Iit mathematics 2

12. MAXIMA & MINIMA (1)

Basic definitions: (i) Stationary points: For a given function f(x), stationary points are those locations of x (let it be 'x0') for which f'(x0) = 0, and functioning value f(x0) is termed as the stationary value of function. (ii) Critical points: For a given function f(x), critical points are those locations of x(let it be 'x') in the domain of f(x) for which f'(x0) = 0 or f'(x0) doesn't exist.

Note: At critical points for a function, local maxima, local minima

or point of inflection may

exist (iii) Points of extremum: critical points of a function at which either maxima or minima exists is termed as points of extremum. (iv) Point of inflection: critical locations of function at which neither maxima nor minima exists is termed as point of in flection, it is also known as point of no extremum. (v) Local maxima/regional maxima: A function f(x) is said to have a local maximum at critical point x= a if f(a)  f(x)  x  (a  h,a + h) , where h is a very small positive arbitrary number.

If x = a is the point of local maxima, then f(a)  lim f(x) and f(a)  lim f(x) . xa

xa

following figure illustrates the point of local maxima in different situations:

(vi) Local minima/regional minima: A function f(x) is said to have local minimum at critical point x = a if f(a)  f(x)  x  (a  h, a + h), where h is a very small positive arbitrary number.

If x = a is the point of local minima, if then f(a)  lim f(x) and f(a)  lim f(x) .   xa

xa

following figure illustrates the point of local minima in different situations:

ma r a h S Note: .

82 K .

6 L 7 7 in the (i) local maxima of a function at x = a is the largest value function 8nearby 5 Er.of 8 02only 1 0 neighbourhood of critical point x = a.

9 8015 (ii) local minima of a function at x = a is the smallest value of function 39 only in the nearby 8 neighbourhood of critical point x = a.

Mathematics Concept Note IIT-JEE/ISI/CMI

Page 97

(iii) If a function f(x) is defined on [a,b], then f(x) is having point of extremum at the boundary points. (iv)for function f(x), if local maxima on local minima doesn't exist at critical point x = a, then point of inflection exists.

Following figure illustrates the point of inflection or point of no extremum

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Test for Maxima/Minima 1. Ist derivative Test First derivative test is applicable only in those function for which the critical points are the points of continuity. (i) If f '(x) changes sign from negative to positive while passing through x = a from left to right then x = a is a point of local minima. (ii) If f '(x) changes sign from positive to negative while passing through x = a from left to right then x = a is a point of local maxima. (iii) If f '(x) does not changes its sign about x = a then x = a is neither a point of maxima nor minima. (i.e. x=a is point of inflection) following figure illustrates the Ist derivative test

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 98

2. IInd derivative Test If f(x) is continuous function in the neighborhood of x = 0 such that f'(x) = 0 and f''(a) exists then we can predict maxima or minima at x = 0 by examining the sign of f''(a) (i) If f''(a) > 0 then x = a is a point of local minima. (ii) If f''(a) < 0 then x = a is a point of local maxima. (iii) If f''(a) = 0 then second derivative test does not give any use conclusive results.

3. nth derivative test Let f(x) be function such that f'(a) = f''(a) = f''(a) = ............. = f n1 (a) = 0 and n th f n (a) = 0, where f denotes the n derivative, then

(i) If n is even and f n (a) < 0, there is a local maximum at a, white if f n (a) > 0, there is a local minimum at a. (ii) If n is odd, there is no extremum at the point a.

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Maxima and Minima of Parametrically defined Functions Let a function y = f(x) be defined parametrically as y = g(t) and x = h(t) where g and h both are twice differentiable functions for a certain interval of t and h'(t)  0.

Now,

dy dy / dt g'(t)    0  g'(t)  0 dx dx / dt h'(t)

Let one of the root of g'(t) = 0 is say t0. d2 y d  g'(t)  d  g'(t)  1 h'(t)  g'(t)h "(t)      .

Now 2 3 dx  g'(t)  dt  h'(t)  dx / dt dx {h'(t)} 

d2 y g "(t)  because g'(t) = 0. Here {h'(t)}2 > 0 , 2 dx {h '(t)}2

Thus sign of

d2 y is same as that of the sign of g"(t) dx2

Now, if  g "(t) t  t0  0, f(x) has a maxm. at x = x0 = h(t0) if  g "(t) t  t0  0, f(x) has a minm. at x = x0 = h(t0) and so on.

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Global Maxima/Minima (or Absolute Maxima/Minima) Global maximum or minimum value of f(x)  x  [a, b] refers to the greatest value and least value of f(x), mathematically (i) If f(c)  f(x) for  x  [a, b] then f(c) is called global maximum or absolute maximum value of f(x).

ma r a h S .

82 K .

6 L 7 Note: Er. 81027 5058 For function f(s) defined on [a,b], is c , c , c , ...... c are 9 the locations 01of critical point, 8 9 then 83 Global maximum value = max{f(a), f(c ), f(c ), ...... f(c ), f(b)} (ii) Similarly if f(d)  f(x)  x  [a,b] then f(d) is called global minimum or absolute minimum value.

1

1

2

3

2

n

n

Global minimum value = min{f(a), f(c1), f(c2), ...... f(cn), f(b)} Mathematics Concept Note IIT-JEE/ISI/CMI

Page 99

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Useful Formulae of Mensuration to Remember : 1.

Volume of a cuboid =  bh.

2.

Surface area of cuboid = 2(  b + bh + h  ).

3.

Volume of cube = a3

4.

Surface area of cube = 6a2

5.

Volume of a cone =

6.

Curved surface area of cone =  r  (  = slant height)

7.

Curved surface of a cylinder = 2  rh.

8.

Total surface of a cylinder = 2  rh + 2  r2.

9.

Volume of a sphere =

10.

Surface area of a sphere = 4  r2.

11.

Area of a circular sector =

12.

Volume of a prism = (area of the base)  (height).

13.

Lateral surface of a prism = (perimeter of the base)  (height)

14.

Total surface of a prism = (lateral surface) + 2(area of the base)

15.

Volume of a pyramid =

16.

Curved surface of a pyramid =

1 r2h.

3 

4  r3.

3

1 2 r , when  is in radians.

2 

1 (area of the base)  (height).

3 1 (perimeter of the base)  (slant height) 2

Curvature of function For continuous function f(x), If f''(x0) = 0 or doesnot exist at points where f'(x0) exists and if f''(x) changes sign when passing through x = x0 , then x0 is called a point of inflection . At the point of inflection the curve changes its concavity i.e. (i) If f''(x) < 0, x  (a,b) then the curve y = f(x) is convex in (a,b)

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839

(ii) If f''(x) > 0, x  (a,b) then the curve y = f(x) is concave in (a,b).

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 100

18. INDEFINITE INTEGRAL If f and g are functions of x such that g'(x) = f(x), then

 f(x)dx = g(x) + c

d  dx {g(x) + c} = f(x), where c is called the constant of integration.

Note: (i) Integral of a function is also termed as primitive or antiderivative (ii) The indefinite integral of a function geometrically represents a family of curves having parallel tangents at their points of intersection with the line orthogonal to the axis of variable of integration.

(1)

Theorems on Integration (i)

c

(ii)

 (f(x) g(x))dx

(iii)

d dx

(iv)  (v)

(2)

f(x).d x = c  f(x ).d x

 f(x)dx   g(x)dx

=

  f(x)dx   f(x)

d (f(x)).dx  f(x)  c . dx

 f(x)dx  g(x)+c



 f(ax+b)dx 

g(ax+b) +c a

Standard Formula :

2

xn1

(i)

 x dx  n  1  C , n  1

(ii)

 x dx  ln |x| + C

(iii)

 e dx  e

1

x

x

C

lna + C

(iv)  ax dx  ax/

( a x  b ) n 1 + c, n  a ( n 1)

(v)  (ax+b)n dx = (vi) 

dx 1 = a ax+b

(vii)  eax+bdx =

ln |(ax + b)| + c 1 ax + b e +c a

1

(viii)  apx+qdx = p (ix)

1

 sin(ax+b)dx

apx  q px+q a / na

lna + c; a > 0

=

1

 a cos (ax + b) + c

Mathematics Concept Note IIT-JEE/ISI/CMI

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Page 101

(x)

1

 cos(ax+b)dx = a

sin (ax + b) + c

(xi)  tan(ax+b)dx =

1 a

ln |sec (ax + b)| + c

(xii)  cot(ax+b)dx =

1 a

ln |sin(ax + b)| + c 1 tan(ax + b) + c a

(xiii)  sec2(ax+b)dx =

(xiv)  cosec2 (ax+b)dx =



1 cot(ax + b) + c a 1 sec (ax + b) + c a

(xv)  sec(ax+b).tan(ax+b)dx =

(xvi)  cosec(ax+b).cot(ax+b)dx = (xvii)  sec x dx =

(xviii)  cosec x dx =

dx

(xx)  (xxi) 

dx 2

=

2

x x a

(xxiii)  (xxiv) 

dx 2

ln |tan 2x |+c or  ln|(cosec x + cot x)| +c

dx a  x2 dx 2

2

x a

1 x sec–1 +c a a

x2  a2  |+ C

=

ln | x

x2 a2 

dx

2

x +c a

ln | x

x a

x2  a2

ln |(sec x – tan x)| + c

=

2

ln |tan  4  2x  | + c

1 x tan–1 +c a a

=

a2  x2

(xxii) 

(xxv) 

= sin–1

a2 x2

or

ln |(cosec x  cot x)| + c

or dx

1 cosec (ax + b) + c a

ln |(sec x + tan x)| + c or –

(xix) 



 

 

=

1 2a

ln aaxx

+c

=

1 2a

ln xx aa

+c

x (xxvi)  a2  x2 dx = 2 x (xxvii)  x2  a2 dx = 2

x (xxviii)  x2 a2 dx = 2

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

|+C

ma r a h S .

82 K .  x x a  6 L 7 ln  a  + E c r.

27 058 0 1   98 8015 839  x  x a 

x a2 –1 a2  x2 + 2 sin a + c

a2 x2  a2 + 2

2

2

2

x a

 a2

2

ln  

2

a

2

2

 + c 

Page 102

(xxix)  eax.sin bx dx =

eax 2

a b

(xxx)  eax.cos bx dx =

(3)

2

 b cos bx) + c or

eax

 b  sin  bx  tan1     C  a  a b  2

2

eax

eax 2

(a sin bx

2

a b

(a cos bx + b sin bx) + c or

  b  cos  bx  tan1     C  a  a b  2

2

Integration by Subsitutions If x =  (t) is substituted in a integral, then  

everywhere x is replaced in terms of t. dx gets converted in terms of dt.



 (t) should be able to take all possible value that x can take.

Note: (i)

 [f(x)n f'(x) dx =

(ii)



(iii)



f '(x) dx  f(x) f '(x) f(x)

(f(x))n+1 +C n+1

ln|f(x)| + C

dx 2 f(x)  C

(iv)  f(x).f '(x)dx  2 f(x) f(x)  C 3

(4)

Integration by Part : If f(x) and g(x) are the first and second function respectively then

 f(x).g(x)dx  f(x). g(x)dx   f '(x). g(x)dx dx 

The choice of f(x) and g(x) is decided by ILATE rule. The function which comes later is taken as second function (integral function).

I  Inverse function L  Logarithmic function A  Algebraic function T  Trigonometric function E  Exponential function

Note:

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(i)

 ex [f(x) f '(x)]dx ex.f(x) c

(ii)

 [f(x) xf '(x)]dx  xf(x) c

ma r a h S .

82 K .

6 L 7 r. 1027 058 E f(x) 15function of a 98 algebraic g(x) defines a rational 0 8 839

Integration of Rational Algebraic Functions by using Partial Fractions : PARTIAL FRACTIONS : If f(x) and g(x) are two polynomials, then rational function of x.

Mathematics Concept Note IIT-JEE/ISI/CMI

Page 103

f(x) If degree of f(x) < degree of g(x), then g(x) is called a proper rational function. f(x) If degree of f(x)  degree of g(x then g(x) is called an improper rational function

f(x) If g(x) is an improper rational function, is divided f(x), by g(x) so that the rational function f(x) (x) g(x) is expressed in the form  (x) + (g)x where  (x) and  (x) are polynomials such that f(x) the degree of  (x) is less than that of g(x). Thus, g(x) is expressible as the sum of a

polynomial and a proper rational function. f(x) Any proper rational function g(x) can be expressed as the sum of rational functions, each

having a simple factor of g(x). Each such fraction is called a partial fraction and the process f(x) of obtaining them is called the resolution or decomposition of g(x) into partial fractions. f(x) The resolution of g(x) into partial fractions depends mainly upon the nature of the factors

of g(x) as discussed below.

CASE I When denominator is expressible as the product of non-repeating linear factors.

Let g(x) = (x  a1) (x  a2)..........(x  an). Then, we assume that A1 A2 A2 f(x) = + +..... + x  a x  a x an g(x) 1 2

where A1, A2, ...... An are constants and can be determined by equating the numerator on R.H.S. to the numerator on L.H.S. and then substituting x = a1, a2, ......., an.

(Refer section 14,15,16,17 in A List of Evaluation Techniques )

Note: In order to determine the value of constants in the numerator of the partial fraction corresponding to the non-repeated linear factor px + q in the denominator of a rational expression, we may proceed as follows : Replace x =

q

 p (obtained by putting px + q = 0) everywhere in the given rational expres-

sion except in the factor px + q itself. For example, in the above illustration the value of A is obtained by replacing x by 1 in all factors of

3x 2 except (x  1) i..e (x 1)(x 2)(x 3)

312 5 A = (12)(13) = 2 Similarly, we have

B=

32 1 = (12)(2 3)

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

 8 and, C =

33 2 11 = (3 1)(3 2) 2

ma r a h S .

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6 L 7 Er. 81027 5058 9 801 839 Page 104

A List of Evaluation Techniques S.NO. Form of Integrate

1.

m

x cosn x dx where

 sin

where m, n  N

Value/Evaluation Techique

If m is odd put cos x = t If n is odd put sin x = t If both m and n are odd, put sin x = t if m  n and cos x = t otherwise.

If both m and n are even, use power reducting formulae sin2 x 

1 (1  cos 2x) 2

1 (1  cos 2x) 2 If m + n is a negative even integer , put tan x = t cos2 x 

2.

a sin x  b cos x  c

 p sin x  q cos x  r dx aex  be x

 pe

x

dx

 qex



a tan x  b

 p tan x  q dx  pe

2x

q



Change to

dx

or

(p , q  0, at least one of a , b  0) 4.

sin x and constant term (or ex, e–x) write the integral as d (DEN) dx  dx   dx dx   DEN DEN

of a, b 0)

ae2x  b

d (DEN) +  dx

Find ,  and by comparing the eoefficients of cos x,

(p, q,  0 and at least one

3.

Write Num =  (DEN) 

dx

 a sin x  b cos x



a sin x  b cos x

 p sin x  q cos x dx

aex  be x

 be

x

 qex

dx

and use (8) above

Use a = r cos  , b = r sin  to put the interal in the form

1 r

(Both a, b  0)

5.



dx Q

or

dx

or

Q

or



L

 Q dx

or

Q or

L

 Q dx

L

Q

where L  px  q, p  0

dx

 sin(x  ) . Now, use formula for  cosec x dx

c  2 b Write Q  a  x  x   a a 

ma r a h S b .

82 K t  x  . and put 6 L 7 2a Er. 81027 5058 9 801 839

and Q = ax2 + bx + c, a  0 Mathematics Concept Note IIT-JEE/ISI/CMI

Page 105

S.NO. Form of Integrate 6.

7.

L1 dx L2



Value/Evaluation Techique Write as L1 dx L1L2

where L1 = ax+b, a  0



L2 = px + q , p  0

to reduce to form



Q dx L

Put L 



L dx Q

1 to reduce to form t



dx Q

where L = px + q, p  0 and Q=ax2 + bx + c, a  0

8.



Q Rewrite the integrand as L Q

Q dx L

where L = px + q, p  0 Q = ax2 + bx + c, a  0

Divide Q by L to obtain Q = L1(L) +  where L1 is a linear expression in x and  is a constant.

The integral reduces to

9.



Q1

dx

Q2

Write Q1   Q2  



L1 Q

dx  

dx

L

Q

.

d [Q2 ]   dx

Find  ,  ,  and write the integral as

where

Q1 = ax2+bx+c , a  0





d dx Q2 dx   [Q2 ]dx  y Q





dx Q2

Q2 = px2+qx+r , p  0 10.

L

dx Put L2 = t2

L2

1

where L1 = ax + b and L2 = px + q, with a , p  0, a  p.

11.

Q

dx

where Q1=ax2+b, Q2=px2+q with a, p  0 , p  0 12.

1 / p, 1 / q,

 R(x

ma r a h x=1/t and simplify then put S .

82 K .

6 p + qt = u L 7 Er. 81027 5058 Let l = lcm (p, q, r, ...) 9 01 8 9 83 and put x = t First put

Q2

1

x

x1 / t ...)dx

where R is a rational function.

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

2

2

l

Page 106

S.NO. Form of Integrate 13.

P(x)

Value/Evaluation Techique

 Q(x) dx

If deg P(x)  deg Q(x) first isolate the quotient A(x)

where P(x) and Q(x) are

when P(x) is divided by Q(x). Now,

polynomials in x

P(x) B(x)  A(x)  Q(x) Q(x) where deg B(x) < deg Q(x) Now use techniques 22 to 29

14.

P(x)

 Q(x) dx

Write

where deg P(x)<deg Q(x)

A1 A2 An P(x)    ...  Q(x) x  1 x   2 x  n

and Q(x) is a product of distinct linear factors, that is Q(x)=A(x- 1 )(x– 2 )...

Evaluate Ai ' s and use

dx

 x  a  log | x  a |

(x– r ) where ai ' s are distinct and A  0 P(x)

15.

dx

 (x  a)

n

Put x–a = t,

where n  1 and P(x) is polynomial in x.

express P(x) in terms of ts and then integrate.

P(x)dx

16.

 (x  a)

m

(x  b)n

where m , n  1, a  b , and deg P(x) < m + n.

Write the integrand as A1 A2 Am B1 B2 Bn   ...    ...  2 m 2 x  a (x  a) x  b (x  b) (x  a) (x  b)n

Evaluate Ai's and Bj' and the integral.

P(x)dx

17.

 (x  a)(x

2

 bx  c)

where b2 – 4c < 0 and deg P(x) < 3 18.

19.

x x

x2  1 4

2

 kx  1

x2  1 4

 kx2  1

dx

dx

Write the integrand as

A Bx  C  x  a x2  bx  c

a k S 2 .

8 K Now, put x–1/x = t L. 2776 8 . r E 810 505 Divide the numerator by9 x to obtain 01 8 9 83 m hax r 1 / x

Divide the numerator by x2 to obtain

2

2

2

(1  1 / x2 )dx

x Mathematics Concept Note IIT-JEE/ISI/CMI

(1  1 / x2 )

2

 1 / x2  k

Now , put x + 1/x = t Page 107

S.NO. Form of Integrate 20.

xP(x2 )

 Q(x ) dx

Value/Evaluation Techique Put x2 = t

2

where P and Q are polynomials in x 21.

