VECTOR MATHS DEMO

MATHS

Circle A circle is a locus of a point in a plane whose distance from a fixed point (called centre) is always constant (called radius).

Equation of a circle in various forms : (a)

The circle with centre as origin & radius ‘r’ has the equation; x 2 + y2 = r2.

(b)

The circle with centre (h, k) & radius ‘r’ has the equation; (x  h)2 + (y  k)2 = r2.

(c)

The general equation of a circle is x 2 + y2 + 2gx + 2fy + c = 0 with centre as (g, f) & radius = g2 f 2 c .

This can be obtained from the equation (x – h)2 + (y – k)2 = r2  x2 + y2 – 2hx – 2ky + h2 + k2 – r2 = 0 Take – h = g, – k = f and h2 + k2 – r2 = c Condition to define circle :g² + f²  c > 0  real circle. g² + f²  c = 0  point circle. g² + f²  c < 0  imaginary circle, with real centre, that is (– g, – f) Note : That every second degree equation in x & y, in which coefficient of x 2 is equal to coefficient of y2 & the coefficient of xy is zero, always represents a circle. (d)

The equation of circle with (x 1, y1) & (x 2, y2) as extremeties of its diameter is: (x  x 1) (x  x 2) + (y  y1) (y  y2) = 0.

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MATHS This is obtained by the fact that angle in a semicircle is a right angle.  (Slope of PA) (Slope of PB) = – 1 

y  y1 y  y2 x  x1 . x  x 2 = – 1



(x – x 1) (x – x 2) + (y – y1) (y – y2) = 0

Note that this will be the circle of least radius passing through (x 1, y1) & (x 2, y2).

Example # 1 : Find the equation of the circle whose centre is (1, –2) and radius is 4.

Solution : The equation of the circle is (x – 1) 2 + (y – (–2))2 = 42  (x – 1)2 + (y + 2)2 = 16  x 2 + y2 – 2x + 4y – 11 = 0 Ans. Example # 2 : Find the equation of the circle which passes through the point of intersection of the lines 3x – 2y – 1 = 0 and 4x + y – 27 = 0 and whose centre is (2, – 3).

Solution :

Let P be the point of intersection of the lines AB and LM whose equations are respectively 3x – 2y – 1 = 0 ..........(i) and 4x + y – 27 = 0 ..........(ii) Solving (i) and (ii), we get x = 5, y = 7. So, coordinates of P are (5, 7). Let C(2, –3) be the centre of the circle. Since the circle passes through P, therefore CP = radius 

( 5  2) 2  ( 7  3 ) 2

 radius = 109 .

Hence the equation of the required circle is (x – 2)2 + (y + 3)2 =

 109 

2

Example # 3 : Find the centre & radius of the circle whose equation is x 2 + y2 – 4x + 6y + 12 = 0 Solution : Comparing it with the general equation x 2 + y2 + 2gx + 2fy + c = 0, we have 2g = – 4  g = –2 2f = 6  f=3 & c = 12  centre is (–g, –f) i.e. (2, –3) and radius =

g2  f 2  c =

( 2)2  (3)2  12 = 1

Example # 4 : Find the equation of the circle, the coordinates of the end points of whose diameter are (–1, 2) and (4, –3) Solution : We know that the equation of the circle described on the line segment joining (x 1, y1) and (x 2, y2) as a diameter is (x – x 1) (x – x 2) + (y – y1) (y – y2) = 0.

Here, x 1 = –1, x 2 = 4, y1 = 2 and y2 = –3.

So, the equation of the required circle is (x + 1) (x – 4) + (y – 2) (y + 3) = 0  x 2 + y2 – 3x + y – 10 = 0.

Self practice problems : (1)

Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of the lines x + y + 1 = 0 and x – 2y + 4 = 0.

(2)

Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6)

(3)

Find the equation of a circle whose radius is 6 and the centre is at the origin.

Answers :

(1) x 2 + y2 + 4x – 2y = 0

(2) x 2 + y2 – 2x – 4y – 20 = 0 (3) x 2 + y2 = 36.

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MATHS Intercepts made by a circle on the axes: The intercepts made by the circle x 2 + y2 + 2gx + 2fy + c = 0 on the coordinate axes are 2 g2 c 2 (on x-axis) & 2 f c (on y-axis) respectively..

If

Example # 5 Solution :

g2 > c



circle cuts the x axis at two distinct points.

g2 = c



circle touches the xaxis.

g2 < c



circle lies completely above or below the xaxis.

AB = 2AD = 2

r 2  CD 2 = 2

r2  f 2 = 2

g2  f 2  c 2  f 2 = 2 g2  c

Find the equation to the circle touching the negative y-axis at a distance 3 from the origin and intercepting a length 8 on the x-axis.

Let the equation of the circle be x 2 + y2 + 2gx + 2fy + c = 0. Since it touches y-axis at (0, –3) and (0, – 3) lies on the circle.  c = f 2 ...(i) 9 – 6f + c = 0 .......(ii) From (i) and (ii), we get 9 – 6f + f 2 = 0  (f – 3)2 = 0  f = 3.

Putting f = 3 in (i) we obtain c = 9.

It is given that the circle x 2 + y2 + 2gx + 2fy + c = 0 intercepts length 8 on x-axis 

2 g2  c = 8

2 g2  9 = 8

 2



g2 – 9 = 16



g=±5

2

Hence, the required circle is x + y ± 10x + 6y + 9 = 0.

Self practice problems : (4)

Find the equation of a circle which touches the positive axis of y at a distance 3 from the origin and intercepts a distance 6 on the axis of x.

(5)

Find the equation of a circle which touches positive y-axis at a distance of 2 units from the origin and cuts an intercept of 3 units with the positive direction of x-axis.

Answers :

(4)

x 2 + y2 ± 6 2 x – 6y + 9 = 0

(5)

x 2 + y2 – 5x – 4y + 4 = 0

Parametric equations of a circle: The parametric equations of (x  h)2 + (y  k)2 = r2 are: x = h + r cos  ; y = k + r sin ;   <    where (h, k) is the centre, r is the radius &  is a parameter.

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MATHS Example # 6 : Find the parametric equations of the circle x 2 + y2 – 4x – 2y + 1 = 0 Solution : We have : x 2 + y2 – 4x – 2y + 1 = 0  (x 2 – 4x ) + (y2 – 2y) = – 1 2 2 2  (x – 2) + (y – 1) = 2 So, the parametric equations of this circle are x = 2 + 2 cos  , y = 1 + 2 sin .

Example # 7 : Find the equations of the following curves in cartesian form. Also, find the centre and radius of the circle x = a + c cos , y = b + c sin  Solution :

We have : x = a + c cos , y = b + c sin 2



x a y b , sin  = c c



cos  =



(x – a)2 + (y – b)2 = c 2

2

xa  y b   +   = cos 2 + sin2  c   c 

Clearly, it is a circle with centre at (a, b) and radius c.

Self practice problems : (6)

Find the parametric equations of circle x 2 + y2 – 6x + 4y – 12 = 0

(7)

Find the cartesian equations of the curve x = –2 + 3 cos , y = 3 + 3 sin 

Answers :

(6)

x = 3 + 5 cos , y = –2 + 5 sin 

(7)

(x + 2)2 + (y – 3)2 = 9

Position of a point with respect to a circle: The point (x 1, y1) is inside, on or outside the circle S  x 2 + y2 + 2gx + 2fy + c = 0. according as S1  x 1² + y1² + 2gx 1 + 2fy1 + c <, = or > 0.

Note : The greatest & the least distance of a point A (lies outside the circle) from a circle with centre C & radius r is AC + r & AC  r respectively.

Example # 8 : Discuss the position of the points (1, 2) and (6, 0) with respect to the circle x 2 + y2 – 4x + 2y – 11 = 0 Solution : We have x 2 + y2 – 4x + 2y – 11 = 0 or S = 0, where S = x 2 + y2 – 4x + 2y – 11.

For the point (1, 2), we have S1 = 12 + 22 – 4 × 1 +2 × 2 – 11 < 0 For the point (6, 0), we have S2 = 62 + 02 – 4 × 6 +2 × 0 – 11 > 0 Hence, the point (1, 2) lies inside the circle and the point (6, 0) lies outside the circle.

