, we get

This represents an ellipse. b) The length of the semi-major aixs = J80/9. The length of the semi-minor axis = (0, 0) is the centre.

The focii are gfven by (0, *c) where c = it., ( 9

$ &) and (0, - JT)

10) Suppose ux + vy + wz = p is a tangent plane to the given ellipsoid

Then, we have the relation p2 = a2u2 + b2v2 + C2

Now, we find the intersection of the plane and the coordinate axes. The plane meets the x-axis, y-axis and z-axis at the point.

(t: b

A -, 0 , 0 , B 0. -, 0 , C 0,0, - respectively.

Let (x,, yo, z,) be the centroid of the AABC. Then

Substituting for u, v and w in (*) we get

This shows that the centroid (x,, yo, zo) lies on the surface a 2 b2 c2 represented by the equation -+ 7 + - = 9. x2 y- z2