22.

23.

24.

P(x2 )

 Q(x ) dx

Put x2 = t

where P and Q are polynomials in x

for partial fraction not for integration

2

1

 (x

2

 a2 )n

dx  An

Begin with An–1 A n 1 

 R(sin x , cos x)dx

Universal substitution

where R is a rational function Special Cases (a) If R(–sinx , cosx) =–R(sinx , cosx) (b) If R(sinx , – cosx) =–R(sinx , cosx) (c) If R(–sinx, –cosx) =R(sinx , cosx)

tan (x/2) = t.

m

x

 1. (x

1

where n > 1

2

 a2 )n 1

dx

Put cos x = t Put sin x = t Put tan x = t

(a  bxn )p dx

where m, n, p are rational numbers, (a) p is a positive integer Expand (a + bxn)n by using binomial theorem (b) p is a negative integer, Put x = tk m

(c)

a c ,n , b d

where k = L CM (b,d)

a, b, c, d  I, b > 0, d > 0

where k = L CM (b,d)

m1 + p is an integer n

Put a + bxn = ts where p = r/s, r, s  I, s > 0

(d)

m 1  p is an integer n

Put a + bxn = xnts where p = r/s, r , s  I, s > 0

25.



Pn (x) 2

dx

ax  bx  c

where Pn(x) is a polynomial of degree n in x

ma r a h P (x) S 2dx . dx  Q (x) ax K  bx c k 8  ax  bx  c  .

6 . L 0277 ax58bx  c r E where Q (x) is polynomial of in x and k 1degree1(n–1) 50w.r.t.

98both is a constant. Differentiate the sides x and 0 8 9 3 multiplying by ax  bx 8 c to get the identiy Write

n

n 1

2

2

2

n–1

2

Pn(x) = Q'n–1 (x)(ax2 + bx+c) + 1/2 Qn–1 (x) (2ax+b) + k Compare coefficients to obtain Qn–1(x) and k.

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

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19. DEFINITE INTEGRAL (1)

Newton-Leibnitz formula Let f(x) be a continuous function defined on [a,b] and

 f(x)dx = F(x) + c, then

b

 f(x)dx = F(b)  F(a) is called definite integral and this formula is known as Newton-

a

Leibnitz formula.

Note: (i) Newton-Leibnitz formula is only applicable iff integrand f(x) is continous in [a,b]. b

(ii) Geometrically definite integral

 f(x)dx represents the algebraic sum of the area bounded a

by integrand f(x) with the x-axis, lines x = a and x = b, bounded area above the x-axis is treated as positive and bounded area below the x-axis is treated as negative.

b

 f(x)dx  A

1

 A 2.

a

(2)

Method of Substitution in a Definite Integral b

If evaluation of definite integral

 f(x)dx ,

if variable x is substituted as (t), where

a

()  a and ()  b, then b





f(x)dx 

a

 f((t)). '(t)dt. 

Note: Substitution x = (t) must satisfy the following conditions.

ma r a h continuous derivative  '(t); S .

82 K .

6 L 7 7 the5limits  with t varying on [, ] the values of the function xr.(t) do not 8 of E 8102leave 0 [a, b]; 9 8015  ()  a and ()  b .

839 

(t) is a continuous single-valued function defined in [, ] and has in this interval a

Mathematics Concept Note IIT-JEE/ISI/CMI

Page 109

(3)

Properties of Definite Integral b

Property:1

b

 f(x)dx =

 f(t)dt

a

a

definite integral is independent of variable of integration. b

Property:2

a

 f(x)dx =

  f(x)dx

b

c

a

Property:3

b

b

 f(x)dx =  f(x)dx + a

a

 f(x)dx, c

where c may lie inside or outside the interval [a,b]. b

Property:4



b

f ( x )d x =

 f(a  b  x)dx

a

a a

a

 f(x)dx =

0

 f(a  x)dx

0

Note: b

(i) If I 

 f(x)dx , than by property –4, a

b

I

b

 f(a  b  x)dx  2I   {f(x)  f(a  b  x)}dx a

a

b

(ii) If I 

f(x)

 f(x)  f(a  b  x) dx, then I  a

a

Property:5



a

f(x)dx =

a

a

2a

Property:6



0

 (f(x)  f(x))dx

0

a



ba .

2

f(x)dx 

 a  2 f(x)dx ; if f(x)  f(x) ( i.e. f(x) is even function)  0  ; if f(x)  f(x) (i.e. f(x) is odd function)  0



a

f(x)dx =  (f(x)  f(2a  x))dx 0

a m r a  h S  .

82 K .

6 L 7 Er. 81027 5058 9 801 Note: 9 is symmetrical about (i) If function f satisfy the condition f(2a–x) = f(x), then graph 8of3f(x) 2a

0

 a  2 f(x)dx ; if f(2a  x)  f(x) f(x)dx   0  ; if f(2a  x)  f(x)  0

the line x = a.

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

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(ii) If function f satisfy the condition f(2a–x) = –f(x), then graph of f(x) is symmetrical about point (a, 0)  /2

(iii)

 /2

loge (cos x)dx 

 0

0

Property:7 nT

(i)

If f(x) is a periodic function with period T and m, n  I, then T



0

anT

T

f(x) dx = n  f(x) dx, n  z, a  R



a

(iii)

0

nT

T



f(x) dx = (n  m)  f(x) dx ,m, n  z

mT

0

a

anT

(iv)

f(x) dx =



z, a  R

b

b nT

f(x) dx =



a nT

(4)

 f(x) dx, n 

0

nT

(v)

 loge 1 / 2 2

f(x)dx = n  f(x)dx, n  z

0

(ii)

loge (sin x) dx 



 f(x)dx n  z, a, b  R a

Standard Results for Definite Integral b

(i) If f(x)  0  x  [a,b] , then

 f(x)dx  0 a

b

(ii) If f(x)  0  x  [a,b] , then

 f(x)dx  0 a

(iii) If  (x)  f(x)   (x) for a  x  b, then



b

b

b

(x) dx 



f(x) dx 

a

a

 (x) dx

a

b

(iv)If m  f(x)  M for a  x  b, then m(b  a) 

 f(x) dx

 M(b  a)

a

Further if f(x) is monotonically decreasing in (a,b) then b

 f(x) dx

ma r a b h S .

82 K f(a)(b  a) <  f(x) dx < f(b) (b  a). .

6 L 7 a Er. 81027 5058 9 801 9 of equality holds if (v) If f(x) is continous in [a,b], then  | f(x) | dx   f(x)dx ; 8 the3sign f(b) (b  a) <

< f(a)(b  a) and if f(x) is monotonically increasing in (a,b) then

a

b

b

a

a

f(x) is non-negative or non-positive for all x  [a,b] .

Mathematics Concept Note IIT-JEE/ISI/CMI

Page 111

(vi)If f(x) is continous and increasing function in [a,b] and have concave graph, then b

(b  a)f(a) 

 f(a)  f(b)   2 

 f(x)dx  (b  a)  a

(vii) If f(x) is continous and increasing function in [a,b] and have convex graph, then  f(a)  f(b)  (b  a)   2  

b

 f(x)dx  (b  a)f(b). a

(viii) If f(x) and g(x) are continous function for all x  [a,b] and f2(x), g2(x) are integrable b

functions, then



b b   f2 (x)dx   g2 (x)dx     a a 



f(x). g(x) dx 

a



b

(ix) If f(x) is continuous on the interval [a,b], then

 f(x)dx  f(c)(b  a)

; where a < c < b.

a

(x) Leibnitz Rule: (x)

If F(x) 



f(t)dt , then F'(x) = f(ψ(x)).ψ'(x) - f(φ(x)). φ'(x).

(x)

(5)

Definite Integral as a Limit of Sum. If f(x) is the continuous function in the interval [a,b] where a and b are finite and b > a and if the interval [a,b] be divided into n equal parts, each of width h so that n h = b – a , [h{f(a)  f(a  h)  f(a  2h)  .........  f(a  (n  1)h}] is called the definite integral of Then hlim 0 f(x) between limits a and b, b



 f(x)dx  a

lt h[{f(a)  f(a  h)  f(a  2h)  ......  f(a  (n  1)h)]

h0

 n 1   lim h. f(a  rh) h 0   r 0 



If should be noted that as h  0 , 0   nh = b – a Putting a = 0 b = 1, so that h = 1/n.

We get



1

0

f(x) dx  lim

n 

1 . n

n 1

r 

 f  n 

r 0

Working Rule

ma r a h S .

82 K .

6 L 7 .

27 058 1  r  Er 0 lim . f 1 (ii) The limit when n   is its sum  n  n  Replace9r/n8by x, 1/n 1by5dx and 0 8 39 8 lim  by the sign of integration .  (i) Express the given series in the form of

1

r 

 n f  n 

n 1

n

r 0

n

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

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(iii) The lower and upper limits of integration will be the value of r/n for the first and last term (or the limits of these values respectively).

Note: qn  b

1 lim n n

(6)

r  f  n r  pn  a  



q

 f(x)dx. p

Important Results:  /2

If

In =



 /2 n

sin xdx =



0

n

cos xdx , then

0

 n1   n3   n5       ....... I0 or I1  n   n2   n 4 

In = 

according as n is even or odd , I0 =

 , I1 = 1 2

n5   1    n1   n3  ; n is even ... .  n  n2  n   4   2  2 Hence In =   n1  n3  n5   2    n  n2  n 4 ... 3 .1 ; n is odd        /2

If Im,n =



n

n

sin x cos xdx m, n  N then,

0

 (n1)(n3)(m5)......(n1)(n3)(n5).....   (m1)(mn2)(mn 4)......

2 Im,n =  (m1)(m3)(m5)......(n1)(n3)(n5).....  (mn)(mn2)(mn 4)...... 

;

when both m,n are even

;

when both m,n are not even

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics Concept Note IIT-JEE/ISI/CMI

Page 113

20. AREA BOUNDED BY CURVES Following cases illustrates the method of computation of bounded area under different conditions:

b

(i)

A

 f(x)dx. a

b

(ii)

A   f(x)dx.

 a

A  A1  A2 

(iii)

c

b



 f(x)dx

A   f(x)dx  a

c

b

(iv)

A

  f(x)  g(x) dx. a

ma r a h S A    f(x)  g(x)  dx. .

82 K .

6 L 7 Er. 81027 5058 9 801 839 b

(v)

a

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 114

b

(vi)

A

  f(x)  g(x) dx. a

b

(vii)

A

  f(x)  g(x) dx. a

b

(viii)

A

 f(y)dy a

b

(ix)

A   f(y)dy

 a

ma r a A  A  A .S h 82 K .

6 L 7 58 ErA. 8 f(y)dy 027 f(y)dy.

1 0 5 9 801 839 1

(x)

Mathematics Concept Note IIT-JEE/ISI/CMI

2

c

b

a

c

Page 115

b

(xi)

A

  f(y)  g(y) dy a

b

(xii)

A

  f(y)  g(y) dy a

b

(xiii)

A

  f(y)  g(y) dy a

b

(xiv)

A

  f(y)  g(y) dy a

ma r a h Curve Tracing : S .

82 K .

6 L 7 To find the approximate shape of a non-standard curve, must 8 be 5 Er. the81following 027 steps 0 checked .

9 8015 (1)Symmetry about x-axis : 839 If all the powers of 'y' in the equation are even then the curve is symmetrical about the x-axis.

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 116

(2)Symmetry about y-axis : If all the powers of 'x' in the equation are even then the curve is symmetrical about the y-axis.

(3) Symmetry about both axis : If all the powers of 'x' and 'y; in the equation are even, the curve is symmetrical about the axis of 'x' as well as 'y'.

(4) Symmetry about the line y = x : If the equation of the curve remains unchanged on interchanging 'x' and 'y', then the curve is symmetrical about the line y = x.

a>0 E.g.: x2 +y3= 3axy.

(5)Symmetry in opposite quadrants : If the equation of the curve remains unaltered when 'x' and 'y' are replaced by  x and  y respectively then there is symmetry in opposite quadrants.

(6) Intersection points of the curve with the x-axis and the y-axis.

(7) Locations of maxima and minima of the curve.

(8) Increasing or decreasing behaviour of the curve.

(9) Nature of the curve when x   or x  .

(10) Asymptotes of the curve.

Asymptoto(s) is (are) line(s) whose distance from the curve tends to zero as point on curve moves towards infinity along branch of curve. (i) If (ii) If

Lt f(x) =

x a



or xLt f(x) = a

  then x= a is asymptote of y = f(x)

Lt f(x) = k or Lt f(x) = k, then y = k is asymptote of y = f(x) x

x

f(x) (iii) If xLt = m1 , xLt f(x) = m1(x) = c, then y = m1x + c1 is an asymptote.   x (inclined to right)

(iv) If

Lt f(x) = m , Lt (f(x  m x) = c , then y = m x + c is an asymptote 2 2 2 2 2 x  x

x 

(inclined to left).

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics Concept Note IIT-JEE/ISI/CMI

Page 117

21. DIFFERENTIAL EQUATION (1)

Introduction : An equation involving independent and dependent variables and the derivatives of the dependent variables is called a differential equation. There are two kinds of differential equation .

Ordinary Differential Equation : If the dependent variables depend on one independent variable x, then the differential equation is said to be ordinary. For example:

d3y

dy + xy = sin x, dx

3

dx

+2

dy + y = ex, dx

3/2

2 2   dy  dy  dy k 2 = 1 + dx  , y = x + k 1 + dx  dx dx    

 

d2y

 

Partial differential equation : If the dependent variables depend on two or more independent variables, then it is known as partial differential equation. For example  2z 2z z  2z y +y = ax, + = xy.. x y y2 x2 2

(2)

Linear and Non-Linear Differential Equations: A differential equation is a linear differential equation if it is expressible in the form P0

dny n

dx

 P1

dn 1y n 1

dx

 P2

dn  2 y n2

dx

 ...  Pn 1

dy  Pn y  Q dx

where P0, P1, P2, ... , Pn–1 , Pn and Q are either constants of function of independent variable x.

Thus, if a differential equation when expressed in the form of a polynomial involves the derivatives and dependent variable in the first power and there are no product of these, and also the coefficient of the various terms are either constants or function of the independent variable, then it is said to be linear differential equation. Otherwise, it is a non linear differential equation.

If follows from the above definition that a differential equation will be non-linear differential equation if (i) its degree is more than one. (ii) any of the differential coefficient has exponent more than one (iii) exponent of the dependent variable is more than one. (iv)products containing dependent variable and its differential coefficients are present.

(3)

Order and Degree of a Differential Equation :

ma r a h S 2 .

8 K .

6 Degree: degree of differential equation is the highest exponent with which the highest L 7 27 0equation order derivative is raised in the differential equation, provided 58 is Er. 8the10differential 5 expressed in the polynomial form.

9 801 Note: 839

Order: Order of differential equation is the order of highest order derivative which is present in the equation.

If differential equation can't be expressed in the polynomial form, then its degree is not defined.

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 118

d2 y

2

 d3 y   x  3   y  0 is differential equation of 3 order and 2 degree.

Example: dx2  dx 

(4)

Geometrical Significance of Differential Equation: Geometrically, differential equation represents a family of curves which is having as many number of parameters as that of the order of differential equation. For example (i)

dy  ex represents y  ex  c. dx

(ii) y

(iii)

(5)

dy  x  0 represents x2  y2  c. dx

d2 y dx2

 ex represents y  ex  c1x  c2 .

Formation of Differential Equation: In order to formulate a differential equation for a given family of curves of n parameter (independent arbitrary constants), following steps is used.

Step (1): differentiate the family of curves 'n' times to obtain n set of equations.

Step (2): from the set of equations of step(1) and the equation of curve, all the arbitrary constants are eliminated to get the differential equation.

Note: Differential equation for a family of curves should not have any arbitrary constant and its order must be same as that of the number of parameters in family of curves.

(6)

Solution of a Differential Equation : Finding the dependent variable from the differential equation is called solving or integrating it. The solution or the integral of a differential equation is , therefore, a relation between dependent and independent variables (free from derivatives) such that it satisfies the given differential equation

Note: The solution of the differential equation is also called its primitive, because the differential equation can be regarded as a relation derived from it.

There can be three types of solution of a differential equation : (i) General solution (or complete integral or complete primitive) :A relation in x and y satisfying the given differential equation and involving exactly same number of arbitrary constants as that of the order of differential equation. (ii) Particular solution : A solution obtained by assigning values to one or more than one arbitrary constant of general solution of a differential equation.

(7)

ma r a h S .

82 K .

6 L 7 Variables separable : If the differential equation can be r form,7f(x) dx = 8  (y) dy 2 05each E put. in8the1by0 we say that variables are separable and solution can be obtained integrating side 9 8015 separately.

39 8 f(x) dx  (y) dy A general solution of this will be  =  + c,

Elementary Types of First Order and First Degree Differential Equations :

where c is an arbitrary constant.

Mathematics Concept Note IIT-JEE/ISI/CMI

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Equations Reducible to the Variables Separable form : If a differential equation can be reduced into a variables separable form by a proper substitution, then it is said to be "Reducible to the variables separable type". Its general form is dy = f(ax + by + c) where a, b  0. dx

substitution, ax + by + c = t. reduces the general form in to variable separable form Homogeneous differential Equations : f(x,y) dy = g(x,y) where f and g are homogeneous dx function of x and y and of the same degree, is called homogeneous differential equation and can be solved by putting y = vx.

A differential equation of the from

Equations Reducible to the Homogeneous form Equations of the form

ax by  c dy = Ax By C dx

.......... (1)

can be made homogeneous (in new variables X and Y) by substituting x = X + h and y = Y + k, where h and k are constants, we get

aX bY (ahbk  c) dY = AX BY (AhBk  C) ...........(2) dX

Now, h and k are chosen such that ah + bk + c = 0, and Ah + Bk + C = 0; the differential equation can now be solved by putting Y = vX.

Note: If the homogeneous equation is of the form : yf(xy) dx + xg(xy) dy = 0, the variables can be separated by the substitution xy = v.

Exact Differential Equation : The differential equation M(x,y)dx + N(x,y)dy = 0 ..........(1) where M and N are functions of x and y is said to be exact if it can be derived by direct differentiation of an equation of the form f(x,y) = c

Note: M N (i) The necessary condition for exact differential equation is y = x

(ii) For finding the solution of exact differential equation, following exact differentials must be remembered: (a) xdy + ydx = d(xy)

(b)

xdy ydx y = d ln x xy

(e)

(d)

 

xdy ydx x2

xdy ydx 2

2

x y



y =d x



(c) 2(xdx + ydy) = d(x2 + y2)

1

= d tan

y x



(f)

xdy  ydx = d(ln xy) xy

a m r a xy x y h S .

82 K .

6 L 7 r. 1027 058 E Linear differential equations of first order and first degree:8 9 8015 dy The differential equation + Py = Q, is linear in y..

839 dx xdy  ydx

(g)

2

2

 

1 = d  xy

(h)

xdx  ydy 2

2

1   d  ln(x2  y2 )  2 

where P and Q are functions of x.