Self practice problem : (8)

How ar e the points (0, 1) ( 3, 1) and ( 1, 3) s ituated with res pec t to the c irc le x 2 + y2 – 2x – 4y + 3 = 0?

Answer : (8)

(0, 1) lies on the circle ; (3, 1) lies outside the circle ; (1, 3) lies inside the circle.

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MATHS Line and a circle: Let L = 0 be a line & S = 0 be a circle. If r is the radius of the circle & p is the length of the perpendicular from the centre on the line, then: (i) p>r  the line does not meet the circle i. e. passes out side the circle. (ii) p=r  the line touches the circle. (It is tangent to the circle) (iii) p<r  the line is a secant of the circle. (iv) p=0  the line is a diameter of the circle.

Also, if y = mx + c is line and x 2 + y2 = a2 is circle then (i) c 2 < a2 (1 + m 2) the line is a secant of the circle.

2 2 2 (ii) c = a (1 + m )  the line touches the circle. (It is tangent to the circle) (iii) c 2 > a2 (1 + m 2)  the line does not meet the circle i. e. passes out side the circle.

These conditions can also be obtained by solving y = mx + c with x2 + y2 = a2 and making the discriminant of the quadratic greater than zero for secant, equal to zero for tangent and less the zero for the last case.

Example # 9 : For what value of c will the line y = 2x + c be a tangent to the circle x 2 + y2 = 5 ?

Solution :

We have : y = 2x + c or 2x – y + c = 0 ......(i) and x 2 + y2 = 5 ........(ii) If the line (i) touches the circle (ii), then length of the  from the centre (0, 0) = radius of circle (ii)

20  0  c



2  ( 1) 2

2

=

c



5

=±

5

5





c 5

=

5

c=±5

Hence, the line (i) touches the circle (ii) for c = ± 5 Self practice problem : (9)

For what value of , does the line 3x + 4y =  touch the circle x 2 + y2 = 10x.

Answers :

(9)

40, –10

Slope form of tangent : y = mx + c is always a tangent to the circle x 2 + y2 = a2 if c 2 = a2 (1 + m 2). Hence, equation

 a 2m a 2  , . c c 

of tangent is y = mx ± a 1  m2 and the point of contact is  

Point form of tangent : (i) The equation of the tangent to the circle x 2 + y2 = a2 at its point (x 1, y1) is, x x 1 + y y1 = a². (ii) The equation of the tangent to the circle x 2 + y2 + 2gx + 2fy + c = 0 at its point (x 1, y1) is: xx 1 + yy1 + g (x+x 1) + f (y+y1) + c = 0.

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MATHS Note : In general the equation of tangent to any second degree curve at point (x 1, y1) on it can be obtained by replacing x 2 by x x 1, y2 by yy1, x by xy by

x  x1 y  y1 , y by , 2 2

x1y  xy1 and c remains as c.

2

Parametric form of tangent : The equation of a tangent to circle x 2 + y2 = a2 at (a cos , a sin ) is x cos  + y sin  = a.  a cos    a sin      2 , 2  NOTE : The point of intersection of the tangents at the points P() & Q() is      cos cos 2  2  Example # 10 : Find the equation of the tangent to the circle x 2 + y2 – 30x + 6y + 109 = 0 at (4, –1).

Solution :

Equation of tangent is  y  ( 1)  x4  + 109 = 0  +6  4x + (–y) – 30  2    2 

or 4x – y – 15x – 60 + 3y – 3 + 109 = 0 or –11x + 2y + 46 = 0 or 11x – 2y – 46 = 0 Hence, the required equation of the tangent is 11x – 2y – 46 = 0 Example # 11 : Find the equation of tangents to the circle x 2 + y2 – 6x + 4y – 12 = 0 which are parallel to the line 4x + 3y + 5 = 0 Solution : Given circle is x 2 + y2 – 6x + 4y – 12 = 0 .......(i) and given line is 4x + 3y + 5 = 0 .......(ii) Centre of circle (i) is (3, –2) and its radius is 5. Equation of any line 4x + 3y + k = 0 parallel to the line (ii) .......(iii) If line (iii) is tangent to circle, (i) then

| 4.3  3( 2)  k | 42  32

= 5 or |6 + k| = 25

or 6 + k = ± 25  k = 19, – 31 Hence equation of required tangents are 4x + 3y + 19 = 0 and 4x + 3y – 31 = 0 Self practice problem : (10)

Find the equation of the tangents to the circle x 2 + y2 – 2x – 4y – 4 = 0 which are (i) parallel, (ii) perpendicular to the line 3x – 4y – 1 = 0

Answer.

(i) 3x – 4y + 20 = 0 and 3x – 4y – 10 = 0 (ii) 4x + 3y + 5 = 0 and 4x + 3y – 25 = 0

Normal : 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 x 2 + y2 + 2gx + 2fy + c = 0 at (x 1, y1) is; y  y1 =

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y1  f (x  x 1). x1  g

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MATHS Example # 12 : Find the equation of the normal to the circle x 2 + y2 – 5x + 2y – 48 = 0 at the point (5, 6).

Solution :

5  Since normal is line joining centre  ,  1 and (5, 6) 2  

Slope =

14 5

Hence, the equation of the normal at (5, 6) is y – 6 = (14/5) (x – 5)  14x – 5y – 40 = 0 Self practice problem : (11)

Find the equation of the normal to the circle x 2 + y2 – 2x – 4y + 3 = 0 at the point (2, 3).

Answer :

(11)

x–y+1=0

Pair of tangents from a point : The equation of a pair of tangents drawn from the point A (x 1, y1) to the circle x 2 + y2 + 2gx + 2fy + c = 0 is : SS1 = T².

Where S  x 2 + y2 + 2gx + 2fy + c ; S1  x 1² + y1² + 2gx 1 + 2fy1 + c T  xx 1 + yy1 + g(x + x 1) + f(y + y1) + c.

Example # 13 Find the equation of the pair of tangents drawn to the circle x 2 + y2 – 2x + 4y = 0 from the point (0, 1) Solution : Given circle is S = x 2 + y2 – 2x + 4y = 0 .......(i) Let P  (0, 1) For point P, S1 = 02 + 12 – 2.0 + 4.1 = 5 Clearly P lies outside the circle and T  x . 0 + y . 1 – (x + 0) + 2 (y + 1) i.e.

T  –x +3y + 2.

Now equation of pair of tangents from P(0, 1) to circle (1) is SS1 = T 2 or 5 (x 2 + y2 – 2x + 4y) = (– x + 3y + 2) 2 or 5x 2 + 5y2 – 10x + 20y = x 2 + 9y2 + 4 – 6xy – 4x + 12y or 4x 2 – 4y2 – 6x + 8y + 6xy – 4 = 0 or 2x 2 – 2y2 + 3xy – 3x + 4y – 2 = 0 .......(ii) Separate equation of pair of tangents : From (ii), 2x 2 + 3(y – 1) x – 2(2y2 – 4y + 2) = 0

 3( y  1)  9( y  1)2  8(2y 2  4 y  2) 4



x=

or

4x + 3y – 3 = ±



Separate equations of tangents are 2x – y + 1 = 0 and x + 2y – 2 = 0

25 y 2  50 y  25 = ± 5(y – 1)

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MATHS Self practice problems : (12)

Find the joint equation of the tangents through (7, 1) to the circle x 2 + y2 = 25.

Answer :

(12)

12x 2 – 12y2 + 7xy – 175x – 25y + 625 = 0

Length of a tangent and power of a point : The length of a tangent from an external point (x 1, y1) to the circle 2 2 S  x 2 + y2 + 2gx + 2fy + c = 0 is given by L = x1  y1  2gx1  2f1y  c = S1 .

AP = length of tangent AP2 = AD . AE Square of length of the tangent from the point A is also called the power of point w.r.t. a circle.

Power of a point w.r.t. a circle remains constant.

Power of a point P is positive, negative or zero according as the point ‘A’ is outside, inside or on the circle respectively.

Example # 14 Find the length of the tangent drawn from the point (5, 1) to the circle x 2 + y2 + 6x – 4y – 3 = 0 Solution :

Given circle is x 2 + y2 + 6x – 4y – 3 = 0 Given point is (5, 1). Let P = (5, 1) Now length of the tangent from P(5, 1) to circle (i) =

.........(i) 5 2  12  6.5  4.1  3 = 7

Self practice problems : (13)

Find the area of the quadrilateral formed by a pair of tangents from the point (4, 5) to the circle x 2 + y2 – 4x – 2y – 11 = 0 and a pair of its radii.