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

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Integrating Factor (I.F.) : It is an expression which when multiplied to a differential equation converts it into an exact form.

I.F for linear differential equation = e  Pdx (constant of integration need not to be considered)  after multiplying above equation by I. F it becomes ; dy . e  Pdx + Py.. e  Pdx = Q.. e  Pdx dx d  Pdx  Pdx ) = Q.. e  dx (y.. e



y.. e

 Pdx

=

 Q.e Pdx

+C

(General solution).

Bernoulli's equation : Equations of the form

dy + Py = Q.yn, n  0 and n  1 dx

where P and Q are functions of x, is called Bernoulli's equation 1

To convert Bernoulli's equation to linear form

n 1

y

is substituted as t.

dy  dy  f  is termed as clairaut's dx  dx  equation. To get the solution of this form , the equation is differentiated as , explained in the next step:

Clairaut's Equation: Differential equation of the form y  x

dy d2 y dy  dy  d2 y x   f '  . 2 dx dx dx2  dx  dx   

d2 y   dy   x  f '    0.

2  dx   dx   d2 y 2

dx

0

or

dy  Constant=c dx



 general solution is given by y = cx + f(c).

 dy  x  f '   0.  dx   dy  f '   x . gives  dx 

another solution, which is termed as singular solution

Differential equation reducible to the linear differential equation of first order and first degree. (i)

dx  P(y)x  Q(y). dy

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 39 8 dt  P(x)t  Q(x).

I.F.  e  P(y)dy and general solution is given by:

x(I.F.)  (ii) f '(y)

 Q(y)I.F.dy  c

dy  P(x).f(y)  Q(x) dx

Put f(y)  t



f '(y)

Mathematics Concept Note IIT-JEE/ISI/CMI

dy dt  dx dx



dx

Page 121

(iii) f '(x)

dx  P(y)f(x)  Q(y). dy

Put f(x)  t  

(8)

f '(x)

dx dt  dy dy

dt  P(y)t  Q(y). dy

Orthogonal Trajectory : An orthogonal trajectory of a given system of curves is defined to be a curve which cuts every member of the given family of curve at right angle.

Steps to find orthogonal trajectory : (i) Let f(x,y,c) = 0 be the equation of the given family of curves, where 'c' is an arbitrary constant. (ii) Differentate the given equation with respect to x and then eliminate c. (iii) Replace

dy by dx

dx

 dy in the equation obtained in (ii).

(iv)Solving the differenetial equation obtained in step (iii) gives the required orthogonal trajectory.

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

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22. BASICS OF 2 Dimensional GEOMETRY In 2-dimensional coordinate system any point P is represented by (x,y), where |x| and |y| are the distances of 'P' from the y-axis and x-axis respectively

Note: (i) If both x and y  I , point 'P' is termed as integral point. (ii) If both x and y  Q , point 'P' is termed as rational point. (iii) If atleast one of x, y  Q , point 'P' is termed as irrational point.

(1) Distance Formula : The distance between the points A(x1,y1) and B(x2,y2) is

(x1  x2 )2 (y1  y2 )2

(2) Section Formula : If P(x,y) divides the line joining A(x1,y1) and B(x2,y2) internally in the ratio m : n, then :



x 



mx2 nx1 my 2 ny1 ;y  and for external division P(x, y) is m n m n



x 

mx2 nx1 my2 ny1 ;y  m n m n



Note:  x1  x2 y1  y2  , (i) The mid-point of AB is  .

2 2  

(ii) If P divides AB internally in the ratio m : n and Q divides AB externally in the ratio m : n then P and Q are said to be harmonic conjugates to each other with respect to A and B. Mathematically, 1 2 1 = + AQ AB AP

( AP, AB & AQ are in H.P.) .)

(3) Standard points of triangle.

If A(x1,y1), B(x2,y2), C(x3,y3) are the vertices of triangle ABC, whose sides BC, CA, AB are of lengths a, b, c respectively, then the coordinates of the special points of triangle ABC are as follows : (i) Centroid G 



x1  x2  x3 y1  y2  y3 , 3 3

(iii) Excentre I1 



 (ii)Incentre I   

ax1  bx 2  cx 3 ay1  by 2  cy 3 , ab  c ab  c



ax1 bx2  cx3 ay1 by2  cy3 ,  ab  c  ab  c

x (iv) Orthocentre H   1 

x (v) Circumcentre C   1 

Mathematics Concept Note IIT-JEE/ISI/CMI

ma r a hC 2 tan A  x tanB  x tan C y tan y tanB .yStan , K  8 .  tan7C 76 tan A  tanB  tan C tan A.L tanB  Er 8102 5058 sin2A  x sin2B  x sin2C y sin2A 9  y sin2B 01y sin2C  , 8 9 sin 2A  sin 2B  sin C sin 2A 3 8  sin 2B  sin2C  2

2

3

1

3

2

1

3

2

3

Page 123

Note: (i) Orthocentre, Centroid and Circumcentre are collinear and centroid divides the line joining orthocentre and circumcentre in the ratio 2 : 1. (ii) In an isosceles triangle G, H, I and C lie on the same line and in an equilateral triangle all these four points coincide with each other. (iii) Incentre and excentre are harmonic conjugate of each other with respect to the angle bisector on which they lie. (iv) In right angle triangle, orthocentre lies at the vertex of right angle and circumcentre lies at midpoint of hypotenuse.

(4) Area of a Triangle : If A(x1,y1), B(x2,y2), C(x3,y3) are the vertices of triangle ABC, then its area is equal to x1 1 x  ABC = 2 2 x3

y1 1 y2 1 y3 1

Note: Area of n-sided polygon formed by points (x1,y1); (x2,y2) ; ........ (xn,yn) is given by  x1 x2  x2 x3 y y + ..........  1 2 y2 y3 (5) Change of Axes: 1 2

xn 1 xn yn 1 yn



xn x1 yn y2

  

(i) Rotation of Axes: If the axes are rotated through an angle  in the anticlockwise direction keeping the origin fixed, then the coordinates (X,Y) of point P(x,y) with respect to the new system of coordinate are given by X = x cos  +y sin  and Y = y cos  – x sin  . (ii) Translation of Axes: The shifting of origin of axes without rotation of axes is called Translation of axes. If the origin (0, 0) is shifted to the point (h,k) without rotation of the axes then the coordinates (X,Y) of a point P(x,y) with respect to the new system of coordinates are given X = x – h, Y = y – k.

(6) Slope Formula : If  is the angle at which a straight line is inclined to the positive direction of x-axis and

0    ,    / 2 , then the slope of the line denoted by m is defined by m = tan  .

If A(x1,y1) and B(x2,y2), x1  x2, are points on a straight line , then the slope m of the line is  y y 

given by  x2  x 1  .  2 1

(7) Condition of collinearity of three points : Points A(x1,y1), B(x2,y2), C(x3,y3) are collinear if (i) mAB = mBC = mCA

(ii)

x1 x2 x3

y1 1 y2 1 = 0 y3 1

ma r a h Locus: S 2 .

8 K .

6 The locus of a moving point is the path traced out by that point under one or more given L 7 condition.

Er. 81027 5058 1 points say Approach to find the locus of a point: Let (h,k) be the co-ordinates 9 of the 0moving 8 P. Now apply the geometrical condition on h, k. This gives a relation between h and k. Now 9 3 replace h by x and k by y in th eleminant and resulting equation8 would be the equation of the (iii) AC = AB + BC or |AB  BC|

(8)

locus.

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

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23.STRAIGHT LINE (1) Equation of a straight Line in various forms : (i) Point-Slope form : y  y1 = m(x  x1) is the equation of a straight line whose slope is m and which passes through the point (x1,y1). (ii)  Slope-intercept form : y = mx + c is the equation of a straight line whose slope is m and which makes an intercept c on the y-axis. (iii) Two point form : y  y1 =

  y  y1 2

x2  x1 (x  x1) is the equation of a straight line which

passes through the points (x1,y1) and (x2,y2). (iv) Determinant form : Equation of line passing through (x1,y1) and (x2,y2) is x

y

x1 x2

y1 1 = 0 y2 1

1

(v) Intercept form :

x y + = 1 is the equation of a straight line which makes intercepts a b

a and b on X-axis and Y-axis respectively. (vi) Perpendicular/Normal form : xcos  +ysin  = p (where p  0, 0    2 ) is the equation of the straight line where the length of the perpendicular from the origin on the line is p and this perpendicular makes an angle  with positive x -axis. x x

y y

(vii)Parametric form : cos 1 = sin 1 = r or P(r)=(x,y)  (x1 + rcos  , y1 + r sin  ) is the equation of the line in parametric form , where 'r' is the parameter whose absolute value is the distance of any point (x, y) on the line from the fixed point (x1,y1) on the line. 'r' is take as positive for co-ordinate in increasing direction of 'y' and negative for decreasing direction of 'y'.  is angle of inclination of the line and   [0, ) (viii) General Form : ax + by + c = 0 is the equation of a straight line in the general form a in this case, slope of line =  b

(2) Angle between two straight lines: If m1 and m2 are the slopes of two intersecting straight lines (m1m2   1) and  is the acute m m2

1 angle between them, then tan  = 1+m .

1m2

Note: (i) Let m1, m2, m3 are the slopes of three lines L1 = 0; L2 = 0; L3 = 0 where m1 > m2 > m3 then the interior angles of the  ABC formed by these lines are given by,,

ma r a h S 2 line . (ii) The equation of lines passing through point (x , y ) and making angle 8the  with K .

6 L 7 y = mx + c are given by : r. 1027 058 E (y  y ) = tan(    )(x  x ) and (y  y ) = tan (  +  ) (x  x ), tan15= m.

98where 0 8 (iii) When two straight lines are parallel their slopes are equal. Thus any line parallel to 839 y = mx + c is of the type y = mx + k, where k is a parameter. m m2

m m3

1

m3 m1

2

tan A = 1m m ; tan B = 1m m & tan C = 1m m 3 1 1 2 2 3 1

1

Mathematics Concept Note IIT-JEE/ISI/CMI

1

1

1

1

Page 125

(iv) Two lines ax + by + c = 0 and a'x + b'y + c' = 0 are parallel if

a b c = . a' b'  c'

Thus any line parallel to ax + by + c = 0 is of the type ax + by + k = 0, where k is a parameter. (v) The distance between two parallel lines with equations ax + by + c1 = 0 and ax + by + c2 = 0 is

c1  c2 a2  b2

.

(vi) The area of the parallelogram =

p1p2 , where p1 & p2 are sin 

distances between two pairs of opposite sides &  is the angle between any two adjacent sides. (vii)Area of the parallelogram bounded by the lines y = m1x + c1, y = m1x + c2 and y = m2x + d1, y = m2x + d2 is given by

(c1  c2 )(d1  d2 ) . m1  m2

(viii) When two lines of slopes m1 and m2 are at right angles, the product of their slopes is  1, i.e. m1 m2 =  1. Thus any line perpendicular to y = mx + c is of the form y=–

1 x + d where d is any parameter.. m

(ix) Two lines ax + by + c = 0 and a'x + b'y + c' = 0 are perpendicular if aa' + bb' = 0.

Thus any line perpendicular to ax + by + c = 0 is of the from bx  ay + k = 0, where k is any parameter.

(3) Position of the point (x1, y1) with respect to the line ax + by + c = 0 : If ax1 + by1 + c is of the same sign as c, then the point (x1,y1) lie on the origin side of ax + by + c = 0. But if the sign of ax1 + by1 + c is opposite to that of c, the point (x1, y1) will lie on the non-origin side of ax + by + c = 0.

In general two points (x1, y1) and (x2, y2) will lie on same side or opposite side of ax + by + c = 0 according as ax1 + by1 + c and ax2 + by2 + c are of same or opposite sign respectively.

Note:

If b is positive in line ax+by+c = 0, then point P(x1,y1) lies above the line if ax1 + by1 + c > 0 and point P lies below the line if ax1 + by1 + c < 0.

(4) Length of perpendicular from a point on a line :

The length of perpendicular from P(x1,y1) on ax + by + c = 0 is

ax1  by1  c a2  b2

(5) Reflection of a point about a line :

ma r a h S ax by  c x x y y .

82  K = b =  .  6 a L 7 a b 7 r. is given 58 02by: 1 0 The image of a point (x ,y ) about the line ax + by E + c = 08 5 9 801 ax by  c x x y y 39 8 = = 2    a b a b

(i) Foot of the perpendicular from a point (x1 , y1 ) on the line ax+by+c = 0 is given by: 1

1

1

1

2

(ii)

1

1

1

1

9810277682/8398015058

1

1

2

Mathematics for IIT-JEE Er. L.K.Sharma

2

2

Page 126

(6) Bisectors of the angles between two lines : Equations of the bisectors of angles between the lines ax + by + c = 0 and ax by  c

a'x + b'y + c' = 0, (ab'  a'b) are :

a2  b2

= 

a'x b'y  c' a' 2 b' 2

Note: (i) Equation of straight lines passing through P(x1, y1) and equally inclined with the lines a1x +b1y + c1 = 0 & a2x + b2y + c2 = 0 are those which are parallel to the bisectors between these two lines & passing through the point P. (ii) If  be the angle between one of the lines and one of the bisectors and |tan  | < 1, then this bisector is the acute angle bisector, if |tan  | > 1, then the bisector is obtuse angle bisector. (iii) If aa' + bb' < 0, then the equation of the bisector of this acute angle is ax by  c 2

=+

a'x b'y  c'

(c and c´ constants are positive)

a' 2 b' 2

2

a b

If aa' + bb' > 0, the equation of the bisector of the obtuse angle is : ax by  c

=+

a2  b2

a'x b'y  c'

(c and c´ constants are positive)

a' 2 b' 2

(iv) To discriminate between the bisector of the angle containing the origin and that of the angle not containing the origin. the equations are written as ax + by + c = 0 and a'x + b'y + c' =0 such that the constant terms c, c' are positive. Then ax by  c

=+

a2  b2

a'x b'y  c'

ax by  c

origin &

gives the equation of the bisector of the angle containing the

a' 2 b' 2

a2  b2

= 

a'x b'y  c' a' 2 b' 2

gives the equation of the bisector of the angle not

containing the origin. In general equation of the bisector which contains the point (  ,  ) is

ax by  c a2  b2

=+

a'x b'y  c'

or

a' 2 b' 2

ax by  c

= 

a2  b2

a'x b'y  c' a' 2 b' 2

according as a  +b  +c

and a'  +b'  +c' having same sign or otherwise.

(7) Condition of Concurrency : Three non parallel lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 & a3x + b3y + c3 = 0 are a1 b1 c1 a concurrent if 2 b2 c2 = 0 a3 b3 c3

 a1 a a   2  3  b2 b3   b1

Alternatively : If three constants A, B and C (not all zero) can be found such that A(a1x + b1y + c1) + B(a2x + b2y + c2) + C(a3x + b3y + c3)=0, then the three straight lines are concurrent.

(8)

a m r a Family of Straight Lines : h S .

82 K .

6 L 7 7 The equation of a family of straight lines passing through of r.the point 2intersection E 0 1 0L5=8of0 the lines. L  a x + b y + c = 0 and L  a x + b y + c = 0 9 is 8 given by L 1+ i.e.

5  0 8 (a x + b y + c ) +  (a x + b y + c ) = 0, where  is an arbitrary real number. .

39 8 Note: 1

1

1

1

1

1

1

2

2

2

2

2

2

1

2

2

(i) If u1 = ax + by + c, u2 = a'x + b'y + d, u3 = ax + by + c', u4 = a'x + b'y + d' Mathematics Concept Note IIT-JEE/ISI/CMI

Page 127

u 3= 0

C 0

0

D

A (ii)

u2 =

u4 =

then u1 = 0; u2 = 0; u3 = 0; u4 = 0 form a parallelogram.

The diagonal BD can be given by u2u3  u1u4 = 0.

u1= 0

B

The diagonal AC is given by u1 +  u4 = 0 and u2 +  u3 = 0, if the two equations are identical for some real  and  .[For getting the values of  &  compare the coefficients of x, y & the constant terms].

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

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24. PAIR OF STRAIGHT LINES (1) A Pair of Straight lines through origin : (i) A homogeneous equation of two degree ax2 + 2hxy + by2 = 0 always represents a pair of straight lines passing through the origin if : (a) h2 > ab  lines are real and distinct. (b) h2 = ab  lines are coincident. (c) h2 < ab  lines are imaginary with real point of intersection i.e. (0,0)

(ii) If y = m1x and y = m2x be the two equations represented by ax2 + 2hxy + by2 = 0, then; m1 + m2 = -

2h a and m1 m2 = . b b

(iii) If  is the acute angle between the pair of straight lines represented by 2 h2  ab ax + 2hxy + by = 0, then; tan  = . ab 2

2

(iv) The condition that these lines are : (a) At right angles to each other is a + b = 0. i.e. coefficient of x2 + co-efficient of y2 = 0. (b) Coincident is h2 = ab. (c) Equally inclined to the axis of x is h = 0. i.e. coeff. of xy = 0.

Note:

A homogeneous equation of degree n represents n straight lines passing through origin. (v) The equation to the pair of straight lines bisecting the angle between the straight lines, ax2 + 2hxy + by2 = 0 is

xy x2 y2 = . h ab

(2) General equation of second degree representing a pairs of straight lines:

ma r a h S a h g .

82 K .

6 L 7 h b f abc + 2fgh  af  bg  ch = 0, i.e. if = 0. r.

E 81027 5058 g f c 9 801 9 83equation The angle  between the two lines representing by a general is the same

(i) ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of straight lines if :

2

(ii)

2

2

as that between the two lines represented by its homogeneous part only.

Mathematics Concept Note IIT-JEE/ISI/CMI

Page 129

(3) Homogenization : The equation of a pair of straight lines joining origin to the points of intersection of the line L   x + my + n = 0 and a second degree curve.

S  ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 2  x my   x my   x my   + 2fy   + c = 0.   n   n   n 

is ax2 + 2hxy + by2 + 2gx 

The equation is obtained by homogenizing the equation of curve with the help of equation of line.

NOTE: Equation of any curve passing through the points of intersection of two curves C1 = 0 and C2 = 0 is given by  C1 +  C2 = 0 where  &  are parameters.

(4) Some Important Results (i) If ax2 + 2hxy + 2gx + 2fy + c = 0 represent a pair of parallel straight lines, then the distance between them is given by 2

g2  ac f 2  bc or 2 a(a  b) b(a  b)

(ii) Th e l i n e s joi n i n g t h e or igi n t o t h e po i n t s of i n t ersec t i on of t h e cu r ves ax 2 + 2hxy + by 2 + 2gx = 0 and a' x 2 + 2h' xy + b' y 2 + 2' gx = 0 will be mutually perpendicular, if g'(a + b). (iii) If the equation hxy + gx + fy + c = 0 represents a pair of straight lines, then fg = ch. (iv) The pair of lines (a2 – 3b2)x2 + 8abxy + (b2 – 3a2) y2 = 0 with the line ax+by+c = 0 form an equilateral triangle and its area 

c2 3(a2  b2 )

.