(14)

If the length of the tangent from a point (f, g) to the circle x 2 + y2 = 4 be four times the length of the tangent from it to the circle x 2 + y2 = 4x, show that 15f 2 + 15g2 – 64f + 4 = 0 Answer.

(13) 8 sq. units

Director circle : 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.

Proof :

AC = r cosec 45º = r 2

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MATHS Example # 15 Find the equation of director circle of the circle (x – 2) 2 + (y + 1)2 = 2.

Solution : Centre & radius of given circle are (2, –1) & 2 respectively..

Centre and radius of the director circle will be (2, –1) & 2 ×  equation of director circle is (x – 2) 2 + (y + 1)2 = 4  x 2 + y2 – 4x + 2y + 1 = 0.

2 = 2 respectively..

Self practice problems : (15)

Find the equation of director circle of the circle whose diameters are 2x – 3y + 12 = 0 and x + 4y – 5 = 0 and area is 154 square units.

Answer

(15)

(x + 3)2 + (y – 2)2 = 98

Chord of contact : If two tangents PT 1 & PT 2 are drawn from the point P(x 1, y1) to the circle S  x 2 + y2 + 2gx + 2fy + c = 0, then the equation of the chord of contact T 1T 2 is: xx 1 + yy1 + g (x + x 1) + f (y + y1) + c = 0.

Note : Here R = radius; L = length of tangent. (a)

Chord of contact exists only if the point ‘P’ is not inside.

(b)

Length of chord of contact T 1 T 2 =

(c)

Area of the triangle formed by the pair of the tangents & its chord of contact =

(d)

 2RL   Tangent of the angle between the pair of tangents from (x 1, y1) =  2 2  L R 

2 LR R 2  L2

.

R L3

(e)

R 2  L2

Equation of the circle circumscribing the triangle PT 1 T 2 is: (x  x 1) (x + g) + (y  y1) (y + f) = 0.

Example # 16 Find the equation of the chord of contact of the tangents drawn from (1, 2) to the circle x 2 + y2 – 2x + 4y + 7 = 0 Solution :

Given circle is x 2 + y2 – 2x + 4y + 7 = 0 .......(i) Let P = (1, 2) For point P (1, 2), x 2 + y2 – 2x + 4y + 7 = 1 + 4 – 2 + 8 + 7 = 18 > 0 Hence point P lies outside the circle For point P (1, 2), T = x . 1 + y . 2 – (x + 1) + 2(y + 2) + 7 i.e.

T = 4y + 10 Now equation of the chord of contact of point P(1, 2) w.r.t. circle (i) will be 4y + 10 = 0 or 2y + 5 = 0

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MATHS Example # 17 Tangents are drawn to the circle x 2 + y2 = 12 at the points where it is met by the circle x 2 + y2 – 5x + 3y – 2 = 0; find the point of intersection of these tangents.

Solution :

Given circles are S1  x 2 + y2 – 12 = 0 ....... (i) and S2 = x 2 + y2 – 5x + 3y – 2 = 0 ....... (ii) Now equation of common chord of circle (i) and (ii) is S1 – S2 = 0 i.e.

5x – 3y – 10 = 0 ....... (iii) Let this line meet circle (i) [or (ii)] at A and B Let the tangents to circle (i) at A and B meet at P(), then AB will be the chord of contact of the tangents to the circle (i) from P, therefore equation of AB will be

x + y – 12 = 0 ....... (iv) Now lines (iii) and (iv) are same, therefore, equations (iii) and (iv) are identical 

  12 = = 5 3  10



 = 6,  = –

18 5

18    Hence P =  6,  5  

Self practice problems : (16)

Find the co-ordinates of the point of intersection of tangents at the points where the line 2x + y + 12 = 0 meets the circle x 2 + y2 – 4x + 3y – 1 = 0

(17)

Find the area of the triangle formed by the tangents drawn from the point (4, 6) to the circle x 2 + y2 = 25 and their chord of contact.

Answers :

12.

(16)

(1, – 2)

(17)

405 3 sq. unit; 4x + 6y – 25 = 0 52

Pole and polar: (i)

If through a point P in the plane of the circle, there be drawn any straight line to meet the circle in Q and R, the locus of the point of intersection of the tangents at Q & R is called the Polar of

the point P, also P is called the Pole of the Polar. (ii)

The equation to the polar of a point P (x 1, y1) w.r.t. the circle x 2 + y2 = a 2 is given by xx 1 + yy 1 = a 2 , & if the circle is general then the equation of the polar becom es xx 1 + yy1 + g (x + x 1) + f (y + y1) + c = 0 i.e. T = 0. Note that if the point (x 1, y1) be on the circle then the tangent & polar will be represented by the same equation. Similarly if the point (x 1, y1) be outside the circle then the chord of contact & polar will be represented by the same equation.

(iii) (iv) (v)

 Aa 2 Ba 2  . , Pole of a given line Ax + By + C = 0 w.r.t. circle x 2 + y2 = a2 is   C   C If the polar of a point P pass through a point Q , then the polar of Q passes through P. Two lines L1 & L2 are conjugate of each other if Pole of L1 lies on L2 & vice versa. Similarly two points P & Q are said to be conjugate of each other if the polar of P passes through Q & vice-versa.

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MATHS Example # 18 Find the equation of the polar of the point (2, –1) with respect to the circle x 2 + y2 – 3x + 4y – 8 = 0 Solution : Given circle is x 2 + y2 – 3x + 4y – 8 = 0 ............(i) Given point is (2, –1) let P = (2, –1). Now equation of the polar of point P with respect to circle (i)  x  2  y  1  +4   –8=0 x.2 + y(–1) – 3  2    2 

or

4x – 2y – 3x – 6 + 4y – 4 – 16 = 0

or

x + 2y – 26 = 0

Example # 19 Find the pole of the line 3x + 5y + 17 = 0 with respect to the circle x 2 + y2 + 4x + 6y + 9 = 0 Solution : Given circle is x 2 + y2 + 4x + 6y + 9 = 0 and given line is 3x + 5y + 17 = 0 Let P() be the pole of line (ii) with respect to circle (i) Now equation of polar of point P() with respect to circle (i) is x + y + 2(x + ) + 3(y +) + 9 = 0 or ( + 2)x + ( + 3) y + 2 + 3 + 9 = 0 Now lines (ii) and (iii) are same, therefore, 2 3 2  3  9 = = 3 5 17

(i) (ii) (iii) From (i) and (ii), we get 5 + 10 = 3 + 9 or 5 – 3 = – 1 From (i) and (iii), we get 17 + 34 = 6 + 9 + 27 or 11 – 9 = –7 Solving (iv) & (v), we get  = 1,  = 2. Hence required pole is (1, 2).

......(iv) ......(v)

Self practice problems : (18)

Find the co-ordinates of the point of intersection of tangents at the points where the line 2x + y + 12 = 0 meets the circle x 2 + y2 – 4x + 3y – 1 = 0.

(19)

Find the pole of the s tr aight line 2x – y + 10 = 0 with r es pect to the c irc le x 2 + y2 – 7x + 5y – 1 = 0

Answers :

(18)

(1, – 2)

(19)

3 3  ,–  2 2 

Equation of the chord with a given middle point: The equation of the chord of the circle S  x 2 + y2 + 2gx + 2fy + c = 0 in terms of its mid point M (x 1, y1) is xx 1 + yy1 + g (x + x 1) + f (y + y1) + c = x 12 + y12 + 2gx 1 + 2fy1 + c which is designated by T = S1.

Notes : (i) (ii)

The shortest chord of a circle passing through a point ‘M’ inside the circle, is one chord whose middle point is M. The chord passing through a point ' M ' inside the circle and which is at a maximum distance from the centre is a chord with middle point M.

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MATHS Example # 20 Find the equation of the chord of the circle x 2 + y2 + 6x + 8y – 11 = 0, whose middle point is (1, –1) Solution : Equation of given circle is S  x 2 + y2 + 6x + 8y – 11 = 0 Let L  (1, –1) For point L(1, –1), S1 = 12 + (–1)2 + 6.1 + 8(–1) – 11 = –11 and T  x.1 + y (–1) + 3(x + 1) + 4(y – 1) – 11 i.e.