(v) The area of a triangle formed by the lines ax2 + 2hxy + by2 = 0 and lx + my + n = 0 is given by

n2 h2  ab am2  2hlm  bl2

(vi) The lines joining the origin to the points of intersection of line y = mx + c and the circle x2 + y2 = a2 will be mutually perpendicular, if a2(m2 + 1) = 2c2. (vii)If the distance of two lines passing through origin from the point (x1,y1) is d, then the equation of lines is (xy1 – yx1)2 = d2(x2 + y2) (viii) The lines represented by the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 will be equidistant from the origin, if f4 – g4 = c(bf2 – ag2) (ix) The product of the perpendiculars drawn from (x1,y1) on the lines ax2 + 2hxy + by2 = 0 is given by

ax12  2hx1y1  by12 (a  b)2  4h2

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839

(x) The product of the perpendiculars drawn from origin on the lines Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

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c

ax2 + 2hxy + by2 + 2gx + 2fy = 0 is

(a  b)2  4h2

(xi) If the lines represented by the general equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 are perpendicular, then the square of distance between the point of intersection and origin

f 2  g2 f 2  g2 or . h2  b2 h2  a2 (xii) The square of distance between the point of intersection of the lines represented by is

the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 and origin is

c(a  b)  f 2  g2 . ab  h2

(xiii) If one of the line given by the equation ax2 + 2hxy + by2 = 0 coincide with one of these given by a'x2 + 2h'xy + b'y2 = 0 and the other lines represented by them be perpendicular, then

ha'b ' h' ab 1   (aa'bb '). b ' a' b  a 2

(xiv)The straight lines joining the origin to the points of intersection of the straight line kx + hy = 2hk with the curve (x – h)2 + (y – k)2 = c2 are at right angles, if h2 + k2 = c2.

ma r a h S .

82 K .

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25. CIRCLES (1) Equations of a Circle (i) An equation of a circle with centre (h, k) and radius r is (x–h)2 + (y–k)2 = r2 (ii) An equation of a circle with centre (0, 0) and radius r is x2+y2 = r2 (iii) An equation of the circle on the line segment joining (x1, x1) and (x2, y2) as diameter is (x – x1) (x – x2) + (y – y1) (y – y2) = 0 (iv) General equation of a circle is x2 + y2 + 2g x + 2f y + c = 0 where g, f and c are constants.  Centre of this circle is (–g, –f) 

Its radius is

g2  f 2  c, (g2  f 2  c).

(v) The intercepts made by the circle x2 + y2 + 2gx + 2fy + c = 0 on the x-axis and y-axis is given by 2 g2 c and 2 f 2  c respectively.. (vi) General equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 in x and y, represents a circle if and only if  coefficient of x2 equals coefficient of y2, i.e., a = b  0.  coefficient of x y is zero, i.e., h = 0. 

g2 + f2 – ac  0

Note: (i) Equation of circle circumscribing a triangle whose sides are given by L1 = 0; L2 = 0 and L3 = 0 is given by ; L1L2 +  L2L3 +  L3L1 = 0, provided co-efficient of xy = 0 and coefficient of x2 = co-efficient of y2. (ii) Equation of circle circumscribing a quadrilateral whose side in order are represented by the lines L1 = 0, L2 = 0, L3 = 0 and L4 = 0 are u L1 L3 +  L2L4 = 0 where values of u and  can be found out by using condition that co-efficient of x2 = co-efficient of y2 and co-efficient of xy = 0.

(2) Parametric Equations of a circle : Parametric equation of circle (x  h)2 + (y  k)2 = r2 is given by

x h y k   r , where (h, k) cos  sin 

is the centre, r is the radius and  is a parameter (0    2) Any point P() on the circle (x – h)2 + (y – k)2 = r2 can be assumed in parametric form as P  (h  r cos  , k  r sin )

Note: The equation of chord PQ to the circle x2 + y2 = a2 joining two points P(  ) and Q(  ) on it is given by x cos

(3)

   + y sin = a cos .

2 2 2

ma r a Position of a point with respect to a circle : h S .

82 K .

6 L 7 Let equation of the circle is given by S  x + y + 2gx + c2 =7 0, then following 58 Er.+ 2fy 0 1 0 condition illustrates the relative position of point P. 5 8 9 801 (i) Point 'P' lies on the circle if: 839 2

2

x12  y12  2gx1  2fy1  c  0 .

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(ii) Point 'P' lies inside the circle if:

x12  y12  2gx1  2fy1  c  0 (iii) Point 'P' lies outride the circle if:

x12  y12  2gx1  2fy1  c  0

Note: The greatest and the least distance of a point P from a circle with centre C and radius r is PC + r and PC  r respectively..

(4) Line and circle If line y = mx + c intersects the circle x2 + y2 = a2 at point A and B, then the x-co-ordinate of the points of intersection is given by x2 + (mx + c)2 – a2 = 0. ........(1)  (1 + m2) x2 + 2cm x + c2 – a2 = 0 In equation (1), if discriminant 'D' is positive, then two distinct real values of 'x' is obtained which confirms the intersection of line and circle at two distinct points (refer figure no. (1)).

D  0  c2  a2 (1  m2 )  p  a.

If equation (1), if dircriminant 'D' is negative, then line is outside the circle (refer figure (2))

 c2  a2 (1  m2 )  p  a

If equation (1), if discriminant 'D' is zero, then line is tangential to the circle (refer figure(3)).

 c2  a2 (1  m2 )  p  a

(5) Tangent to a Circle: (a) Slope Form : Straight line y = mx + c is tangent to the circle x2 + y2 = a2 if c2 = a2(1 + m2). Hence,  a2m a2 

equation of tangent is y = mx  a 1m2 and the point of contact is   c , c  .  

Note:

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 point 8 = a at its (x0 ,y1 ) is given by 9 3 8

Equation of tangent of slope 'm' to the circle (x – h)2 + (y – k)2 = r2 is given by: (y – k) = m(x – h)  r 1  m2 (b) Point form : (i) The equation of the tangent to the circle x2 + y2 x x1 + y y1 = a2.

2

1

1

(ii) The equation of the tangent to the circle x2 + y2 + 2gx + 2fy + c = 0 at its point Mathematics Concept Note IIT-JEE/ISI/CMI

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(x1, y1) is given by x x1 + y y1 + g(x + x1) + f(y + y1) + c = 0.

Note: In general the equation of tangent to any second degree curve at point (x1, y1) on it can be obtained by replacing x2 by xx1 , y2 by yy1 , x by

x  x1 y  y1 x y  xy1 , y by , xy by 1 2 2 2

and c remains as c. (c) Parametric form : The equation of a tangent x cos  + y sin  = a.

to

circle x2 + y2 = a2 at

(a cos  , a sin  ) is

Note:      a sin  a cos 2 2 , The point of intersection of the tangents at the points P(  ) & Q(  ) is   cos   cos     2 2

  .   

(6) Normal to Circle If a line is normal/orthogonal to a circle then it must pass through the centre of the circle.

Using this fact normal to the circle x2 + y2 + 2gx + 2fy + c = 0 at (x1 y1) is given by: y1  f y  y1 = x  g (x  x1).

1

(7) Pair of Tangents from a Point : The equation of a pair of tangents drawn from the point P(x1, y1) to the circle x2 + y2 + 2gx + 2fy + c = 0 is given by SS1 = T2.

Where S  x2 + y2 + 2gx + 2fy + c ; S1  x12 + y12 + 2gx1 + 2fy1 + c T  xx1 + yy1 + g(x + x1) + f(y + y1) + c.

Note: The locus of the point of intersection of two perpendicular tangents is called the director circle of the given circle. The director circle of a circle is the concentric circle having radius equal to

2 times the original circle.

(8) Length of a Tangent and Power of a Point : The length of a tangent from an external point (x1,y1) to the circle S  x2 + y2 + 2gx + 2fy + c = 0 is given by L =

Note:

x 12  y 12  2g x 1  2 f1y  c =

S1

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 39 8 are drawn from the point P(x ,y ) to the circle S x + y + 2gx

(i) Square of length of the tangent from the point P is called the power of point with respect to a circle. (ii) Power of a point with respect to a circle remains constant i.e. PA.PB = PT2.

(9) Chord of Contact :

If two tangents PT1 and PT2  2 1 1 + 2fy + c = 0, then the equation of the chord of contact T1 T2 is given by: T  0.

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xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0.

Note: (i) Chord of contact exists only if the point 'P' is outside the circle.

2LR

(ii) Length of chord of contact T1T2 =

R 2  L2

(iii) Area of the triangle formed by the pair of the tangents and its chord of contact is given by

RL3 R2 L2

.

(iv) If T1PT2  , then tan  

2RL L  R2 2

(v) Quadrilateral PT1OT2 is cyclic and the equation of the circle circumscribing the triangle PT1T2 is given by (x  x1) (x + g) + (y  y1) (y + f) = 0.

(10) Equation of the Chord with a given Middle Point : The equation of the chord of the circle S  x2 + y2 + 2gx + 2fy + c = 0 in terms of its mid point M(x1,y1) is given by T = S1. xx1 + yy1 + g(x + x1) + f(y + y1) + c = x12 + y12 + 2gx1 + 2fy1 + c

(11) Common Tangents to two Circles : Direct (or external) common tangents meet at a point which divides the line joining centre of circles externally in the ratio of their radii.

Transverse (or internal) common tangents meet at a point which divides the line joining centre of circles internally in the ratio of their radii.

If (x – g1)2 + (y – f1)2 = r12 and (x – g2)2 + (y – f2)2 = r22 are two circles with centres C1(g1, f1) and C2(g2, f2) and radii r1 and r2 respectively, then 'P' and 'Q' are the respective points from which direct and Transverse common tangents can be drawn (as shown in figure).

ma r a h S .

82 K Note: .

6 L 7 Length of an external (or direct) common tangent & internal 58 Er.(or8transverse) 027 common 1 0 5 9 01 tangent to the circles are given by : L = d (r r ) & L = d (r8 r ) , 9 8r3, r are the radii of where d = distance between the centres of the two circles and ext

the two circles.

Mathematics Concept Note IIT-JEE/ISI/CMI

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2

1

2

int

2

2

1

1

2

2

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(12) Relative Position of two Circles Case

Number of Common Tangents

Condition

(i)

4 common tangents

r1 + r2 < c1 c2.

(ii)

3 common tangents.

r1 + r2 = c1 c2.

(iii)

2 common tangents.

|r1  r2| < c1 c2 < r1 + r2

(iv)

1 common tangent

|r1  r2| = c1 c2

(v)

No common tangent

c1 c2 < |r1  r2|

(13) Orthogonality Of Two Circles : Two circles S1 = 0 and S2 = 0 are said to be orthogonal or said to intersect orthogonally if the tangents at their point of intersection include a right angle. The condition for two circles to be orthogonal is : 2g1g2 + 2f1f2 = c1 + c2.

Note: (i) The centre of a variable circle orthogonal to two fixed circles lies on the radical axis of two circles. (ii) The centre of a circle which is orthogonal to three given circles is the radical centre provided the radical centre lies outside all the three circles.

(14) Radical Axis and Radical Centre : The radical axis of two circles is the locus of a point whose powers with respect to the two circles are equal. The equation of radical axis of the two circles S1 = 0 and S2 = 0 is given by S1–S2 = 0 i.e: 2(g1–g2)x + 2(f1–f2)y + (c1–c2) = 0.

ma r a h Note: S .

82 K . are equal.

6 (i) The length of tangents from radical centre to the three.circles L 7 r 27 058 E 0 1 (ii) If two circles intersect, then the radical axis is the common 15two circles.

98chord8of0the (iii) If two circles touch each other then the radical axis is the common 39 tangent of the 8 two circles at the common point of contact.

The common point of intersection of the radical axes of three circles taken two at a time is called the radical centre of three circles.

(iv) Radical axis is always perpendicular to the line joining the centres of the two circles.

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(v) Radical axis will pass through the mid point of the line joining the centres of the two circles only if the two circles have equal radii. (vi) Radical axis bisects a common tangent between the two circles. (vii) A system of circles, every two of which have the same radical axis, is called a coaxial system. (viii) Pairs of circles which do not have radical axis are concentric circles.

(15) Family of Circles : (i) The equation of the family of circles passing through the points of intersection of two circles S1 = 0 and S2 = 0 is : S1 +  S2 = 0 ;   R  {1} (    1 if the co-efficient of x2 and y2 in S1 and S2 are same)

Note:

If Coefficients of x2 and y2 are same in two intersecting circles S1 = 0 and S2 = 0, then the common chord of circles is given by: S1 – S2 = 0 (ii) The equation of the family of circles passing through the point of intersection of a circle S = 0 and a line L = 0 is given by S +  L = 0 ;   R (iii) The equation of a family of circles passing through two given points (x1,y1) and (x2,y2) can be written in the form : x x {(x  x1) (x  x2) + (y  y1)(y  y2)} +  1 x2

y 1 y1 1 = 0 where   R . y2 1

(iv) The equation of a family of circles touching a fixed line ax+by+c = 0 at the fixed point (x1, y1) is {(x  x1)2 + (y  y1)2} +  (ax+by+c) = 0 where  is a parameter.. (v) The equation of family of circles touching a fixed circle S = 0 at fixed point (x1,y1) is given by: {(x-x1 )2 +(y-y1 )2 }+S=0 ;   R .

ma r a h S .

82 K .

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26. PARABOLA (1) Conic Sections : A conic section is the locus of a point which moves in a plane so that its distance from a fixed point (i.e. focus) is in a constant ratio (i.e. eccentricity) to its perpendicular distance from a fixed straight line (i.e. directrix). Let point P be (x,y) then equation of conic is given by:

PF  e = constant ratio  PM

(x  )2  (y  )2  e

px  qy  r p2  q2

 ax2  2hxy  by2  2gx  2fy  c  0

Note: All conic sections can be derived from a right circular cone, as shown in the figure

(2) Recognisation of Conics: The equation of conics is represented by the general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0..............(1) where  = abc + 2fgh – af2 – bg2 – ch2 Different conics can be recognized by the conditions given in the tabular form.

Case I: when   0 In this case equation (1) represents the Degenerate conic whose nature is given in the following table: Condition   0 and h2 –ab = 0   0 and h2 –ab  0   0 and h2 –ab > 0   0 and h2 –ab < 0

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

ma r a h S Nature of Conic .

82 K .

6 L 7 A pair of coincident r. lines 27 058 E 0 1 8 of straight Real or imaginary9 pair 15lines 0 8 9 line.

A pair of intersecting3 straight 8 Point.

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Case II: When   0, In this case equation (1) represents the Non-degenerate conic whose nature is given in the following table: Condition

Nature of Conic

  0, h = 0, a = b

a circle

  0, h2 – ab = 0

a parabola

2

  0, h – ab < 0

an Ellipse or empty set

  0, h2 – ab > 0

a Hyperbola

  0, h2 – ab > 0 and a + b = 0

a Rectangular hyperbola

PARABOLA (3) Basic Definitions A parabola is the locus of a point, whose distance from a fixed point (focus) is equal to perpendicular distance from a fixed straight line (directrix). Standard forms of the parabola are y2 = 4ax; y2 =  4ax; x2 = 4ay; x2 =  4ay

Focal Distance : The distance of a point on the parabola from the focus.

Focal Chord : A chord of the parabola, which passes through the focus. y2 = 4ax.

Double Ordinate : A chord of the parabola perpendicular to the axis of the symmetry Latus Rectum : A double ordinate passing through the focus or a focal chord perpendicular to the axis of parabola is called the latus Rectum (L.R.).

For y2 = 4ax.  Length of the latus rectum = 4a.

 ends of the latus rectum are L(a, 2a) & L' (a,

 2a).

Note: (i) Perpendicular distance from focus on directrix = half the latus rectum. (ii) Vertex is middle point of the focus & the point of intersection of directrix & axis. (iii) Two parabolas are said to be equal if they have the same latus rectum.

(4) Parametric Representation :

ma r a h S .

Note: 82 K .

6 L 7 (i) The equation of a chord joining t and t is 2x  (t + Etr).y +821at0t 2=70. 5058 (ii) If t and t are the ends of a focal chord of the parabola 9 y = 4ax then 01t t =  1. Hence 8 9 2a  a 83 the co-ordinates at the extremities of a focal chord are (at ,2at) and  ,  .

Parametric co-ordinates of a point on the parabola is (at2, 2at) i.e. the equations x = at2 and y = 2at together represents the parabola y2 = 4ax, t being the parameter and t  R

1

2

1

2

1 2

2

1

2

1 2

2

t

2

t 

2

(iii) Length of the focal chord making an angle  with the x-axis is 4acosec  .

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(5) Position of a point Relative to a Parabola : The point (x1,y1) lies outside, on or inside the parabola y2 = 4ax according as the expression y12  4ax1 is positive, zero or negative respectively..

(6) Line and a Parabola : The line y = mx + c meets the parabola y2 = 4ax in two points real, coincident or imaginary according as c 

a a a ,c  or c  respectively. Condition of tangency is c = a/m. m m m

(7) Tangents to the Parabola y2 = 4ax : (i) yy1 = 2a(x + x1) at the point (x1, y1); (ii) y = mx +

 a 2a  a (m  0) at  2 , m  m m

(iii) yt = x + at2 at (at2, 2at).

Note:

Point of intersection of the tangents at the point t1 and t2 is (at1t2, a(t1 + t2))

(8) Normal to the parabola y2 = 4ax : y1 (i) y  y1 = 2a (x  x1) at (x1, y1) ; (ii) y = mx  2am  am3 at (am2,  2am) (iii) y + tx = 2at2 + at3 at (at2 2at).

Note: (i) Point of intersection of normal at t1 and t2 are, (a (t12 + t22 + t1t2 + 2),  at1t2 (t1 + t2)). (ii) If the normals to the parabola y2 = 4ax at the point t1 meets the parabola again at 

2



1

the point t2 then t2=   t1  t  . (iii) If the normals to the parabola y2 = 4ax at the points t1 and t2 intersect again on the parabola at the point 't3' then t1t2 = 2; t3=  (t1 + t2) and the line joining t1 and t2 passes through a fixed point (  2a,0).

(9) Pair of Tangents : The equation to the pair of tangents which can be drawn from any point (x1,y1) to the parabola y2 = 4ax is given by : SS1 = T2 where : S  y2  4ax ; S1 = y12  4ax1 ; T  yy1  2a(x + x1).

Note: (i) From any point on the directrix perpendicular tangents can be drawn to the parabola and their chord of contact is the focal chord. (ii) If tangents are drawn at the extremity of a focal chord of parabola, then these tangents are perpendicular and meet on the line of directrix. (iii) Circle drawn with focal chord as the diameter always touches the directrix of parabola.

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 (10) Chord of Contact : 9 801 39 8 Equation to the chord of contact of tangents drawn from a point P(x ,y ) is 1

1

yy1 = 2a (x + x1).

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Note: The area of the triangle formed by the tangents from the point (x1,y1) and the chord of contact is (y12  4ax1)3/2/2a.

(11) Chord with a given middle point : Equation of the chord of the parabola y 2=4ax whose middle point is (x 1,y 1) is T = S1, where, S1 = y12  4ax1 ; T  yy1  2a(x + x1).

(12) Important Results : (i) If the tangent and normal at any point 'P' of the parabola intersect the axis at T and G then ST=SG=SP where 'S' is the focus. In other words the tangent and the normal at a point P on the parabola are the bisectors of the angle between the focal radius SP and the perpendicular from P on the directrix. From this we conclude that all rays emanating from S will become parallel to the axis of the parabola after reflection. (ii) The portion of a tangent to a parabola cut off between the directrix and the curVe subtends a right angle at the focus. (iii) The tangents at the extremities of a focal chord intersect at right angles on the directrix, and hence a circle on any focal chord as diameter touches the directrix. Also a circle on any focal radii of a point P(at2,2at) as diameter touches the tangent at the vertex and intercepts a chord of length a 1 t2 on a normal at the point P.. (iv) Any tangent to a parabola & the perpendicular on it from the focus meet on the tangent at the vertex. (v) If the tangents at P and Q meet in T, then :  TP and TQ subtend equal angles at the focus S.  ST2= SP. SQ and  The triangles SPT and STQ are similar. (vi) Semi latus rectum of the parabola y2 = 4ax, is the harmonic mean between segments of any focal chord of the parabola. (vii)The area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points. (viii) If normal are drawn from a point P(h,k) to the parabola y2 = 4ax then k = mh  2am  am3 i.e. am3 + m(2a  h) + k = 0.

2ah k ; m1m2m3 =  . a a where m1 m2 & m3 are the slopes of the three concurrent normals. m1 + m2 + m3 = 0; m1m2 + m2m3 + m3m1 =

Note:  

algebraic sum of the slopes of the three concurrent normals is zero. algebraic sum of the ordinates of the three co-normal points on the parabola is zero.

 

Centroid of the  formed by three co-normal points lies on the x-axis.

Condition for three real and distinct normals to be drawn from a point P(h,k) is 4

h > 2a and k2 < 27a (h  2a)3

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839

(ix) Length of subtangent at any point P(x,y) on the parabola y2 = 4ax equals twice the abscissa of the point P and the subtangent is bisected at the vertex. (x) Length of subnormal is constant for all points on the parabola and is equal to the semi latus rectum.

Mathematics Concept Note IIT-JEE/ISI/CMI

Page 141

27. ELLIPSE Ellipse is the locus of a point which moves in such a way so that sum of its distances from two fixed points is always constant

Note: (i) PS + PS' = AA' (ii) Eccentricity (e) < 1

(1) Basic definitions: Standard equation of an ellipse referred to its principal axes along the co-ordinate axes is

x2 a

2

+

y2 b2

= 1, where a > b & b2 = a2 (1  e2)

Eccentricity : e = 1

b2 a2

, (0 < e < 1)

Foci : S  (ae,0) and S'  (  ae, 0).

Equations of Directrix : x =

a a and x =  . e e

Major Axis : The line segment AA' in which the foci S and S' lie is of length 2a and is called the major axis (a > b) of the ellipse. Point of intersection of major axis with directrix is called the foot of the directrix (Z and Z').

Minor Axis : The y-axis intersects the ellipse in the points B'  (0,  b) & B  (0,b). The line segment BB' is of length 2b (b<a) is called the minor axis of the ellipse.

Principal Axis : The major & minor axes together are called principal axis of the ellipse.

Vertices : Point of intersection of ellipse with major axis. A'  (  a,0) & A  (a,0).

Focal distances: The focal distance of the point (x,y) on the ellipse

x2 y2   1 are a+ex a2 b2

and a–ex Focal Chord : A chord which passes through a focus is called a focal chord.

Double Ordinate : A chord perpendicular to the major axis is called a double ordinate.

Latus Rectum : The focal chord perpendicular to the major axis is called the latus rectum.

2

Length of latus rectum (LL') =

(minor axis) 2b2 =2 = 2a(1  e2) major axis a

Centre : The point which bisects every chord of the conic drawn through it, is called the centre of the conic. C = (0,0) the origin is the centre of the ellipse

x2 a2

+

y2 b2

=1.

ma r a h If the equation of the ellipse is given as + =1 and nothing is.mentioned then S 2 the a b 8 K .

6 standard form (a > b) is assumed. r. L 10277 058 E If b > a is given, then the y-axis become major axis and x-axis 5 minor 1the 98 become 0 axis and all other points and lines change accordingly. (as shown 8 if figure) 39 8 x y

Note: (i) (ii)

2

vertical ellipse is given by Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

2

a

x2

y2

2

2

2



2

b

 1 ; (a < b)

Page 142

Vertical Ellipse:

(2) Auxiliary Circle: A circle described on major axis of ellipse as diameter is called the auxiliary circle.

Let Q be a point on the auxiliary circle x2 + y2 = a2 such that line through Q perpendicular to the x-axis on the way intersects the ellipse at P, then P & Q are called as the Corresponding Points on the ellipse and the auxiliary circle respectively. ' θ ' is called the Eccentric Angle of the point P on the ellipse (0    2) .

Note: (i)

Semi minor axis (PN) b = = Semi major axis (QN) a

(ii) If from each point of a circle perpendiculars are drawn upon a fixed diameter then the locus of the points dividing these perpendiculars in a given ratio is an ellipse of which the given circle is the auxiliary circle.

(3) Parametric Representation : The equations x = a cos θ & y = b sin θ together represent the ellipse

x2 a2

+

y2 b2

= 1.

Where θ is a parameter, if P( θ )  (acos θ , b sin θ ) is on the ellipse then, P( θ )  (acos θ , a sin θ ) is on the auxiliary circle.

The equation to the chord of the ellipse joining two points with eccentric angles  &  is given by

(4)

 

 

 

      x y cos 2 + sin 2 =cos 2 a b

a m r a Position of a Point with respect to an Ellipse : h S .

82 K .

6 L 7 7 58 Er. 8102 x y50  9 as 8  01  1 is posiThe point P(x ,y ) lies outside, inside or on the ellipse according b a  9 3 8 tive, negative or zero respectively.

1

1

Mathematics Concept Note IIT-JEE/ISI/CMI

2 1

2 1

2

2

Page 143

(5) Line and an Ellipse : The line y = mx + c meets the ellipse

x2

a2 according as c2 is < = or > a2m2 + b2.

+

y2 b2

Hence y = mx + c is tangent to the ellipse

x2 a2

= 1 in two points real, coincident or imaginary

+

y2 b2

= 1 if c2 = a2m2 + b2.

(6) Tangents to Ellipse: x 2 y2 (a) Slope form : y = mx  a2m2 b2 is tangent to the ellipse 2 + 2 = 1. a b

(b) Point form :

xx 1 2

a

+

yy 1 b

2

= 1 is tangent to the ellipse

x2 a

2

+

y2 b2

= 1 at (x1, y1).

x cos  y sin  x2 y2 + = 1 is tangent to the ellipse 2 + 2 = 1 at the a b a b point (a cos  ,b sin  ).

(c) Parametric form :

Note: (i) There are two tangents to the ellipse having the same slope m, i.e. there are two tangents parallel to any given direction. These tangents touches the ellipse at extremities of a diameter. (ii) Point of intersection of the tangents to the ellipse at the points         a cos   b sin  2  2     '  ' and '  ' is  ,        cos    cos  2     2  

     

(iii) The eccentric angles of the points of contact of two parallel tangents differ by  .

(7) Normals of Ellipse: (i) Equation of the normal at (x1, y1) to the ellipse

x2 a2

+

y2 b2

= 1 is

a2 x b2y = a2  b2.  x1 y1

(ii) Equation of the normal at the point (acos  ,bsin  ) to the ellipse

x2 a2

+

y2 b2

= 1 is

axsec   bycosec  =(a2  b2).

Note: (i) Atmost four normals can be drawn to an ellipse from a point in its plane. (ii) In general, there are four points A, B, C and D on the ellipse the normals at which pass through a given point. These four points A, B, C, D are called the co-normal points.

ma r a h S .

82 K .

6 L 7 The equation to the pair of tangents which can be drawn any 2 point y )8 to the 7 (x , 5 Er.from 0 ellipse is given by : SS = T where : 1 0 98 8015 xx yy x y x y 39 8 S + + + 1 ; S = 1 ; T  1. a b a b a b

(8) Pair of Tangents :

1

2

1

1

2

2

2

2

1

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

2 1

2 1

2

2

1

2

1

2

Page 144

Note: Locus of the point of intersection of the tangents which meet at right angles is called the Director Circle. The equation to this locus is x2 + y2 = a2 + b2 i.e. a circle whose centre is the centre of the ellipse & whose radius is the length of the line joining the ends of the major and minor axes.

(9) Chord of Contact : Equation to the chord of contact of tangents drawn from a point P(x1, y1) to the ellipse

x2 2

a

+

y2 2

b

= 1 is T = 0, where T =

xx1

+

2

a

yy1 b2

1

(10) Chord with a given middle point :

Equation of the chord of the ellipse

where S1 =

x 21 2

a

+

y 21 2

b

1 ; T

xx1 2

a

x2 a2

+

+

yy1 b2

y2 b2

= 1 whose middle point is (x1, y1) is T = S1,

 1.

(11) Important results: Refering to the ellipse 











x2 a2

+

y2 b2

=1

The tangent & normal at a point P on the ellipse bisect the external & internal angles between the focal distances of P. This refers to the reflection property of the ellipse which states that rays from one focus are reflected through other focus & vice-versa. Hence we can deduce that the straight lines joining each focus to the foot of the perpendicular from the other focus upon the tangent at any point P meet on the normal PG and bisects it where G is the point where normal at P meets the major axis.

The product of the lengths of the perpendicular segments from the foci on any tangent to the ellipse is b2 and the feet of these perpendiculars lie on auxiliary circle and the tangents at these feet to the auxiliary circle meet on the ordinate of P and that the locus of their point of intersection is a similar ellipse as that of the original one.

The portion of the tangent to an ellipse between the point of contact and the directrix subtends a right angle at the corresponding focus.

If the normal at any point P on the ellipse with centre C meet the major and minor axes in G and g respectively and if CF be perpendicular upon this normal then (i) PF. PG = b2 (ii) PF. Pg = a2 (iii) PG. Pg = SP. S'P (iv) CG. CT = CS2 (v) locus of the mid point of Gg is another ellipse having the same eccentricity as that of the original ellipse.

[S and S' are the focii of the ellipse and T is the point where tangent at P meet the major axis] The circle on any focal distance as diameter touches the auxiliary circle. Perpendiculars from the centre upon all chords which join the ends of any perpendicular diameters of the ellipse are of constant length.

If the tangent at the point P of a standard ellipse meets the axes in Q and R and CT is the perpendicular on it from the centre then, (i) QR. PT = a2  b2 (ii) least value of QR is a + b.

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839

Mathematics Concept Note IIT-JEE/ISI/CMI

Page 145

28. HYPERBOLA Hyperbola is the locus of a point which moves is such a ways so that difference of its distances from two fixed points is always constant.

Note: (i) |PS–PS'|=AA'; AA' is the length of transverse axis. (ii) eccentricity (e) > 1.

(1) Basic Definitions: Standard equation of the hyperbola is

x2 2

a



y2 b2

= 1,where b2 = a2(e2  1)

Eccentricity (e) :

e  1

b2 a2

Foci : S  (ae, 0) & S'  (  ae, 0).

Equations Of Directrix : x=

a a and x =  e e

Transverse Axis : The line segment AA' of length 2a in which the foci S and S' both lie is called the transverse axis of the hyperbola.

Conjugate Axis : The line segment BB' between the two points B'  (0,  b) and B  (0, b) is called as the conjugate axis of the hyperbola.

Principal Axes : The transverse and conjugate axis together are called Principal Axes of the hyperbola.

Vertices : A  (a, 0) and A'  (–a, 0) Focal distance: the focal distance of any point (x,y) on the hyperbola

x2 y2   1 are ex– a2 b2

a and ex+a.

Focal Chord : A chord which passes through a focus is called a focal chord.

Double Ordinate : A chord perpendicular to the transverse axis is called a double ordinate.

Latus Rectum (  ) : The focal chord perpendicular to the transverse axis is called the latus rectum.

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 39 y 8 = 1. 

2b2 = 2a (e2  1) = 2e (distance from focus to directrix) a Centre : The point which bisects every chord of the conic drawn through it is called the centre of the conic.

=

C  (0,0) the origin is the centre of the hyperbola Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

x2 a2

2

b2

Page 146

Note: Since the fundamental equation to the hyperbola only differs from that to the ellipse in having  b2 instead of b2 it is found that many propositions for the hyperbola are derived from those for the ellipse by simply changing the sign of b2.

(2) Conjugate Hyperbola : Two hyperbolas such that transverse and conjugate axes of one hyperbola are respectively the conjugate & the transverse axes of the other are called Conjugate Hyperbolas of each other. for example:

x2 a2



y2 b2

= 1 and

x2 a2

+

y2 b2

= 1 are conjugate hyperbolas of each other..

Note: (i) If e and e' are the eccentricities of the hyperbola & its conjugate then 1/e2 + 1/e'2 = 1. (ii) The foci of a hyperbola and its conjugate are concyclic and form the vertices of a square. (iii) Two hyperbolas are said to be similar if they have the same eccentricity. (iv) Two similar hyperbolas are said to be equal if they have same latus rectum. (v) If a hyperbola is equilateral then the conjugate hyperbola is also equilateral.

(3) Auxiliary Circle : A circle drawn with centre C and transverse axis as a diameter is called the Auxiliary Circle of the hyperbola. Equation of the auxiliary circle is x2 + y2 = a2.

In the figure P and Q are called the "Corresponding Points" on the hyperbola and the auxiliary circle.

(4) Parametric Representation : y a m ara b h S 2 38 K.  7, 6  .[0, 2)  7 where  is a parameter, which is termed as eccentric angle,. L  Er 8102  2 520 58 9 point If P(  )  (asec  , btan  ) is on the hyperbola then corresponding 01 8 9 Q(  )  (acos  , asin  ) is on the auxiliary circle.

83 The equations x = a sec  and y = b tan  together represents the hyperbola

x2 2

2



2

=1

The equation to the chord of the hyperbola joining two points with eccentric angles  &  Mathematics Concept Note IIT-JEE/ISI/CMI

Page 147

 

 

 

    x y cos 2  sin 2 = cos 2 . a b

is given by

(5) Position of Point 'P' with respect to Hyperbola :  x12 y12  The quantity S1   2  2  1 is positive, zero or negative according as the point (x1,y1) b   a lies inside, on or outside the curve respectively.

(6) Line And A Hyperbola :

The straight line y=mx+c is a secant, a tangent or passes outside the hyperbola

x2 2

a



y2 b2

=1

according as c2 > or = or < a2m2  b2 respectively..

(7) Tangent of Hyperbola (i) Slope Form : y = mx  a2m2 b2

x2

hyperbola

a2



y2 b2

can

be

taken

xx1

yy1



a2

b2

the

x2 a2



y2 b2

to

the

= 1 at the point (x1, y1)

= 1.

(iii) Parametric Form : Equation of the tangent to the hyperbola point.

tangent

= 1 having slope 'm'.

(ii) Point Form : Equation of tangent to the hyperbola

is

as

(a sec  , b tan  ) is

x2 a2



y2 b2

= 1 at the

x sec  y tan  = 1.  a b

Note:         a cos  2  b sin  2     ,    (i) Point of intersection of the tangents at '  ' and '  ' is        cos  cos      2   2 

(ii) If |    | =  , then tangents at these points '  ' and '  ' are parallel.

(8) Normal of Hyperbola:

ma r a h a b S .

82 K .

6 L 7 Er. 81027 5058 9 801 point P(acos  , btan  ) on the hyperbola 839

(i) The equation of the normal to the hyperbola

is

a2 x b2y + = a2 + b2. x1 y1

(ii) The

equation

x2 2

a



y2 2

b

of the normal at the

= 1 is

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

x2 2



y2 2

= 1 at the point P(x1,y1) on it

ax by + = a2 + b2 or sec  tan

Page 148

     

(9) Pair of Tangents : The equation to the pair of tangents which can be drawn from any point (x1,y1) to the

x2

hyperbola

S 

x2 2

a





a2 y2

y2

= 1 is given by : SS1 = T2 where :

b2

1

2

b

;

S1 =

x 21 2

a

y 21



1

2

b

; T 

xx1 a2



yy1 b2

 1.

Note: The locus of the intersection point of tangents which are at right angles is known as the Director Circle of the hyperbola. The equation to the director circle is : x2 + y2 = a2  b2.

If b2 < a2 this circle is real.

If b2 = a2 (rectangular hyperbola) the radius of the circle is zero & it reduces to a point circle at the origin. In this case the centre is the only point from which the tangents at right angles can be drawn to the curve.

If b2 > a2, the radius of the circle is imaginary, so that there is no such circle & so no pair of tangents at right angle can be drawn to the curve.

(10) Chord of Contact : Equation to the chord of contact of tangents drawn from a point P(x1,y1) to the hyperbola

x2 2

a



y2 2

b

= 1 is T = 0, where T 

xx1



2

a

yy1 b2

1

(11) Chord with a given middle point : Equation of the chord of the hyperbola

where S1 

x 21 2

a



y 21 2

b

1 ; T

xx1 2

a



yy1 b2

x2 a2



y2 b2

= 1 whose middle point is (x1, y1) is T = S1,

 1.

(12) Diameter : The locus of the middle points of a system of parallel chords with slope 'm' of a hyperbola is called its diameter. It is a straight line passing through the centre of the hyperbola and has the equation y = 

b2x a2m

, all diameters of the hyperbola passes through its centre.

(13) Asymptotes : Definition : If the length of the perpendicular let fall from a point on a hyperbola to a straight line tends to zero as the point on the hyperbola moves to infinity along the hyperbola, then the straight line is called the Asymptote of the hyperbola.

ma r a Equations of Asymptote : h S .

82 K .

6 L 7 x b x y x y r. 1027 058 + = 0 and =0  =0   E a b a b a a 98 8015 Note: 39 8 (i) A hyperbola and its conjugate have the same asymptote.

2

2

2

2

(ii) The equation of the pair of asymptotes differs from the equation of hyperbola Mathematics Concept Note IIT-JEE/ISI/CMI

Page 149

(or conjugate hyperbola) by the constant term only. (iii) The asymptotes pass through the centre of the hyperbola & are equally inclined to the transverse axis of the hyperbola. Hence the bisector of the angles between the asymptotes are the principle axes of the hyperbola. (iv) The asymptotes of a hyperbola are the diagonals of the rectangle formed by the lines drawn through the extremities of each axis parallel to the other axis.

(14) Rectangular Or Equilateral Hyperbola : The hyperbola in which the lengths of the transverse & conjugate axis are equal is called an Equilateral Hyperbola, the eccentricity of the rectangular hyperbola is Rectangular Hyperbola (xy = c2) :

2.

It is referred to its asymptotes as axis of co-ordinates.