T  4x + 3y – 12 Now equation of the chord of circle (i) whose middle point is L(1, –1) is T = S1 or 4x + 3y – 12 = –11 or 4x + 3y – 1 = 0 Second Method : Let C be the centre of the given circle, then C  (–3, –4). L  (1, –1) slope of CL = 

4  1 3 =  3 1 4

Equation of chord of circle whose middle point is L, is y + 1 = –

( chord is perpendicular to CL)

or

4 (x – 1) 3

4x + 3y – 1 = 0

Self practice problems : (20)

Find the equation of that chord of the circle x 2 + y2 = 15, which is bisected at (3, 2)

(21)

Find the co-ordinates of the middle point of the chord which the circle x 2 + y2 + 4x – 2y – 3 = 0 cuts off on the line y = x + 2.

Answers :

(20)

3x + 2y – 13 = 0

(21)

 3 1  ,   2 2

Equation of the chord joining two points of circle : The equation of chord PQ to the circle x 2 + y2 = a2 joining two points P() and Q() on it is given by the equation of a straight line joining two point  &  on the circle x 2 + y2 = a2 is x cos

     + y sin = a cos .

2 2 2

Common tangents to two circles: Case

(i)

Number of Tangents

Condition

4 common tangents (2 direct and 2 transverse)

r 1 + r 2 < c 1 c 2.

(ii)

3 common tangents.

r 1 + r 2 = c 1 c 2.

(iii)

2 common tangents.

r1  r2 < c 1 c 2 < r1 + r2

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MATHS (iv)

1 common tangent.

r1  r2 = c 1 c 2.

(v)

No common tangent.

c 1 c 2 < r1  r2.

(Here C1C2 is distance between centres of two circles.)

Notes :(i)

(ii)

The direct common tangents meet at a point which divides the line joining centre of circles externally in the ratio of their radii.

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

Length of an external (or direct) common tangent & internal (or transverse) common tangent to the two circles are given by: Lext = d2  (r1  r2 )2 & Lint = d2  (r1  r2 )2 , where d = distance between the centres of the two circles and r 1, r2 are the radii of the two circles. Note that length of internal common tangent is always less than the length of the external or direct common tangent.

Example # 21 Examine if the two circles x 2 + y2 – 2x – 4y = 0 and x 2 + y2 – 8y – 4 = 0 touch each other externally or internally.

Solution : Given circles are x 2 + y2 – 2x – 4y = 0 ...........(i) and x 2 + y2 – 8y – 4 = 0 ...........(ii) Let A and B be the centres and r 1 and r2 the radii of circles (i) and (ii) respectively, then A  (1, 2), B  (0, 4), r1 = 5, r2 = 25 Now AB =

(1  0)2  (2  4)2 = 5

and

r1 + r2 = 3 5 , |r1 – r2| =

5

Thus AB = |r1 – r2|, hence the two circles touch each other internally.

Self practice problems : (22)

Find the position of the circles x 2 + y2 – 2x – 6y + 9 = 0 and x 2 + y2 + 6x – 2y + 1 = 0 with respect to each other.

Answer :

(22)

One circle lies completely outside the other circle.

Orthogonality of two circles: Two circles S1= 0 & 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: 2 g 1 g 2 + 2 f 1 f 2 = c 1 + c 2.

Proof :

 

(C1C2)2 = (C1P)2 + (C2P)2 (g1 – g2)2 + (f1 – f2)2 = g12 + f12 – c1 + g22 + f22 – c2 2g1g2 + 2f1f2 = c1 + c2

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MATHS Notes : (a)

The centre of a variable circle orthogonal to two fixed circles lies on the radical axis of two circles.

(b)

If two circles are orthogonal, then the polar of a point 'P' on first circle w.r.t. the second circle passes through the point Q which is the other end of the diameter through P. Hence locus of a point which moves such that its polars w.r.t. the circles S1 = 0, S2 = 0 & S3 = 0 are concurrent in a circle which is orthogonal to all the three circles.

(c)

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.

Example # 22 Obtain the equation of the circle orthogonal to both the circles x 2 + y2 + 3x – 5y+ 6 = 0 and 4x 2 + 4y2 – 28x +29 = 0 and whose centre lies on the line 3x + 4y + 1 = 0.

Solution.

Given circles are x 2 + y2 + 3x – 5y + 6 = 0 ...........(i) and 4x 2 + 4y2 – 28x + 29 = 0

29 = 0.

4 Let the required circle be x 2 + y2 + 2gx + 2fy + c = 0 Since circle (iii) cuts circles (i) and (ii) orthogonally or

x 2 + y2 – 7x +

..........(ii) ..........(iii)



3  5 2g   + 2f    = c + 6 2  2

or

3g – 5f = c + 6

...........(iv)

and

 7 29 2g    + 2f.0 = c +  2 4

or

– 7g = c +

29 4

...........(v)

From (iv) & (v), we get 10g – 5f = –

5 4

or 40g – 20f = – 5.

Given line is 3x + 4y = – 1 Since centre (– g, – f) of circle (iii) lies on line (vii),  – 3g – 4f = – 1 1 Solving (vi) & (viii), we get g = 0, f = 4  

..........(vi) ..........(vii) .........(viii)

29 4 from (iii), required circle is from (5), c = –

x 2 + y2 +

1 29 y– =0 2 4

or

4(x 2 + y2) + 2y – 29 = 0

Self practice problems : (23)

For what value of k the circles x 2 + y2 + 5x + 3y + 7 = 0 and x 2 + y2 – 8x + 6y + k = 0 cut orthogonally.

(24)

Find the equation to the circle which passes through the origin and has its centre on the line x + y + 4 = 0 and cuts the circle x 2 + y2 – 4x + 2y + 4 = 0 orthogonally.

(23) – 18 (24) 3x 2 + 3y2 + 4x + 20y = 0

Ans.

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MATHS Radical axis and radical centre: The radical axis of two circles is the locus of points whose powers w.r.t. the two circles are equal. The equation of radical axis of the two circles S1 = 0 & S2 = 0 is given by S1  S2 = 0 i.e. 2 (g1  g2) x + 2 (f 1  f 2) y + (c 1  c 2) = 0.

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. Note that the length of tangents from radical centre to the three circles are equal.

Notes : (a)

If two circles intersect, then the radical axis is the common chord of the two circles.

(b)

If two circles touch each other, then the radical axis is the common tangent of the two circles at the common point of contact.

(c)

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

(d)

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.

(e)

Radical axis bisects a common tangent between the two circles.

(f)

A system of circles, every two which have the same radical axis, is called a coaxial system.

(g)

Pairs of circles which do not have radical axis are concentric.

Example # 23 Find the co-ordinates of the point from which the lengths of the tangents to the following three circles be equal.

3x 2 + 3y2 + 4x – 6y – 1 = 0 2x 2 + 2y2 – 3x – 2y – 4 = 0 2x 2 + 2y2 – x + y – 1 = 0 Solution : Here we have to find the radical centre of the three circles. First reduce them to standard form in which coefficients of x 2 and y2 be each unity. Subtracting in pairs the three radical axes are 17 5 x–y+ =0 6 3

–

;

–x–

3 3 y– =0 2 2

11 1 5 x+ y– = 0.

6 6 2

 16 31    which satisfies the third also. This point is solving any two, we get the point    21 63 

called the radical centre and by definition the length of the tangents from it to the three circles are equal.

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MATHS Self practice problem : (25)

Find the point from which the tangents to the three circles x 2 + y2 – 4x + 7 = 0, 2x 2 + 2y2 – 3x + 5y + 9 = 0 and x 2 + y2 + y = 0 are equal in length. Find also this length.

Answer :

(25)

(2, – 1) ; 2.

Family of Circles: This article is aimed at obtaining the equation of a group of circles having a specific characteristic.

For example, the equation x2 + y2 + 4x + 2y +  = 0 where  is arbitrary, represents a family of circles with fixed centre (–2, –1) but variable radius. We have the following results for some other families of circles. (a)

The equation of the family of circles passing through the points of intersection of two circles S1 = 0 & S2 = 0 is : S1 + K S2 = 0 (K  1, provided the coefficient of x 2 & y2 in S1 & S2 are same)

(b)

The equation of the family of circles passing through the point of intersection of a circle S = 0 & a line L = 0 is given by S + KL = 0.

(c)

The equation of a family of circles passing through two given points (x 1, y1) & (x 2, y2) can be written in the form:

x (x  x 1) (x  x 2) + (y  y1) (y  y2) + K x1 x2

y 1 y1 1 = 0, where K is a parameter.. y2 1

(d)

The equation of a family of circles touching a fixed line y  y1 = m (x  x 1) at the fixed p o i n t (x 1, y1) is (x  x 1)2 + (y  y1)2 + K (y  y1  m (x  x 1)) = 0, where K is a parameter.

(e)

Family of circles 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 coefficient of x 2 = coefficient of y2.

(f)

Equation of circle circumscribing a quadrilateral whose side in order are represented by the lines L1 = 0, L2 = 0, L3 = 0 & L4 = 0 are u L1L3 + L2L4 = 0 where values of u &  can be found out by using condition that coefficient of x 2 = coefficient of y2 and coefficient of xy = 0.

Example # 24 Find the equations of the circles passing through the points of intersection of the circles x 2 + y2 –2x – 4y – 4 = 0 and x 2 + y2 – 10x – 12y + 40 = 0 and whose radius is 4.

Solution : Any circle through the intersection of given circles is S1 + S2 = 0 or (x 2 + y2 – 2x – 4y – 4) + (x 2 + y2 – 10x – 12y + 40 ) = 0 or

(x 2 + y2) – 2 r= 

(1  5 ) ( 2  6 ) 40  4 x–2 y+ =0 1  1  1 

...........(i)

g2  f 2  c = 4, given

16 =

(1  5 )2 (1   )

2

+

( 2  6 ) 2 (1   )

2

–

40  4 1 

16(1 + 2 +  ) = 1 + 10 + 252 + 4 + 24 + 362 – 402 – 40 + 4 + 4 or 16 + 32 + 162 = 212 – 2 + 9 or 52 – 34 – 7 = 0  ( – 7) (5 + 1) = 0   = 7, – 1/5 Putting the values of  in (i) the required circles are 2x 2 + 2y2 – 18x – 22y + 69 = 0 and x 2 + y2 – 2y – 15 = 0 2

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MATHS Example # 25 Find the equations of circles which touches 2x – y + 3 = 0 and pass through the points of intersection of the line x + 2y – 1 = 0 and the circle x 2 + y2 – 2x + 1 = 0.

Solution : The required circle by S + P = 0 is x 2 + y2 – 2x + 1 +  (x + 2y – 1) = 0 or x 2 + y2 – x (2 – ) + 2y + (1 – ) = 0 2  ,   centre (– g, – f) is   2 

r=

g2  f 2  c

=

(2   )2  2  (1   ) 4

=

1 2

52 =

5 ||.

2

Since the circle touches the line 2x – y + 3 = 0 therefore perpendicular from centre is equal to

2.((2   ) / 2)  (  )  3 radius

5

=

|| 2

5.

 =±2 Putting the values of  in (i) the required circles are x 2 + y2 + 4y – 1 = 0 x 2 + y2 – 4x – 4y + 3 = 0.

Example # 26 Find the equation of circle passing through the points A(1, 1) & B(2, 2) and whose radius is 1.

Solution : Equation of AB is x – y = 0  equation of circle is (x – 1) (x – 2) + (y – 1) (y – 2) + (x – y) = 0 or x 2 + y2 + ( – 3)x – ( + 3)y + 4 = 0 radius =

(   3) 2 (   3) 2  4 4 4

But radius = 1 (given)  or 

(   3) 2 (   3) 2  4 = 1 4 4 ( – 3)2 + ( + 3)2 – 16 = 4  =±1 equation of circles are x 2 + y2 – 2x – 4y + 4 = 0 & x 2 + y2 – 4x – 2y + 4 = 0

Example # 27 Find the equation of the circle passing through the point (2, 1) and touching the line x + 2y – 1 = 0 at the point (3, – 1).

Solution :

Equation of circle is (x – 3) 2 + (y + 1)2 + (x + 2y – 1) = 0 Since it passes through the point (2, 1), 1 + 4 +  (2 + 2 – 1) = 0  

circle is (x – 3)2 + (y + 1)2 –



3x 2 + 3y2 – 23x – 4y + 35 = 0

=–

5 3

5 (x + 2y – 1) = 0 3

Example # 28 Find the equation of circle circumcscribing the triangle whose sides are 3x – y – 9 = 0, 5x – 3y – 23 = 0 & x + y – 3 = 0.

Solution :

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MATHS

L1L2 + L2L3 + µL1L3 = 0 (3x – y – 9) (5x – 3y – 23) + (5x – 3y – 23) (x + y – 3) + µ (3x – y – 9) (x + y – 3) = 0 (15x 2 + 3y2 – 14xy – 114x + 50y + 207) + (5x 2 – 3y2 + 2xy – 38x – 14y + 69) + µ (3x 2 – y2 + 2xy – 18x – 6y + 27) = 0 2 2 (5 + 3µ + 15)x + (3 – 3 – µ)y + xy (2 + 2µ – 14) – x (114 + 38 + 18µ) + y(50 – 14 – 6µ) + (207 + 69 + 27µ) = 0 ...........(i) 2 2 coefficient of x = coefficient of y  5 + 3µ + 15 = 3 – 3 – µ 2 + µ + 3 = 0 ...........(ii) coefficient of xy = 0  + µ – 7 = 0 ..........(iii) Solving (ii) and (iii), we have  = – 10, µ = 17 Puting these values of  & µ in equation (i), we get 2x 2 + 2y2 – 5x + 11y – 3 = 0

Self practice problems : (26)

Find the equation of the circle passing through the points of intersection of the circles x 2 + y2 – 6x + 2y + 4 = 0 and x 2 + y2 + 2x – 4y – 6 = 0 and with its centre on the line y = x.

(27)

Find the equation of circle circumscribing the quadrilateral whose sides are 5x + 3y = 9, x = 3y, 2x = y and x + 4y + 2 = 0.

Answers :

(26)

7x 2 + 7y2 – 10x – 10y – 12 = 0

(27)

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9x 2 + 9y2 – 20x + 15y = 0.

18

MATHS

STRAIGHT LINE A french mathematician and a greatest philosopher named Rene Descartes, pioneered the use of algebra in Geometry. He suggested methods to study geometry by algebraic methods without making direct reference to the actual figures This geometry was called co-ordinate geometry or analytical geometry and it is the branch of geometry in which algebraic equations are used to denote points, lines and curves.

Rectangular cartesian co-ordinate systems : We shall right now focus on two-dimensional co-ordinate geometry in which two perpendicular lines called co-ordinate axes (x-axis and y-axis) are used to locate a point in the plane.

O is called origin. Any point P in this plane can be represented by a unique ordered pair (x, y), which are called co-ordinates of that point. x is called x co-ordinate or abscissa and y is called y co-ordinate or ordinate. The two perpendicular lines xox and yoy divide the plane in four regions which are called quadrants, numbered as shown in the figure.

Let us look at some of the formulae linked with points now.

Distance Formula : The distance between the points A(x 1,y1) and B(x 2,y2) is =

x1  x 2 2  y1  y 2 2

.

Example # 1 : Find the value of x, if the distance between the points (x, –1) and (3, 2) is 5 Solution :

Let P(x ,–1) and Q(3, 2) be the given points. Then PQ = 5 (given)

( x  3 ) 2  ( 1  2 ) 2 = 5



(x – 3)2 + 9 = 25



x = 7 or x = – 1

Self practice problems : (1)

Show that four points (0, –1), (6, 7) (–2, 3) and (8, 3) are the vertices of a rectangle.

(2)

Find the co-ordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, –2) and (2, –2). Also find its circumradius.

Answers :

(2)

(5, 2), 5

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MATHS Section Formula : If P(x, y) divides the line joining A(x 1, y1) & B(x 2, y2) in the ratio m : n, then; x=

Notes : (i)

(ii)

If

mx 2  nx1 my 2  ny1 ;y= mn mn .

m m is positive, the division is internal, but if is negative, the division is external. n n

If P divides AB internally in the ratio m : n & Q divides AB externally in the ratio m : n then P & Q are said to be harmonic conjugate of each other w.r.t. AB. Mathematically,

2 1 1   i.e. AP, AB & AQ are in H.P..