Vertices: (c, c) and (  c,  c) ; Foci: ( 2 c,

2 c) and (  2 c,  2 c)

Directrix: x + y = 

2c

Latus Rectum:  = 2 2 c Parametric equation x = ct, y = c/t , t  R  {0}

Note: (i) Equation of a chord joining the points P(t1) and Q(t2) is x + t1t2y = c(t1+t2). x

y

x

(ii) Equation of the tangent at P(x1,y1) is x + y = 2 and at P(t) is +ty = 2c. t 1 1 (iii) Equation of the normal at P(t) is xt3  yt = c(t4  1). (iv) Chord with a given middle point as (h,k) is kx + hy = 2hk.

(15) Important Results : 





Locus of the feet of the perpendicular drawn from focus of the hyperbola



Light Ray

The ellipse

x2

+

y2

YTangent Q  P

= 1 and the hyperbola

Xa S rm a x y h S = 1 (a > k > b > 0) are confocal  .

82 K a k k b .

6 L 7 and therefore orthogonal.

Er. 81027 5058 9tangent8meets The foci of the hyperbola and the points P and Q in which any 01 the tangents at the vertices are concyclic with PQ as diameter of the circle. 39 8 If from any point on the asymptote a straight line be drawn perpendicular to the transverse 2

2



y2

= 1 upon a2 b2 any tangent is its auxiliary circle i.e. x2+y2=a2 and the product of these perpendiculars is b2.

The portion of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus.

The tangent & normal at any point of a hyperbola bisect the angle between the focal radii. This spells the reflection property of the hyperbola as "An incoming light ray" aimed towards one focus is reflected from the outer surface of the hyperbola towards the other focus. It follows that if an ellipse and a hyperbola have the same foci, they cut at right angles at any of their common point.

Note:



x2

2

a

b2

S’

2

2

2



2

axis, the product of the segments of this line, intercepted between the point & the curve is Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 150



always equal to the square of the semi conjugate axis.

Perpendicular from the foci on either asymptote meet it in the same points as the corresponding directrix & the common points of intersection lie on the auxiliary circle.



The tangent at any point P on a hyperbola

x2 a2



y2 b2

= 1 with centre C, meets the asymp-

totes in Q and R and cuts off a  CQR of constant area equal to ab from the asymptotes & the portion of the tangent intercepted between the asymptote is bisected at the point of contact. This implies that locus of the centre of the circle circumscribing the  CQR in case of a rectangular hyperbola is the hyperbola itself & for a standard hyperbola the locus would be the curve: 4(a2x2 + b2y2) = (a2 + b2)2.

x2

y2



If the angle between the asymptote of a hyperbola



of the hyperbola is sec  .

A rectangular hyperbola circumscribing a triangle also passes through the orthocentre of 

this triangle. If  ct1 , 

a2



b2

= 1 is 2  then the eccentricity

c   i = 1, 2, 3 be the angular points P, Q, R then orthocentre is t 1 

 c  , ct1 t 2 t 3  .   t1t 2 t3  

If a circle and the rectangular hyperbola xy = c2 meet in the four points t1, t2, t3 & t4,then (i) t1 t2 t3 t4 = 1 (ii) the centre of the mean position of the four points bisects the distance between the centres of the two curves. (iii) the centre of the circle through the points t1, t2 & t3 is: 1  c  c   2  t1  t2  t3  , t1t2 t3  2  



1 1 1    t1  t 2  t3 t1 t2 t3

 

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29. VECTORS (1)

Basic Definitions : Vector is a physical quantity having magnitude and definite direction. A directed line seg ment AB represents a vector having initial point 'A' and the terminal point 'B'.   Magnitude of vector AB is represented by | AB |.

In 3-dimensional space if a point P is represented by (x1 , y1 , z1) , then position vector of 'P'   If P(x , y , z ) and Q(x , y , z ) are two points in space , is given by OP  x1i  y1j  z1k.

1 1 1 2 2 2 then :    and OQ  x i  y j  z k  OP  x1i  y1j  z1k 2 2 2  PQ = (position vector of Q)–(position vector of P)    (z  z )k   PQ  (x2  x1 )i  (y2  y1 )j 2 1   | PQ |  (x2  x1 )2  (y2  y1 )2  (z2  z1 )2

(i) Zero Vector/Null Vector : A vector having zero magnitude and indeterminate direction is termed as zero vector.   For example : AA  0 (i.e. vector having same initial and terminal point). (ii) Unit Vector :

 A vector having unit magnitude is termed as unit vector. Unit vector in the direction of a is

represented by  a.    a   a  or a | a |  a. | a| (iii) Parallel Vector or Collinear Vectors : A set of vectors is termed as parallel (or collinear) vectors if they have common line of support or parallel line of support.  Parallel vectors are termed as like vectors if they have same direction and further if like vectors have same magnitude , then vectors are termed as equal vectors.  Parallel vectors are termed as unlike vectors if they have opposite sense of direction and   negative of vector a is represented by – a

ma r a h S .

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       Vectors a , b , c , d , e and f      Vectors a , e , d and f are like vectors.

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     Vectors a and b are unlike vectors , similarly b and d are unlike vectors.     Let a and b are two non-zero collinear vectors , then a  b , where   R .   I f a(a1 , a2 , a3 ) and b (b1 , b2 , b3 ) are col l i n ear or paral l el vect ors , t hen .    (b i  b j  b k)  (a1i  a2j  a3 k) 1 2 3



a1 a a  2  3. b1 b2 b3

Note:

   Three points A(a) ,B(b) and c(c) are collinear , if there exists scalars x , y and z such that     xa  yb  zc  0 , where x + y + z = 0 and x , y , z are not simultaneously zero.

(iv) Coplanar Vectors : Vectors which lie one the same plane of support or parallel plane of support are termed as coplanar vectors.

Note:

    Two vectors are always coplanar and if three vectors a , b and c are coplanar then any y one of the vector can be expressed as a linear combination of other tow vectors.

       If a , b and c are coplanar vectors , then (a  b).c  0     If a(a1 , a2 , a3 ) , b (b1 , b2 , b3 ) and c (c1 , c2 , c3 ) are coplanar , then

a1 a2 a3    a b c   b b b  0 1 2 3   c1 c2 c3

ma r a    h  2 S .  Three points in space are always coplanar and if A(a) , B(b) are four 8 , C(c) and6 D(d) K .

L 7 7 .     r 8 002and x5,0 coplanar points , then xa  yb  zc  wd  0 , where x +E y + z + w1= y ,5 z , w are 8 9 801 scalars which are not simultaneously zero.

39     8   If A , B , C and D are four point lying on a plane , then AB AC AC  0 . 

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(v) Co-initial Vectors : Vectors which are having same initial points are termed as co-initial vectors.    For example : AB , AC , AD are coinitial vectors. (vi) Coterminous Vectors : Vectors which have same terminal point are termed as coterminous vectors.    For example : PA , QA ,RA are coterminous vectors.

(2)

Linear Combination of Vectors :  Linearly Independent Vectors :    A set of vectors a1 , a2 , ..., an is said to be linearly independent iff    x1 a1  x2 a2  ...  xn an  0  x1  x2  ...  xn  0  Linearly Dependent Vectors:    A set of vectors a1 , a2 , ..., an is said to be linearly dependent iff there exists scalars x1 , x2 , ... , xn not all zero such that    x1 a1  x2 a2  ...  xn an  0

Note:

     If two non-zero vectors a and b are linearly dependent , then a and b are collinear or   parallel (  a  b  0 )      If two non-zero vectors a and b are linearly independent , then a  b  0 and   xa  yb  0  x  y  0.   0  x  y  0.

For example : xi  yj     If three non-zero vectors a , b and c are linearly dependent , then any one of the

   vector can be other two vectors which further implies that a , b and c are coplanar..    (  [a b c]  0)         If three non-zero vectors a , b and c are linearly independent , then a b c   c and    hence the vectors a , b and c are non-coplanar..

 Any set of four or more non-orthogonal system of vectors is always linearly dependent

(3)

Addition of Vectors :

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6 L 7 Er. 81027 5058 9 801     8 3c)9 (a  b)  c  a  (b

      If the vectors a and b are represented by AB and AD , then a  b is the vector addition  which is represented by AC , where AC is the diagonal of parallelogram ABCD.

Note:

     ab  b a       a o  o a  a

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

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    a  (a)  o

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Triangle inequality:      | a  b |  | a|  |b |

     | a  b |  | a|  |b |

  Sign of equality holds if a and b are like vectors.        | a  b |  | a  b |  Angle between a and b is 90o.    |a b| 

(4)

    | a |2  | b |2  2 | a | | b | cos 

Multiplication of a Vector by Scalar :   Scalar multiplication of vector a and scalar quantity k , where K  R , is represented by Ka .   Ka is vector which is parallel to a and having magnitude |K| times the magnitude of vector  a.

Note: If P , K  R , then     K(a)  (a)K  K(a)     (K  P)a  Ka  Pa

(5)

    K(Pa)  P(Ka)  PK(a)      K(a  b)  Ka  Kb.

Basic Application of Vectors to Geometry:   (i) If a and b are the position vectors of two points A and B , then position vector of points  A and B , then position vector of point C (r) which divides AB in the ratio m : n is given by    mb  ba r mn    mb  ba r mn

(for internal division). (for external division).

   (ii) If A(a) , B(b) and C(c) are the vertices of ABC , then :

   ab c  P.V. of Controid  3          |b  c | a  | c  a| b  | a  b | c  P.V. of incentre        | a  b |  |b  c |  | c  a|     (iii) If A(a) , B(b) , C(c) and D(d) are the vertices of a tetrahedron , then P.V. of centroid of      ab c d tetrahedron is given by . r  4

(6)

ma r a h S .

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6 L 7 Scalar or Dot Product : Er. 81027 5058 9 801     9 Scalar product of two vector a and b is denoted by a.b and8 is3 defined as :       a.b | a | | b | cos  , where  is the angle (0    ) between vectors a and b .

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 In OBL , OL | b | cos        a.b     a.b    OL   (Projection of b on a )  OL    2  a and LB  b | a |  | a|    Similarly , In OAM , OM  | a | cos    a.b    OM   (Projection of a on b ) |b|     a.b  OM     | b |2 

     a.b   b and MA  a      | b |2  

  b.  

Properties of dot Product:  



i.i  j.j   k.k   1

i.j  j.k   k.i   0    and b   b i  b j  b k  , then a.b  a b  a b  a b .

If  a  a1i  a2j  a3 k 1 2 3 1 1 2 2 3 3   Angle  between vectors a and b is given by : 

 a.b

  cos      | a||b |  

a1b1  a2b2  a3b3



a12  a22  a32

 b

2 1

 b22  b32



    

If a1b1  a2b2  a3b3  0, then  is Acute angle.

If a1b1  a2b2  a3b3  0, then  is obtux angle.

If a1b1  a2b2  a3b3  0, then  is 90o.

If

a1 a a  2  3 , then   0 or . b1 b2 b3

        a.a | b |2  a.b  b.a          (b.j)j    (c.k)k a  a.(b  c)  a.b  a.c  (a.i)i             | a || b |  a.b  | a || b |  | a  b |2  | a |2  | b |2  2 a.b           | a  b  c |2 | a |2  | b |2  | c |2 2(a.b  b.c  c.a)

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Vector or Cross Product :     Vector product of two vectors a and b is denot ed by a  b an d is defin ed as:         is the unit a  b | a || b | sin  n , where  is the angle between vector a and b , and n   vector perpendicular to both a and b . (direction of  n is determined by right handed screw





rule)

  Geometrically | a  b | represents the area of parallelogram whose two adjacent sides are   represented by a and b

Properties of Cross Product :   0  i i  j  j  k.k    a a  0

 , j  k   i and k  i  j  i  j  k

i      and b  b i  b j  b k  , then a  b  a  If a  a1i  a2j  a3k 1 2 3 1

j

 k

a2

a3

b1 b2 b3      a  b  b  a      a  (kb)  k(a  b)

        a  (b  c)  (a  b)  a  c      a  b  0  a and b are parallel

  a b     Unit vector perpendicular to both a and b is given by : n     | a b|       | a  b |2 (a.b)2  | a |2 | b |2 (Lagrange ' s Idantity)     If A(a) , B(b) and C(c) are the vertices of ABC , then area of

1            a  b  c  c  a a  b  b  c  c  a  0  A , B and C are collinear.

2      If A(a) , B(b) , C(c) and D(d) are the vertices of a quadrileteral , then area of Q uad ABC 

(ABCD)  

1         a b  b  c  c  d  d a 2 1   AC  BD 2

Mathematics Concept Note IIT-JEE/ISI/CMI

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8.

Scalar Triple Product :          The scalar triple product of three vectors a , b and c is defined as : a  b.c  | a || b || c |      sin θ cos  where θ is the angle between a and b and  is the angle between a  b and c .    It is also written as [a b c]

Scalar triple product geometrically represents the volume of the parallelopiped whose three coterminous edges are       represented by a , b and c i.e. v  [a b c]

Note: a1 a2 a3       (i) If a  a1i  a2 j  a3k ;b  b1i  b2 j  b3k and c  c1i  c2 j  c3k then [a b c]  b1 b2 b3 c1 c2 c3

(ii) In a scalar triple       a .(b  c)  (a  b).c or

product the position of dot and cross can be interchanged i.e.          a b c   b c a  c b a                    (iii) a.(b  c)  a,(c  b) i.e. [a b c]  [a c b]       (iv) If a , b , c are coplanar  [a b c]  0. (v) Scalar product of three vectors, two of which are equal or parallel is 0.          (vi) If a , b , c are non-coplanar then [a b c]  0 for right handed system and [a b c]  0       for left handed system [K a b c]  [K a b c] (vii) The volume of the tetrahedron OABC with O as origin and the pv's of A, B and C being    1    a , b and c respectively is given by V = 6 [a b c]

The position vector of the centroid of a tetrahedron if the pv's of its vertices are        1  a , b , c and d are given by [a  b  c  d] 4

ma r a h vertices2 to the Note that this is also the point of concurrency of the lines joining the S .

K . tetrahedron.

68In case the centroids of the opposite faces and is also called the centre ofL the 7 7 . r the four 8 tetrahedron is regular it is equidistant from the vertices faces of the5tetraheEand 02 1 0 5 8 dron.

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Vector Triple Product :       Vector quantity a  (b  c) or (a  b)  c is termed as the vector triple product for any three    vectors a , b and c .                a  (b  c)  (a.c)b  (a.b)c  (a  b)  c  (a.c)b  (b.c)a         Let a  (b  c)  R , then by definition of cross product , R is perpendicular to both a and b  c .       a.R  0 and R.(b  c)  0       R is normal to a and R , b and c are coplanar.      a  (b  c) represents a vector which is normal to a and lies in the plane containing   b and c .

Note:      a  (b  c)     Unit vector normal to a and lying in plane of b and c    | a  (b  c) |        In general , a  (b  c)  (a  b)  c.

(10) Scalar Product of four Vectors :         If a , b , c , d are four vectors , the products (a  b).(c  d) is called scalar products of four vectors.         a.b b .c i.e. , (a  b).(c  d)      a.d b .d

(11) Vector product of four Vectors :         If a , b , c , d are four vectors , the products (a  b)  (c  d) is called scalar products of four vectors.         i.e. , (a  b)  (c  d)  [ab d]c  [ab c]d         (a  b)  (c  d)  [a c d]b  [a c c]a.

(12) Reciprocal System of Vector :  If a  a' ,  a.b '

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839

     , b , c are three non coplanar vector such that [a b c]  0, then the system of vectors   b ' , c whi ch sat isfy t h e con dit i on t h at a.a' = b.b' = c.c' = 1 an d       a.c '  b.a'  b.c '  c.a '  c.b '  0.

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   bc a'     [ab c]

      ca ab b '     and c '     [ab c] [ab c]

Properties of reciprocal system     (a) [ab c] [a'b ' c]  1         b ' c ' c ' a'  a' b ' b    c    (b) a     [a'b ' c '] [a'b ' c '] [a' b ' c ']

 is its own reciprocal (c) The system of unit vectors i , j , k '  k  i.e. i' i j'  j and k    Not that any vector can be expressed in terms of a' , b ' and c ' as they also constitute a system of non coplanar vectors.

ma r a h S .

82 K .

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30. THREE DIMENSIONAL GEOMETRY Coordinates of any point 'P' in the space is given by (x, y, z) and the point is located with reference to three mutually perpendiculer corordinates axes ox, oy and oz.

 Position vector of point P is given by OP  xi  yj  zk .

(1)

Distance formula : Distance between any two points A(x1,y1,z1) and B(x2,y2,z2) is given by:

(x1  x2 )2  (y1  y2 )2  (z1  z2 )2

Note:

  If position vector of two points A and B are given as OA and OB , then 



AB = | OB  OA |

(2)



AB = |(x2i + y2j + z2k)  (x1i + y1j + z1k)|



AB =

(x1  x2 )2  (y1  y2 )2  (z1  z2 )2

Section Formula : If point P divides the distance between the points A(x1, y1, z1) and B(x2, y2, z2) internally in the ratio of m : n , then coordinates of P are given by :  mx2  nx1 my2  ny1 mz2  nz1  , ,   mn mn   mn

(3)

Centroid of a Triangle ABC :  x  x2  x3 y1  y2  y3 z1  z2  z3  G 1 , ,  3 3 3  

(4)

Incentre of Triangle ABC :

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 39  cz  8

 ax  bx2  cx3 ay1  by2  cy3 az1  bz2 I 1 , , ab c ab c ab c 

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(5)

Centroid of a tetrahedron : A (x1, y1, z1) , B(x2, y2, z2) , C (x3, y3, z3) and D(x4, y4, z4) are the vertices of a tetrahedron then coordinate of its centroid (G) is given by :  x  x2  x3  x 4 y1  y2  y3  y 4 z1  z2  z3  z 4  G 1 , ,  4 4 4  

(6)

Direction Cosines And Direction Ratios : Direction cosines: Let α , β , γ be the angles which a directed line makes with the positive directions of the axes of x, y and z respectively, then cos α , cos β , cos γ are called the direction cosines of the line. The direction cosines are usually denoted by (  , m, n).



  cos  , m  cos  , n  cos .

Note: (i) If  , m, n be the direction cosines of a line, then  2 + m2 + n2 = 1. (ii) If  , m, n are direction cosines of line L then  ˆi  mˆj  nkˆ is a unit vector parallel to the line L. Direction ratios : Let a, b, c be proportional to the direction cosines  , m , n then a, b, c are called the direction ratios. 

 m m   a b c

Note: (i) If  , m, n be the direction cosines and a, b , c be the direction ratios of a vector,, then  a b c   ,m  ,n   a2  b2  c2 a2  b2  c2 a2  b2  c2 

   

(ii) If the coordinates of P and Q are (x1, y1, z1) and (x2, y2, z2) respectively, then the direction ratios of line PQ can be assumed as: a = x2  x1, b = y2  y1 and c = z2  z1 , and the direction cosines of line PQ are

 = 

x 2  x1 |PQ|

, m= 

y 2  y1

and n = 

|PQ|

z2 z1 |PQ|

.