AB AP AQ

Example # 2 : Find the co-ordinates of the point which divides the line segment joining the points (6, 3) and (– 4, 5) in the ratio 3 : 2 (i) internally and (ii) externally.

Solution : (i)

Let P (x, y) be the required point.

For internal division : x=

3  4  2  6 35  23 21 and y = or x = 0 and y = 32 32 5

 21  So the co-ordinates of P are  0,   5 

(ii)

For external division

x=

3  4  2  6 35  23 and y = 32 32

or x = –24 and y = 9 So the co-ordinates of P are (–24, 9) Example # 3 : Find the co-ordinates of points which trisect the line segment joining (1, – 2) and (– 3, 4).

Solution :

Let A (1, –2) and B(–3, 4) be the given points. Let the points of trisection be P and Q. Then AP = PQ = QB =  (say)

 



PB = PQ + QB = 2 and AQ = AP + PQ = 2 AP : PB =  : 2 = 1 : 2 and AQ : QB = 2 :  = 2 : 1 So P divides AB internally in the ratio 1 : 2 while Q divides internally in the ratio 2 : 1  1 3  2  1 1 4  2  2   1  ,  or   , 0  the co-ordinates of P are  1  2 1  2    3 

 2  3  1 1 2  4  1 ( 2)   5  ,  or   , 2  and the co-ordinates of Q are  2 1 2 1    3   1   5  Hence, the points of trisection are   , 0  and   , 2  .

3    3 

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MATHS Self practice problems : (3)

In what ratio does the point (–1, –1) divide the line segment joining the points (4, 4) and (7, 7).

(4)

The three vertices of a parallelogram taken in order are (–1, 0), (3, 1) and (2, 2) respectively.

Find the co-ordinates of the fourth vertex.

Answers :

(3)

5 : 8 externally

(4)

(–2, 1)

Centroid, Incentre & Excentre : If A (x 1, y1), B(x 2, y2), C(x 3, y3) are the vertices of triangle ABC, whose sides BC, CA, AB are of lengths a, b, c respectively, then the co-ordinates of the special points of triangle ABC are as follows :  x 1  x 2  x 3 y1  y 2  y 3  ,  Centroid G   3 3  

 ax 1 bx 2  cx 3 ay 1 by 2  cy 3 , Incentre I   a  b  c ab c 

  ,and 

  ax 1  bx 2  cx 3  ay 1  by 2  cy 3 , Excentre (to A) I1   abc abc  Notes : (i)

  and so on. 

Incentre divides the angle bisectors in the ratio, (b + c) : a; (c + a) : b & (a + b) : c.

(ii)

Incentre and excentre are harmonic conjugate of each other w.r.t. the angle bisector on which they lie.

(iii)

Orthocentre, Centroid & Circumcentre are always collinear & centroid divides the line joining orthocentre & circumcentre in the ratio 2 : 1.

(iv)

In an isosceles triangle G, O,  & C lie on the same line and in an equilateral triangle, all these four points coincide.

In a right angled triangle orthocentre is at right angled vertex and circumcentre is mid point of hypotenuse In case of an obtuse angled triangle circumcentre and orthocentre both are out side the triangle.

(v) (vi)

Example # 4 : Find the co-ordinates of (i) centroid (ii) in-centre of the triangle whose vertices are (0, 6), (8, 12) and (8, 0).

Solution : (i) We know that the co-ordinates of the centroid of a triangle whose angular points are  x 1  x 2  x 3 y1  y 2  y 3  ,  (x 1, y1), (x 2, y2) (x 3, y3) are  3 3  

So the co-ordinates of the centroid of a triangle whose vertices are (0, 6), (8, 12) and  0  8  8 6  12  0   16  ,  or  , 6  .

(8, 0) are  3 3    3 

(ii)

Let A (0, 6), B (8, 12) and C(8, 0) be the vertices of triangle ABC. Then c = AB =

and

a = BC =

(0  8)2  (6  12)2 = 10, b = CA =

(0  8)2  (6  0 )2 = 10

(8  8)2  (12  0)2 = 12.

 ax 1  bx 2  cx 3 ay 1  by 2  cy 3  ,  The co-ordinates of the in-centre are  abc abc  

or

 12  0  10  8  10  8 12  6  10  12  10  0  ,   12  10  10 12  10  10  

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3

MATHS or

 160 192  ,   or (5, 6)  32 32 

Self practice problems : (5)

Two vertices of a triangle are (3, –5) and (–7, 4). If the centroid is (2, –1), find the third vertex.

(6)

Find the co-ordinates of the centre of the circle inscribed in a triangle whose vertices are (– 36, 7), (20, 7) and (0, – 8) Answers :

(5)

(10, – 2)

(6)

(–1, 0)

Area of a Triangle : If A(x 1, y1), B(x 2, y2), C(x 3, y3) are the vertices of triangle ABC, then its area is equal to

x1 1 x2  ABC = 2 x3

y1 1 y 2 1 , provided the vertices are considered in the counter clockwise sense. The y3 1

above formula will give a  ve area if the vertices (x i, yi), i = 1, 2, 3 are placed in the clockwise sense.

Note:

Area of n-sided polygon formed by points (x 1, y1) ; (x 2, y2); ........;(x n, yn) is given by

1 2

 x1 x 2 x   2 y y y2 2  1

x3 y3

 .......... .....

x n1 x n y n1 y n



xn yn

x1   y1  .

Here vertices are taken in order.

Example # 5 : If the co-ordinates of two points A and B are (3, 4) and (5, –2) respectively. Find the coordinates of any point P if PA = PB and Area of PAB = 10.

Solution : 

Let the co-ordinates of P be (x, y). Then PA = PB  PA2 = PB2 2 2 (x – 3) + (y – 4) = (x – 5)2 + (y + 2)2 

1 2

x 3

y 4

1 1

x – 3y – 1 = 0

Now,

Area of PAB = 10 

 

6x + 2y – 46 = 0 or 6x + 2y – 6 = 0 3x + y – 23 = 0 or 3x + y – 3 = 0 Solving x – 3 y – 1 = 0 and 3x + y – 23 = 0 we get x = 7, y = 2. Solving x – 3y – 1 = 0 and 3x + y – 3 = 0, we get x = 1, y = 0. Thus, the co-ordinates of P are (7, 2) or (1, 0)

5 2 1

= ± 10  6x + 2y – 26 = ± 20

Self practice problems : (7)

The area of a triangle is 5. Two of its vertices are (2, 1) and (3, –2). The third vertex lies on y = x + 3. Find the third vertex.

(8)

The vertices of a quadrilateral are (6, 3), (–3, 5), (4, –2) and (x, 3x) and are denoted by A, B, C and D, respectively. Find the values of x so that the area of triangle ABC is double the area of triangle DBC. Answers

(7)

 7 13   3 3  ,  or   ,  2 2    2 2

(8)

x=

11 3 or – 8 8

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MATHS Slope Formula :  is the angle at which a straight line is inclined to the positive direction of x  axis, & 0°  < 180°,  90°, then the slope of the line, denoted by m, is defined by m = tan . If  is 90°, m does not exist, but the line is parallel to the yaxis. If  = 0, then m = 0 & the line is parallel to the x -axis.

If A (x 1, y1) & B (x 2, y2), x 1  x 2, are points on a straight line, then the slope m of the line is given by : If

 y1  y 2  m =  x  x  .

2   1 Example # 6 : What is the slope of a line whose inclination with the positive direction of x-axis is : (i) 0º (ii) 90º (iii) 120º (iv) 150º Solution :

(i)

Here  = 0º Slope = tan  = tan 0º = 0.

(ii) 

Here  = 90º The slope of line is not defined.

(iii)

Here = 120º



Slope = tan  = tan 120º = tan (180º – 60º) = – tan 60º = –

(iv)

Here  = 150º



Slope = tan  = tan 150º = tan (180º – 30º) = – tan 30º = –

3.

1 3

Example # 7 : Find the slope of the line passing through the points : (i) (1, 6) and (– 4, 2) (ii) (5, 9) and (2, 9) Solution :

(i)

Let

A = (1, 6) and B = (– 4, 2)



Slope of AB =

(ii)

Let



Slope of AB =

4 4 26 = = 5 5  4 1

 y  y1   Using slope  2  x 2  x1  

A = (5, 9), B = (2, 9) 99 0 = =0 25 3

Self practice problems : (9)

Find the value of x, if the slope of the line joining (1, 5) and (x, –7) is 4.

(10)

What is the inclination of a line whose slope is (i) 0

Answers :

(ii) 1 (9)

–2

(iv) –1/ 3

(iii) –1 (10)

(i) 0º,

(ii) 45º,

(iii) 135º,

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(iv) 150º

5

MATHS Condition of collinearity of three points : Points A (x 1, y1), B (x 2, y2), C(x 3, y3) are collinear if (i)

 y  y3  y  y2   =  2 m AB = m BC = m CA i.e.  1  x2  x3  x1  x 2 

x1 (ii)

   

y1 1

 ABC = 0 i.e. x 2 y 2 1 = 0 x3

y3 1

(iii)

AC = AB + BC or AB ~ BC

(iv)

A divides the line segment BC in some ratio.

Example # 8 : Show that the points (1, 1), (2, 3) and (3, 5) are collinear.

Solution :

Let (1, 1) (2, 3) and (3, 5) be the co-ordinates of the points A, B and C respectively.

53 3 1 = 2 and Slope of BC = =2 32 2 1 Slope of AB = slope of BC AB & BC are parallel A, B, C are collinear because B is on both lines AB and BC.

Slope of AB =   

Self practice problem : (11)

Prove that the points (a, 0), (0, b) and (1, 1) are collinear if

1 1 + =1 a b

Equation of a Straight Line in various forms : Now let us understand, how a line can be represented with the help of an algebraic equation.

A moving point (point with variable co-ordinates) is assumed on the line and a link is established between its co-ordinates with the help of some given parameters. There are various forms of lines depending on the specified parameter Point - Slope form : y  y1 = m (x  x 1) is the equation of a straight line whose slope is m & which passes through the point (x 1, y1).

Example # 9 : Find the equation of a line passing through (2, –3) and inclined at an angle of 135º with the positive direction of x-axis.

Solution : Here, m = slope of the line = tan 135º = tan (90º + 45º) = – cot 45º = –1, x 1 = 2, y1 = –3 So, the equation of the line is y – y1 = m (x – x 1) i.e. y – (–3) = –1 (x – 2) or y + 3 = – x + 2 or x + y + 1 = 0

Self practice problem : (12)

Find the equation of the perpendicular bisector of the line segment joining the points A(2, 3) and B (6, –5).

Answer : x–2y–6=0

Slope-intercept form : y = mx + c is the equation of a straight line whose slope is m & which makes an intercept c on the y axis.

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MATHS Example # 10 : Find the equation of a line with slope –1 and cutting off an intercept of 4 units on negative direction of y-axis.

Solution : Here m = –1 and c = – 4. So, the equation of the line is y = mx + c i.e. y = – x – 4 or x + y + 4 = 0 Self practice problem : (13)

Find the equation of a straight line which cuts off an intercept of length 3 on y-axis and is parallel to the line joining the points (3, –2) and (1, 4).

Answer : 3x + y – 3 = 0

y 2  y1

Two point form : y  y1 = x  x (x  x1) is the equation of a straight line which passes through the pointss 2 1 (x 1, y1) & (x 2, y2).

Example # 11 : Find the equation of the line joining the points (– 1, 3) and (4, – 2) Solution :

Here the two points are (x 1, y1) = (–1, 3) and (x 2, y2) = (4, –2).

So, the equation of the line in two-point form is y–3=

3  ( 2 ) (x + 1)  1 4



y–3=–x–1



x+y–2=0

Self practice problem : (14)

Find the equations of the sides of the triangle whose vertices are (–1, 8), (4, –2) and (–5, –3).

Also find the equation of the median through (–1, 8) Answer.

2x + y – 6 = 0, x – 9y – 22 = 0, 11x – 4y + 43 = 0, 21x + y + 13 = 0

x

y

1

Determinant form : Equation of line passing through (x1, y1) and (x2, y2) is x1 y1 1  0 x2

y2 1

Example 12 : Find the equation of line passing through (2, 4) & (– 1, 3).

Solution :

x 2

y 1 4 1

1 3 1

=0



x – 3y + 10 = 0

Self practice problem : (15)

Find the equation of the passing through (– 2, 3) & (– 1, – 1).

Answer.

4x + y + 5 = 0

Intercept form :

x y  = 1 is the equation of a straight line which makes intercepts a & b on OX & OY a b

respectively.

Example 13 : Find the equation of the line which passes through the point (3, 4) and the sum of its intercepts on the axes is 14.

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MATHS Solution :

Let the equation of the line be

x y + =1 a b

This passes through (3, 4), therefore It is given that a + b = 14



Putting b = 14 – a in (ii), we get  

....(i)

3 4 + =1 a b

....(ii)

b = 14 – a.

3 4 + =1 a 14  a

a2 – 13a + 42 = 0 (a – 7) (a – 6) = 0  a = 7, 6 For a = 7, b = 14 – 7 = 7 and for a = 6, b = 14 – 6 = 8.

Putting the values of a and b in (i), we get the equations of the lines x y x y + = 1 and + =1 6 8 7 7

or

x + y = 7 and 4x + 3y = 24

Self practice problem : (16)

Find the equation of the line through (2, 3) so that the segment of the line intercepted between the axes is bisected at this point.

Answer.

3x + 2y = 12.

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 O on the line is p and this perpendicular makes an angle  with positive x axis.

Example # 14 : Find the equation of the line which is at a distance 3 from the origin and the perpendicular from the origin to the line makes an angle of 30º with the positive direction of the x-axis.

Solution : Here p = 3,  = 30º  Equation of the line in the normal form is x cos 30º + y sin 30º = 3 or x

y 3 + = 3 or 2 2

3x+y=6

Self practice problem : (17)

The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150º with the positive direction of y-axis. Find the equation of the line.

Answer.

3 x + y – 14 = 0

Parametric form : x  x 1 y  y1 P (r) = (x, y) = (x 1 + r cos , y1 + r sin ) or cos  sin = r 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 (x 1, y1) on the line.

Remark :

The above form is derived from point-slope form of line. y – y1 = m (x – x 1) where m = tan  

y – y1 =

sin  (x – x 1) cos 

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MATHS Example # 15 : Find the equation of the line through the point A(2, 3) and making an angle of 45º with the x-axis. Also determine the length of intercept on it between A and the line x + y + 1 = 0 Solution :

The equation of a line through A and making an angle of 45º with the x-axis is x2 y3 y 3 x2 = or = 1 cos 45 º sin 45 º 1 2 2

or

x–y+1=0

Suppose this line meets the line x + y + 1 = 0 at P such that AP = r. Then the co-ordinates of P are given by x2 y3 = =r cos 45 º sin 45 º





r x=2+

x = 2 + r cos 45º, y = 3 + r sin 45º

r

2

,y=3+

2

 r r   ,3 Thus, the co-ordinates of P are  2  2 2  r Since P lies on x + y + 1 = 0, so 2 + 

2 r = – 6  r = –3 2

r +3+

2

2

+1=0

 length AP = | r | = 3 2

Thus, the length of the intercept = 3 2 .

Self practice problem : (18)

A straight line is drawn through the point A

 3, 2 making an angle of /6 with positive direction

of the x-axis. If it meets the straight line

3 x – 4y + 8 = 0 in B, find the distance between A

and B. Answer.

6 units

General Form : ax + by + c = 0 is the equation of a straight line in the general form In this case, slope of line = –

x - intercept = –

c , a

a b

y - intercept = –

c b

Example # 16 : Find slope, x-intercept & y-intercept of the line 2x – 3y + 5 = 0.

Solution :

Here, a = 2, b = – 3, c = 5 

slope = –

a 2 = b 3

x-intercept = – y-intercept =

c 5 =– a 2

5 3

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MATHS Self practice problem : (19)

Find the slope, x-intercept & y-intercept of the line 3x – 5y – 8 = 0.