(iii) If a, b, c are the direction ratios of any line L then a ˆi  bˆj  ckˆ is a vector parallel to the line L. (iv) Direction cosines of axes: Direction cosines of x-axis are (1, 0,0) Direction cosines of y-axis are (0,1,0) Direction cosines of z-axis are (0,0,1)

(7)

ma r a h S .

82 K .

6 L 7 Angle Between Two Line Segments : Er. 81027 5058 9 8 If two lines have direction ratios a , b , c and a , b , c respectively then 01vectors parallel 9 to the lines are a i + b j + c k and a i + b j + c k and angle between them is given as : 83 1

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cos  

a1a2  b1b2  c1c2 a12  b12  c12 a22  b22  c22

Note: (i) The line will be perpendicular if a1a2 + b1b2+ c1c2 = 0 a1 b1 c1 (ii) The lines will be parallel if a  b  c 2 2 2

(iii) Two parallel lines have same direction cosines i.e.  1 =  2 , m1 = m2 , n1 = n2.

A LINE (8)

Equation Of A Line (i) A straight line in space is characterised by the intersection of two planes which are not parallel and therefore, the equation of a straight line is a solution of the system constituted by the equations of the two planes, a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0.

This form is also known as non-symmetrical form. (ii) The equation of a line passing through the point (x1, y1, z1) and having direction ratios a, b, c is

x  x1 y  y1 z  z1     . This form is called symmetric form. A a b c

general point on the line is given by (x1  a , y1  b , z1  c) , where  is a scalar quantity (iii) Vector equation: Vector equation of a straight line passing through a fixed point with      position vector a and parallel to a given vector b is r  a  b , where λ is a scalar quantity. (iv) The equation of the line passing through the points (x1, y1, z1) and (x2, y2, z2) is x  x1 y  y1 z  z1    x2  x1 y2  y1 z2  z1

(v) Vector equation of a straight line passing through two points with position vectors       a and b is r  a  (b  a) . (vi) Reduction of cartesion form of equation of a line to vector form & vice versa x  x1 y  y1 z  z1   a b c



   (ai  bj  ck)  r  (x1i  y1 j  z1k)

Note: (a) If lines

x  x1 y  y1 z  z1   a b c

and

x  x2 y  y2 z  z2   intersect each others at a' b' c'

ma r a h a b c 0 unique point, then : S .

82 K .

6 a' b' c' L 7 r. 1027 058 E       98  b0 .1a5 a   0. (b) If lines r  a  b and r  a  b are intersecting, then : b 8 839 x2  x1 y2  y1 z2  z1

1

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(9)

Foot, Length Of perpendicular From A point To A Line : Let equation of the line be

x  x1 y  y1 z  z1    a b c

.....(i)

and A(x' , y' , z') be the point.

Any point on line (i) is P (a  x1 , b  y1 , c  z1 )

....(ii)

If P is the foot of the perpendicular from A on the line, then AP is perpendicular to the line.

(a  x1  x ')a  ( b  y1  y ')b  ( c  z1  z ')c  0



From above equation, value of '  ' is obtained and putting this value of  in (ii), we get the foot of perpendicular from point A on the given line , since foot of perpendicular P is known, then the length of perpendicular is given by AP.

(10) To find image of a point w.r.t a line :

Image of point P(x' , y' , z') w.r.t. line

x  x1 y  y1 z  z1   . a b c

Let image of P(x' , y' , z') be Q( ,  , ) , hence PQ is perpandiclor to the given line a(  x ')  b(  y ')  c(r  z ')  0



......(i)

Now mid-point of PQ lies on the line   x'   y' r  z'  x1  y1  z1 2 2   2  a b c



......(ii)

from above equation, point Q( ,  , ) can be calculeted in terms of  , which further satisfy the equation (i) and hence value of '  ' can be obtained After getting the value of '  ' point Q( ,  , ) cam ne calculeted from equation (ii).

(11) Skew Lines : The straight lines which are non-parallel and non-intersecting are termed as skew lines.

If

x  x1 y  y1 z  z1   a b c

and

x  x2 y  y2 z  z2   a' b' c'

are

skew-li n es

,

th en

x2  x1 y2  y1 z2  z1 a

b

c

a'

b'

c'

0

and the shortest distance (S.D.) between the skew-lines is

x2  x1 a

y2  y1 z2  z1 b c

a m r a given by : S.D.  . h S .

82 K . (bc ' b ' c)  (ac ' a' c)  (ab ' a'b) 6 L 7 Er. 81027 5058 Note: 9 801       (i) If r  a  b and r  a  b are the skew-lines in vector 39form, then 8     a'

b'

2

1

b

1

1



c'

2

2

2

2



 b2 . a2  a1  0 and the shortest distance between them is given by :

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    b1 b2 a2  a1      S.D.  | b1  b2 |

(ii)

   Shortest distance 'd' between the two parallel lines r  a1  λb and 





   (a2  a1 )  b  . r  a2  μb is given by : d = |b |

A PLANE If line joining any two points on a surface is perpendicular to some fixed straight line , then this surface is called a plane and the fixed line is called the normal to the plane.

(12) Equation Of A Plane (i) Normal form of the equation of a plane is  x + my + nz = p, where,  , m, n are the direction cosines of the normal to the plane and p is the distance of the plane from the origin. (ii) General form : ax + by + cz + d = 0 is the equation of a plane, where a, b , c are the direction ratios of the normal to the plane.

(iii) The equat ion of a plane passing t hrough t he point (x 1, y1, z1) is given by a(x  x1) + b(y  y1) + c(z  z1) = 0 where a, b, c are the direction ratios of the normal to the plane. (iv) Plane through three points: The equation of the plane through three non-collinear points(x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is given by : x  x1

y  y1

z  z1

x2  x1 y2  y1 z2  z1  0 x3  x1 y3  y1 z3  z1

(iv) Intercept Form : The equation of a plane cutting intercept a, b, c on the axes is

x y z   = 1. a b c (v) Vector form : The equation of a plane passing through a point having position vector          a normal to vector n is (r  a).n  0 or r .n  a.n (vi) Vector equation of a plane normal to unit vector n and at a distance d from the origin   is r .n  d

Note: (a) Coordinate planes (i) Equation of yz-plane is x = 0 (ii) Equation of xz-plane is y = 0 (iii) Equation of xy-plane is z = 0.

ma r a h S .

82 K (b) Planes parallel to the axes : .

6 L 7 r.of the10plane 27 parallel If a = 0, the plane is parallel to x-axis i.e. equation 58to the E 0 5 8 x-axis is by + cz + d = 0.

1 9 8l0 S i mi l arl y, equ at i on of pl anes parall el to y-axi s an d paral el t o z-axi s are 9 83 ax + cz + d = 0 and ax + by + d = 0 respectively. (c) Plane through origin : Equation of plane passing through origin is ax + by + cz = 0.

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(d) Any plane parallel to the given plane ax + by + cz + d = 0 is ax + by + cz + k = 0.

Distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2=0 | d1  d2 |

is given as

a2  b2  c2

(e) Equation of a plane passing through a given point & parallel to the given vectors:  The equation of a plane passing through a point having position vector a and       parallel to b and c is r  a  b  c (parametric form) where λ and μ are scalars.       or r . (b  c)  a.(b  c) (non parametric form) (f) A plane ax + by + cz +d = 0 divides the line segment joining (x1, y1, z1) and  ax1  by1  cz1  d  (x2, y2, z2) in the ratio   ax  by  cz  d  2 2 2  

(g) The xy-plane divides the line segment joining the points (x1, y1, z1) and (x2, y2, z2) in the ratio 

x z1 y . Similarly yz-plane in  1 and zx-plane in  1 x2 z2 y2

(h) Coplanarity of four points The points A(x1 , y1 , z1), B(x2 , y2 , z2) C(x3 , y3 , z3) and D(x4 , y4 , z4) are coplanar , then x2  x1

y2  y1

z2  z1

x3  x1

y3  y1

z2  z1  0

x4  x1 y4  y1 z4  z1

(13) Volume Of A Tetrahedron : Volume of a tetrahedron with vertices A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and D(x4, y4, z4) is given by V 

x2  x1 y2  y1 z2  z1 1 x3  x1 y3  y1 z2  z1 6 x4  x1 y4  y1 z4  z1

(14) Sides of a plane: A plane divides the three dimensional space in two equal parts.Two points A (x1 , y1 , z1) and B(x2 , y2 , z2) are on the same side of the plane ax + by + cz + d = 0 if ax1 + by1 + cz1 + d and ax2 + by2 + cz2 + d are both positive or both negative and are opposite side of plane if both of these values are in opposite sign.

(15) A Plane and A Point :

ma r a h S . ax  by  cz  d 82 K .

6 p L 7 . a b c Er. 81027 5058 9 80 1 (ii) The length of the perpendicular from a point having position3 9 a to plane 8 vector

(i) Distance of the point (x1 , y1 , z1) from the plane ax + by + cz + d = 0 is given by : 1

1

2

1

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  r .n  d is given by:

  a.n  d  p |n|

(iii) The coordinates of the foot of perpendicular from the point (x1 , y1 , z1) to the plane ax + by + cz + d = 0 are given by : x  x1 y  y1 z  z1  (ax1  by1  cz1  d)       a b c a2  b2  c2  

(iv) Reflection of a point w.r.t. a plane.

The coordinate of the image of point (x1 , y1 , z1) w.r.t. the plane ax + by + cz + d = 0 are given by : x  x1 y  y1 z  z1  (ax1  by1  cz1  d)     2   a b c a2  b2  c2  

(16) Angle Between Two Planes : (i) If two planes are ax + by + cz + d = 0 and a'x + b'y + c'z + d' = 0, then angle between these planes is the angle between their normals. Since direction ratios of their normals are

a,b , c and a', b ', c ' respectively, hence , '  ' the angle betweeen cos  

them, is given by

aa'  bb'  cc' a2  b2  c2 a'2  b'2  c'2

Planes are perpendicular if aa' + bb' + cc' = 0 and planes are parallel if

a b c = = a' b' c'

    (ii) The angle θ between the planes r .n1  d1 and r . n2  d2 is given by, y,    n1 .n2    cos      | n |.| n |  2   1

    Planes are perpendicular if n1 .n2  0 and planes are parallel if n1   n2 .

(17) Angle Between A Plane And A Line : (i) If θ is the angle between line

x  x1 y  y1 z  z1   and the plane  m n

 a  bm  cn  ax + by + cz + d = 0 , then sin θ =  2 2 2 2 2 2  (a  b  c )   m  n

  . 

     (ii) Vector form : If θ is the angle between a line r  a  b and r .n  d, then    b.n  sin       | b || n |

ma r a h S .

82 K .

6  L 7 m n   .

27 058 (iii) Condition for perpendicularity : = = or b E n r0 0 a 1 b c 8 015   9 (iv) Condition for parallel case : a  + bm + cn = 0 or b.n  0 398 8 Mathematics Concept Note IIT-JEE/ISI/CMI

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(18) Condition For A Line To Lie in A Plane : x  x1 y  y1 z  z1   will lie in plane ax + by + cz + d = 0,  m n if ax 1 + by1 + cz1 + d = 0 and a  + bm + cn = 0. holds simultanously          (ii) Vector form :Line r  a  λb will lie in plane r .n  d if b.n  0 and a.n  d holds simultanously.

(i) Cartesian form: Line

(19) Angle Bisectors : (i) The equations of the planes bisecting the angle between two given planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are : a1x  b1y  c1z  d1



a12  b12  c12

a2x  b2y  c2z  d2 a22  b22  c22

(ii) Equation of bisector of the angle containing origin: If both the constant terms d1 and d2 are positive ,then the positive sign in a1x  b1y  c1z  d1 a12



b12



c12



a2x  b2y  c2z  d2 gives the bisector of the angle which contains

a22  b22  c22

the origin. (iii) Bisector of acute/obtuse angle: First make both the constant terms positive , then a1a2 + b1b2 + c1c2 > 0  origin lies on obtuse angle a1a2 + b1b2 + c1c2 < 0  origin lies in acute angle

(20) Reduction Of Non-Symmetrical Form To Symmetrical Form : Let equation of the line in non-symmetrical form be a1x + b1y + c1z + d1 = 0, a2x + b2y + c2z + d2 = 0. To find the equation of the line in symmetrical form, we must know (i) its direction ratios (ii) coordinate of any point on it.

Direction ratios : Let  , m, n be the direction ratios of the line. since the line lies in both the planes, it must be perpendicular to normals of both planes. So a1  + b1m + c 1 n = 0 , a2  + b2m + c2n = 0. From these equations, proportional values of  , m, n can be calculated by cross - multiplication as

 m n   b1c2  b2c1 c1a2  c2a1 a1b2  a2b1

Alternative method i j The vector a1 b1 a2 b2

k c1

= i(b1c2–b2c1) + j(c1a2–c2a1) + k(a1b2–a2b1) will be parallel to the

c2

line of intersection of the two given planes.

ma r a h S Point on the line : As  , m, n cannot be zero simultaneously, so at least one coordinate 2 .

8 K 6 amust be non-zero. Let a b –a b  0, then the line cannot L be.parallel7 to7xy plane, so it intersect it. Let it intersect xy-plane in (x , y , 0). Then 0 an d Era.x 8+1b0y2 + 5d 0=58 a x + b y + d = 0. solving these, we get a point on the line. equation becomes.

1 9Then its 80 9 3 xx yy z0 8 = =   : m : n = (b1c2–b2c1) : (c1a2–c2a1) : (a1b2 –a2b1)

1

2

2

1

1

2

1

2

1

1

1

1

1

1

1

2

1

b1c2  b2 c1

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

1

c1a2  c2a1

a1b2  a2b1

Page 168

If   0, take a point on yz-plane as (0, y1, z1) and if m  0, take a point on xz-plane as (x1, 0, z1).

(21) Family of Planes : (i) Any plane passing through the line of intersection of non-parallel planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given by : (a1x + b1y + c1z + d1) + λ (a2x + b2y + c2z + d2) = 0 (ii ) The equ at i on of pl an e passin g t h rou gh t h e i nt ersect ion of t h e pl anes        r .n1  d1 and r . n2  d2 is given by r .(n1  n2 )  d2  d1 where λ is arbitrary scalar quantity.

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics Concept Note IIT-JEE/ISI/CMI

Page 169

31. TRIGONOMETRIC RATIO AND IDENTITIES An angle is the amount of revolution which a line OP revolving about the point O has undergone in passing from its initial position OA into its final position OB.

If the rotation is in the clockwise sense, the angle measured is negative and it is positive if the rotation is in the anti-clockwise sense.

The two commonly used systems of measuring an angle are 1. Sexagesimal system is which 1 right angle = 90 degrees (90º) 1 degree = 60 minutes (60') 1 minute = 60 seconds (60") 2. Circular systems in which the unit of measurement is the angle subtended at the centre of a circle by an are whose length is equal to the radius and is called a radian.

Relation between degree and radian  radian = 180 degrees 1 radian =

180 degress 

= 57º 17' 45" (approximately) The triagonometrical ratios of an angle are numberical quantities. Each one of them represent the ratio of the length of one side to another of a right angled triangle.

1.

3.

Basic Trigonometric Identities : (a)

sin2  + cos2  = 1

;

 1  sin   1;  1  cos   1    R

(b)

sec2   tan2  = 1

;

   |sec  |  1    R   2 n  1  2 , n  I   

(c)

coses2   cot2  = 1 ;

|cosec  |  1    R  {n  , n  I}

Trigonometric Functions Of Allied Angles : If  is any angle, then   , 90   , 180   , 270   , 360   etc. are called ALLIED ANGLES. (a)

sin(   ) =  sin 

;

cos(   ) = cos 

(b)

sin(90o   ) = cos 

;

cos(90o   ) = sin 

(c)

sin(90o +  ) = cos 

;

(d)

sin(180o   ) = sin 

;

(e)

sin(180o +  ) =  sin 

;

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

ma r a h S 2 .

K cos(90 + L )=  sin  768 .

Er. 81027 5058 cos(180 9  ) =  cos0 1 8 9 3 cos(180 +8  ) =  cos  o

o

o

Page 170

(f)

sin(270o   ) =  cos 

;

cos(270o   ) =  sin 

(g)

sin(270o +  ) =  cos 

;

cos(270o +  ) = sin 

(h)

tan(90o   ) = cot 

;

cot(90o -  ) = tan 

(4) Graphs of Trigonometric Function : (i)

(ii) y  cos x ; x  R ; y  [1, 1]

y  sin x ; x  R ; y  [1, 1]

(iii) y  tan x ; x  R  (2n  1) / 2, n  I;y  R

(iv) y  cot x ; x  R  n, n  I;y  R

(v) y  cosec x ;

(vi) y  sec x ;

x  R  n, n  I;y  R /(1, 1)

5.

x  R  (2n  1) / 2, n  I;y  R /(1, 1)

Trigonometric Functions of Sum or Difference of Two Angles : (a)

sin (A  B) = sinA cosB  cosA sinB

(b)

cos(A  B) = cosA cosB  sin A sin B

(c)

sin2A  sin2B = cos2B  cos2A = sin(A+B). sin(A  B)

(d)

(e)

a m r a cos A  sin B = cos B  sin A = cos (A+B). cos(A  B) h S .

82 K .

6 L 7 tan A  tan B Er. 81027 5058 tan(A  B) = 1  t an A tan B 9 801 839 2

2

Mathematics Concept Note IIT-JEE/ISI/CMI

2

2

Page 171

6.

7.

8.

(f)

cot A cot B 1 cot(A  B) = cot B  cot A

(g)

tan(A+B+C) =

tan A  tanB  tanC  tan A tanB tanC 1 tan A tanB  tanB tanC  tanC tan A

Factorisation of the Sum of Difference of Two Sines or Cosines : C D C D cos 2 2

(a)

sinC + sinD = 2sin

(b)

sinC  sinD = 2cos

(c)

cosC + cosD = 2cos

(d)

cosC  cosD =

C D C D sin 2 2 C D C D cos 2 2 C D

C D

 2sin 2 sin 2

Transformation of Products into Sum or Difference of Sines & Cosines: (a)

2 sinA cosB = sin(A+B) + sin(A  B)

(b)

2 cosA sinB = sin(A+B)  sin(A  B)

(c)

2 cosA cosB = cos(A+B) + cos(A  B)

(d)

2 sinA cosB = cos(A  B)  cos(A + B)

Multiple and Sub-multiple Angles :   cos 2 2

(a)

sin 2A = 2sinA cosA ; sin  = 2 sin

(b)

cos 2A = cos2A  sin2A = 2cos2A  1 = 1  2 sin2A ; 2cos2

  = 1 + cos  2 sin2 = 1  cos  .

2 2

(c)

 2 tan 2A = ; tan  =  1 tan2 1 tan2 A 2

(d)

sin 2A =

(e)

sin 3A = 3 sin A  4 sin3A

(f)

cos 3A = 4 cos3A  3 cos A

(g)

tan 3A =

2 tan

2 tan A

2 tan A 1 tan2 A

3tan A tan3 A 13tan2 A

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

, cos 2A =

1tan2 A 1 tan2 A

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Page 172

9.

Important Trigonometric Ratios : (a)

sin n  = 0 ; cos n  = (  1)n ; tan n  = 0, where n  I

(b)

sin 15o or sin

3 1  5 = = cos 75o or cos ; 12 12 2 2

cos 15o or cos

tan 15o =

(c)

10.

sin

3 1  5 = = sin 75o or sin 12 12 2 2

3 1 = 2  3 = cot 75o; tan 75o = 3 1

 or sin 18o = 10

5 1 4

& cos 36o or cos

3 1 = = cot 15o 3 1 2  3 5 1

 = 5

4

Conditional Identities : If A + B + C =  then :

11.

(i)

sin2A + sin2B + sin2C = 4 sinA sinB sinC

(ii)

sin A + sin B + sinC = 4 cos

(iii)

cos 2A + cos 2B + cos 2C =

(iv)

cos A + cos B + cos C = 1 + 4 sin

(v)

tanA + tanB + tanC = tanA tanB tanC

(vi)

tan

A B B C C A tan + tan tan + tan tan =1 2 2 2 2 2 2

(vii)

cot

A B C A B C + cot + cot = cot . cot . cot 2 2 2 2 2 2

(viii)

cot A cot B + cot B cot C + cot C cot A = 1

(ix)

A+B+C=

 2

A B C cos cos 2 2 2

 1  4 cos A cos B cos C A B C sin sin 2 2 2

then tan A tanB + tanB tanC + tanC tanA = 1

Range of Trigonometric Expression: E = a sin  + b cos  E= =

b

a 2  b 2 sin (  +  ), where tan  = a a2  b 2

a cos (    ), where tan  = b

Hence for any real value of  ,

Mathematics Concept Note IIT-JEE/ISI/CMI



a2 b2  E 

a2 b2

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Page 173

12.

Sine and Cosine Series : n 2 n1   sin  + sin(  +  ) + sin (  + 2  ) + .......+ sin(  + n 1  ) =  sin   2   sin   2

sin

n 2 n1   cos  + cos(  +  ) + cos(  +2  ) +............+ cos(  + r 1  ) =  cos   2   s in   2 s in

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 174

32. TRIGONOMETRIC EQUATIONS Trigonometric Equations Principal Solutions : The solutions of a trigonometric equation which lie in the interval [0,2  ) are called Principal solutions.

General Solution : The expression involving an integer 'n' which gives all solutions of a trigonometric equation is called General solution.

General solution of some standard trigonometric equations are given below : (i)

   If sin  = sin    = n  + (  1)n  where     2 ,2  , n  I.  

(ii)

If cos  = cos    = 2n    where   [0,  ], n  I.

(iii)

   If tan  = tan    = n  +  where     2 ,2  , n  I.  

(iv)

If sin2  = sin2 a   = n   

(v)

(vi)

tan2  = tan2n    = n   

[Note :  is called the principal angle]

cos2  = cos2    = n    .

Types of Trigonometric Equations : v

(i)

Solutions of equations by factorising . Consider the equation : (2sinx  cosx) (1 + cos x) = sin2 x.

(ii)

Solutions of equations reducible to quadratic equations. Consider the equation ; 3 cos2x  10 cos x + 3 = 0.

(iii)

Solving equations by introducing an Auxiliary argument. Consider the equation : sin x + cos x = 2

&

3 cos x + sin x = 2.

(iv)

Solving equations by Transforming a sum of Trigonometric functions into a product. consider the example; cos 3x + sin 2x  sin 4x = 0.

(v)

Solving equations by transforming a product of trigonometric functions into a sum Consider the equation, sin5x. cos 3x = sin 6x. cos 2x.

(vi)

Solving equations by change of variable : (a)

Equations of the form P(sin x  cos x , sin x, cos x) = 0, where P(y z) is a polynomial, can be solved by the change. cos x  sin x = t  1  2 sin x . cos x = t2.

Consider the equation; sinx + cos x = 1 + sin x . cos x.

(b)

Equations of the form of a . sinx + b . cos x + d = 0, where a , b & d are real number s and a, b  0 can be solved by changing sin x & cos x into their corresponding tangent of half the angle. consider the equation 3 cos x + 4 sin x = 5.

(c)

ma r a h S 2 .

K .variable7e.g.

6the8equation, L 7 Many equations can be solved by introducing a.new r 02 5058 sin 2x + cos 2x = sin 2x . cos 2x changesE to 1 8 9 801  1 y  2 (y + 1)   = 0 by substituting, sin 2x . cos 8392x = y..

2  4

(vii)

4

Solving equations with the use of the Boundness of the functions sinx & cos x .

Mathematics Concept Note IIT-JEE/ISI/CMI

Page 175

Trigonometric Inequalitities : Solutions of elementary trigonometric inequalitities are obtained from graphs _____________________________________________________________ Inequality Set of solutions of Inequality (n  z) _____________________________________________________________ sin x > a (|a| sin x < a (|a| cos x > a (|a| cos x < a (|a| tan x > a

< < < <

1) 1) 1) 1)

tan x < a

x  (sin  1a + 2  n,   sin  1a + 2  n) x  (    sin  1a + 2  n, sin  1a + 2  n) x  (  cos  1a + 2  n, cos  1 a + 2  n) x  (cos  1 a + 2  n, 2   cos  1a + 2  n) x  (tan  1a +  n,  /2 +  n)    1 x   2  n,tan an  



___________________________________________________________ Inequalities of the form R(y) > 0, R(y)< 0, where R is a certain rational function and y is a trigonometric function (sine, cosine or tangent), are usually solved in two stages: first the rational inequality is solved for the unknown y and then follows the solution of an elementary trigonometric inequality.

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 176

33. PROPERTIES OF TRIANGLE The angles of a triangle are denoted by capital letters A, B, C and the sides opposite to them are denoted by small letters a, b, c In triangle A+B+C=  also a + b > c and |a – b| < c b + c > a and |b – c| < a c + a > b and |c – a| < b

(1) Sine Rule : In any triangle ABC, the sines of the angles are proportional to the opposite sides i.e.

a b c = = = 2R ; 'R' is circum-radius of ABC sin A sinB sinC

(2) Cosine Formula :

(i) cos A =

b2  c2  a2 or a2 = b2 + c2  2bc cosA = b2 + c2 + 2bc cos(B + C) 2bc

(ii) cos B =

c2  a2 b2 2ca

a2 b2 c2 2ab

(iii)

cos C =

(ii)

b = c cosA + a cos C

(ii)

tan

(3) Projection Formula : (i) a = b cosC + c cos B (iii) c = a cosB + b cosA

(4) Napier's Analogy - tangent rule: (i) tan

B C bc A = cot 2 b c 2

(iii) tan

A B ab C = cot 2 a b 2

C A ca B = cot 2 c a 2

(5) Trigonometric Functions of Half Angles :

(i) sin

A = 2

(ii)

cos

(iii) (iv)

B C (sb)(s c) (s  c)(s  a) ; sin = ; sin = 2 2 bc ca

A = 2

B s(s  a) ; cos = 2 bc

C s(s b) ; cos = 2 ca

(s  a)(s b) ab s(s  c) ab

ma r a h  (s b)(s  c) A a b  c S 2 . tan = = s(s  a) where s = is semi perimeter of8 triangle.

K .

6 2 s(s  a) 2 L 7 r. 1027 058 E 2 2 98 8015 sin A = = bc s(sa)(sb)(s c) bc 839

Mathematics Concept Note IIT-JEE/ISI/CMI

Page 177

(6) Area of Triangle (  ) 1

1

1

 = 2 ab sin C = 2 bc sin A = 2 ca sin B = s(s a)(s b)(s c)

(7) m - n Rule : (m + n) cot  = m cot   n cot  = n cot B  m cotC

A   B

 m

D

C

n

(8) Radius of Circumcircle : R=

a b c abc = = = 2 sin A 2 sinB 2 sinC 4

(9) Radius of The Incircle : (i) r =

 s

(ii) r = (s  a) tan

asin B sin C

(iii) r =

2

2

A

cos 2

A B C = (s  b)tan = (s  c)tan 2 2 2

(iv) r = 4R sin

A B C sin sin 2 2 2

(ii)

A B C ; r2 = s tan ; r3 = s tan 2 2 2

(10) Radius of The Ex-Circles : (i) r1 =

   ,r = ,r = s a 2 s b 3 s c

r1 = s tan

acos B cos C

(iii) r1

2

2

(iv) r1 = 4 R sin

A

cos 2

A B C . cos . cos 2 2 2

(11) Length of Angle Bisectors, Medians and Altitudes : A

ma

Aa a

B

D

E

F

C

(i) Length of an angle bisector from the angle A =  a (ii) Length of median from the angle A = ma = Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

1 2

ma r a h S .

82 K .

6 L 7 Er2bc. cos81027 5058 9 = 01 b  c 98 83 A 2

2b2 2c2  a2

Page 178

(iii) Length of altitude from the angle A = Aa = 2

Note:

2

2 a

3 2 (a + b2 + c2) 4

2

m a + mb + m c =

A and l1a = r1 2

(iii) Excentre (l1)

:

I1 A = r1 cosec

(iv) Orthocentre(H)

:

HA = 2R cos A and Ha = 2R cos B cos C

(v) Centroid (G)

:

1 GA = 3

2

2b2 2c2  a2 and Ga = 3a

(13) Orthocentre and Pedal Triangle : The triangle which is formed by joining the feet of the altitudes is called the Pedal Triangle. (i) Its angles are   2a,   2B and   2C. (ii) Its sides are a cosA = R sin 2A, b cosB = R sin 2B and c cosC = R sin 2C (iii) Circumradii of the triangles PBC, PCA, PAB and ABC are equal.

(14) Excentral Triangle : The triangle formed by joining the three excentres I1, I2 and I3 of  ABC is called the excentral or excentric triangle.

I3

I2

A

B

C

B/2  B 2 2

90

o

I1

(i)

 ABC is the pedal triangle of the  I1 I2 I3 ,

(ii) Its angles are

 2

A 

B



C

 2 , 2  2 and 2  2 .

(iii) Its sides are 4R cos 4R cos

A , 2

B C and 4R cos 2 2

(iv) I I1 = 4R sin

A B C ; I I2 = 4R sin ; I I3 = 4R sin .

2 2 2

Mathematics Concept Note IIT-JEE/ISI/CMI

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Page 179

34. INVERSE TRIGONOMETRIC FUNCTION (1) Domain and Range of Inverse Trigonometric/Circular Functions: No.

Domain

Range

1 x1

 2 y 2

(ii) y = cos–1x

1 x1

0y 

(iii) y = tan–1x

x R

(i)

Function y = sin–1x









 2 <y< 2 



(iv) y = cosec–1x

x

 1 or x  1

 2 y 2 , y 0

(v) y = sec–1x

x

 1 or x  1

 0y ; y 2

(vi) y = cot–1x

x R

0<y< 

Note: (a) 1st quadrant is common to the range of all the inverse functions. (b) 3rd quadrant is not used in inverse functions. (c) 4th quadrant is used in the clockwise direction i.e.



 2  y  0.

(d) No inverse function is periodic.

(2) Properties of Inverse Trigonometric Functions : Property - (i)  , –1  x  1 2

(i)

sin–1 x + cos–1 x =

(ii)

tan–1 x + cot–1 x =

(iii)

cosec–1 x + sec–1 x =

 , x R 2  , |x|  1 2

Property - (ii) (i) sin(sin–1x)= x,  1  x  1 (iii) tan(tan–1x)= x, x  R (v) sec(sec–1x) = x, x   1, x  1

(ii) (iv) (vi)

cos(cos–1x) = x, –1  x  1 cot(cot–1x) = x, x  R cosec(cosec–1x) = x, x  1, x  1

  (i) sin–1(sin x) = x,   x  2 2

(ii)

cos–1(cos x) = x,

  (iii) tan–1(tan x)= x,  <x< 2 2

(iv)

Property - (iii)

(v) sec–1(sec x) = x, 0  x   , x 

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

 2

(vi)

ma r a h S .

82 K .

6 L 7 cot (cot <  8 5 Er.x) =8x,10270 <5x 0 9 801   cosec (cosec x)9 = x; x  0,   x  2 2 3 8 0x 

–1

–1

Page 180

Property - (iv) (i) sin–1(–x) = –sin–1x, –1  x  1 (iii) cos–1(–x) =  –cos–1x,  1  x  1

tan–1(–x) = –tan–1 x, x R –1 –1 cot (–x) =  –cot x, x  R

(ii) (iv)

Note: The functions sin–1x, tan–1x and cosec–1x are odd functions and rest are neither even nor odd Property-(v) (i) cosec–1x = sin–1 (ii) sec–1 x = cos–1

1 ;x–1 , x  1 x

1 ; x  –1, x  1 x

tan1 1 ; x 0  x (iii) cot–1 x =  1 1   tan x ; x  0 Property - (vi) (i) sin(cos–1 x) = cos (sin–1 x) = , (ii) tan(cot–1 x) = cot (tan–1x) =

1  x2 –1  x  1

1 , x  R, x  0 x

|x|

(iii) cosec (sec–1 x) = sec (cosec–1 x) =

x2  1

, |x| > 1

(3) Identities of Addition and Substraction: Property-(i) (i) sin–1x + sin–1y = sin–1 [x 1 y2 + y 1 x2 , x  0, y  0 and (x2 + y2)  1 =   sin–1 [x 1 y2 + y 1 x2 ], x  0, y  0 and x2 + y2 >1

Note:  x2 + y2  1  0  sin  1x + sin–1y  2  x2 + y2 > 1  < sin–1x + sin–1y <  2

(ii)

cos–1x + cos–1y = cos–1 [xy  1 x2

(iii)

x y tan–1x + tan–1y = tan–1 1 xy , x > 0, y > 0 & xy < 1

1 y2 ], x  0, y  0

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 39 8  < tan x + tan y < 

x y =  + tan–1 1 xy , x > 0, y > 0 & xy > 1

=

 , x > 0, y > 0 & xy = 1 2

Note:  xy < 1  0 < tan–1x + tan–1y < ; xy > 1 2

Mathematics Concept Note IIT-JEE/ISI/CMI

 2

–1

–1

Page 181

Property-(ii) (i) sin–1x  sin–1y = sin–1 [x 1 y2  y 1 x2 ], x  0, y  0 (ii) cos–1x  cos–1y = cos–1 [xy +

1  x2

1 y2 ], x  0, y  0, x  y

x y (iii) tan–1x  tan–1y = tan–1 1 xy , x  0, y  0

Note: For x < 0 and y < 0 these identities can be used with the help of properties 2(C) i.e. change x and y to  x and  y which are positive.

Property-(iii) 1  2 sin1 x if |x|  2  1 1    2 if x  (i) sin–1  2x 1 x  =    2 sin x 2    (  2 sin1 x) if x   1  2

(ii) cos

–1

 2 cos 1 x if (2x 1) =  1 2  2 cos x if 2–

0  x 1 1  x  0)

(iii) tan–1

 2 tan1 x if |x|  1  2x 1 if x  1 =    2 tan x 1 x2 (  2 tan1 x) if x  1 

(iv) sin–1

 2 tan1 x if |x|  1  2x 1 if x  1 =    2 tan x 1 x2  (  2 tan1 x) if x  1 

(v)

cos

–1

 2 tan1 x if x  0 =  1 1 x2  2 tan x if x  0 1 x2

Property-(iv)

 x  y  z xyz  If tan–1x + tan–1y + tan–1z = tan–1 1xy  yz zx  if, x > 0, y > 0, z > 0 & (xy+yz+zx)<1 



ma r a h S .  82 K . (ii) If tan x + tan y + tan z = then xy + yz + zx = 1 6 L 7 2 Er. 81027 5058 (iii) tan 1 + tan 2 + tan 3 =  9 801 1 1  (iv) tan 1 + tan + tan = 839 2 3 2 Note:

(i) If tan–1 x + tan–1 y + tan–1 z =  then x + y + z = xyz –1

–1

–1

–1

–1

–1

–1

–1

–1

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 182

(4) Graphs of Inverse Trigonometric Function :    (i) y = sin–1 x, |x|  1, y    ,   2 2

(ii) y = cos–1 x, |x|  1, y  [0,  ]

   (iii) y = tan–1 x, x  R, y    ,   2 2

(iv)





    (v) y = sec–1 x, |x|  1, y   0 , 2    2 ,      

y = cot–1 x, x  R, y  (0,  )





    (vi) y = cosec–1 x, |x|  1,y    2 ,0    0 , 2     

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics Concept Note IIT-JEE/ISI/CMI

Page 183

(vii)y = sin (sin–1 x) = cos (cos–1 x) = x, x  [  1, 1], y  [  1, 1]; y is aperiodic

(viii)

y = tan (tan–1 x) = cot (cot–1 x) = x, x  R, y  R; y is aperiodic

(ix) y = cosec (cosec–1 x) = sec (sec–1 x) = x, |x|  1, |y|  1; y is aperiodic

   (x) y = sin–1(sin x), x  R, y    ,  , is periodic with period 2   2 2

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 184

(xi) y = cos–1 (cos x), x  R, y  [0,  ], is periodic with period 2 





    (xii) y = tan–1 (tan x), x  R  (2n  1) 2 , n  I , y    2 ,2  is periodic with period   

(xiii)

y

sec –1

=



(sec

x),

y

is

periodi c

wi th

peri od

2;



       x  R  (2n  1) 2 , n  I ,y   0 , 2    2 ,      

(xiv)

   y = cosec–1(cosec x), y is periodic with period 2 π ; x  R  {n π , n I}, y    2 , 2  



 {0}

O

– 2

– 2

Mathematics Concept Note IIT-JEE/ISI/CMI

 2



x

2 x



ma r a    .Sh 2 82 K 2 .

6 L 7 Er. 81027 5058 9 801 839 y=

x y=

x+



y=

y=

x

 –  2

 (

–  2

y=

2

y  2

x

Page 185

(xv)

  y = cot–1 (cot x), y is periodic with period  ; x  R  {n  , n  I}, y   0,   2

   ,   2



Part - 3(C)

y   –1 2 (i) graph of y = sin

–1



2 x 1 x 2



–1 2

0

–1

(ii) graph of y = cos (2x

2

x

 

y

–1

1

  2

 1) –1

0

1 x

Note: In this graph it is advisable not to check its derivability just by the inspection of the graph because it is difficult to judge from the graph that at x = 0 there is a shapr corner or not.

y

(iii) graph of y = tan–1

2x



–1



0

1

x 

1x 2



(iv)

graph of y = sin–1

2x

1 x 2

ma r a h y .S 82 K .

6  L 7 Er. 81027 5058 9 801 x  –1 1 0  9 3 8 

Mathematics for IIT-JEE Er. L.K.Sharma 9810277682/8398015058

Page 186

y

(v)

graph of y = cos–1

1 x

2

  2

1 x 2

0

x

ma r a h S .

82 K .

6 L 7 Er. 81027 5058 9 801 839 Mathematics Concept Note IIT-JEE/ISI/CMI

Page 187