3 8 8 , ,– 5 3 5

Answer :

Angle between two straight lines in terms of their slopes: If m 1 & m 2 are the slopes of two intersecting straight lines (m 1 m 2  1) &  is the acute angle between them, then tan  =

Notes : (i)

Let m 1, m 2, m 3 are the slopes of three lines L1 = 0;L2 = 0;L3 = 0 where m 1 > m 2 > m 3 , then the tangent of interior angles of the  ABC formed by these lines are given by, tan A =

(ii)

m1  m 2 .

1  m1m 2

m 2 m 3 m 1 m 2 ; tan B = 1 m 1m 2 1 m 2 m 3

& tan C =

m 3 m 1 1 m 3 m 1

The equation of lines passing through point (x 1, y1) and making angle  with the line y = mx + c are given by : (y  y1) = tan ( ) (x  x 1) & (y  y1) = tan ( + ) (x  x 1), where tan  = m.

Example # 17 : The acute angle between two lines is /4 and slope of one of them is 1/2. Find the slope of the other line.

Solution :

m1  m 2 If  be the acute angle between the lines with slopes m 1 and m 2, then tan  = 1  m m 1 2 Let  =

 4

and m 1 =

1 2



1  m2 2  tan = 1 1  m2 4 2



1  2m 2 1 = 2m 2



1  2m 2 2  m 2 = + 1 or – 1

Now

1  2m 2 2  m2 = 1



m2 = –



The slope of the other line is either – 1/3 or 3

1  2m 2 1 and 2  m = – 1 3 2



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m 2 = 3.

10

MATHS Example # 18 : Find the equation of the straight line which passes through the origin and making angle 60º

Solution : 

with the line x +

3 y + 3 3 = 0.

Given line is x +

3 y + 3 3 = 0.

 1   x–3 y =    3 

1



Slope of (1) = –

3

.

Let slope of the required line be m. Also between these lines is given to be 60º. 

  1  m  1/ 3  m   1/ 3

tan 60º =

3m  1 3 m

=



3

3m  1



3 =

3m+1=3–

3m  1



3 m

3 m 1



3m

=± 3

m=

3

1 Using y = mx + c, the equation of the required line is y = i.e.

x–

3 y = 0.

3m  1 3 m   

3

x+0

( This passes through origin, so c = 0)

=– 3



3m+1 =–3+

3m

m is not defined The slope of the required line is not defined. Thus, the required line is a vertical line. This line is to pass through the origin.

The equation of the required line is x = 0

Self practice problem : (20)

A vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2.

Find the equation of the other sides of the triangle.

Answer.

(2 – 3 )x – y + 2 3 – 1 = 0 and (2 +

3 ) x – y – 2 3 – 1 = 0.

Parallel Lines : (i)

When two straight lines are parallel their slopes are equal. Thus any line parallel to y = mx + c is of the type y = mx + d, where ‘d’ is a parameter.

(ii)

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. (iii)

The distance between two parallel lines with equations ax + by + c 1 = 0 & ax + by + c 2 = 0 is =

c 1 c 2 a 2 b 2

.

Note that coefficients of x & y in both the equations must be same. (iv)

p1p 2 The area of the parallelogram = sin , where p1 & p2 are distances between two pairs of opposite sides &  is the angle between any two adjacent sides. Note that area of the parallelogram bounded by the lines y = m 1x + c 1, y = m 1x + c 2 and y = m 2x + d1, y = m 2x + d2 is given by

( c 1 c 2 )( d1 d 2 )  m 1 m 2

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MATHS Example # 19 : Find the equation of the straight line that has y-intercept 4 and is parallel to the straight line 2x – 3y = 7.

Solution : Given line is 2x – 3y = 7 (1)  





3y = 2x – 7

y=

2 7 x– 3 3

Slope of (1) is 2/3 The required line is parallel to (1), so its slope is also 2/3, y-intercept of required line = 4 By using y = mx + c form, the equation of the required line is y=

2 x + 4 or 2x – 3y + 12 = 0 3

Example # 20 : Two sides of a square lie on the lines x + y = 1 and x + y + 2 = 0. What is its area ?

Solution :

Clearly the length of the side of the square is equal to the distance between the parallel lines x + y – 1 = 0 ........(i) and x + y + 2 = 0 ........(ii) Putting x = 0 in (i), we get y = 1. So (0, 1) is a point on line (i).

Now, Distance between the parallel lines

| 0  1 2 |

= length of the  from (0, 1) to x + y + 2 = 0 =

1 1 2

2

3 =

2

Thus, the length of the side of the square is

2

 3   = 9 and hence its area =  2 2 2  

3

Example # 21 :Find the area of the parallelogram whose sides are x + 2y + 3 = 0, 3x + 4y – 5 = 0, 2x + 4y + 5 = 0 and 3x + 4y – 10 = 0

Solution :

Here,



c1 = –

Area =

3 , 2

c2 = –

 3 5   10 5         2 4  4 4  1 3     2 4

5 , 4

=

d1 =

10 5 1 3 , d2 = , m 1 = – , m 2 = – 4 4 2 4

5 sq. units 4

Self practice problem : (21)

Find the area of parallelogram whose sides are given by 4x – 5y + 1 = 0, x – 3y – 6 = 0, 4x – 5y – 2 = 0 and 2x – 6y + 5 = 0 Answer.

51 sq. units 14

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MATHS Perpendicular Lines: (i)

When two lines of slopes m 1 & m 2 are at right angles, the product of their slopes is

1,

i.e. m 1 m 2 = Thus any line perpendicular to y = mx + c is of the form y=

(ii)



1 x + d, where ‘d’ is any parameter.. m

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 form bx

 ay + k = 0, where ‘k’ is any parameter.

Example # 22 : Find the equation of the straight line that passes through the point (3, 4) and perpendicular to the line 3x + 2y + 5 = 0 Solution :



Aliter

The equation of a line perpendicular to 3x + 2y + 5 = 0 is 2x – 3y +  = 0 ...........(i) This passes through the point (3, 4) 3×2–3×4+=0=6 Putting = 6 in (i), we get 2x – 3y + 6 = 0, which is the required equation.

The slope of the given line is –3/2. Since the required line is perpendicular to the given line. So, the slope of the required line is 2/3. As it passes through (3, 4). So, its equation is y – 4 =

2 3

(x – 3) or 2x – 3y + 6 = 0 Self practice problem : (22)

The vertices of a triangle are A(10, 4), B (–4, 9) and C(–2, –1). Find the equation of its altitudes.

Also find its orthocentre.

Answer :

9  x – 5y + 10 = 0, 12x + 5y + 3 = 0, 14x – 5y + 23 = 0,   1,  5 

Position of the point (x 1 , y 1 ) relative of the line ax + by + c = 0 : If ax 1 + by1 + c is of the same sign as c, then the point (x 1, y1) lie on the origin side of ax + by + c = 0.

But if the sign of ax 1 + by1 + c is opposite to that of c, the point (x 1, 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 ax 1 + by1 + c and ax 2 + by2 + c are of same or opposite sign respectively.

Example # 23 : Show that (1, 4) and (0, –3) lie on the opposite sides of the line x + 3y + 7 = 0.

Solution : 

At (1, 4), the value of x + 3y + 7 = 1 + 3(4) + 7 = 20 > 0.

At (0, – 3), the value of x + 3y + 7 = 0 + 3(–3) + 7 = –2 < 0 The points (1, 4) and (0, – 3) are on the opposite sides of the given line.

Self practice problems : (23)

Are the points (3, – 4) and (2, 6) on the same or opposite side of the line 3x – 4y = 8 ?

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MATHS (24)

Which one of the points (1, 1), (–1, 2) and (2, 3) lies on the side of the line 4x + 3y – 5 = 0 on which the origin lies?

Answers :

(23)

Opposite sides

(24)

(–1, 2)

The ratio in which a given line divides the line segment joining two points : Let the given line ax + by + c = 0 divide the line segment joining A(x 1, y1) & B(x 2, y2) in the ratio m : n, then

ax1by1c m  . If A & B are on the same side of the given line then m/n is negative but if A & n ax 2 by 2 c

B are on opposite sides of the given line, then m/n is positive Example # 24 : Find the ratio in which the line joining the points A (1, 2) and B(– 3, 4) is divided by the line x + y – 5 = 0.

Solution : 

Let the line x + y = 5 divides AB in the ratio k : 1 at P   3k  1 4k  2  ,  co-ordinate of P are   k 1 k 1 

Since P lies on x + y – 5 = 0   Aliter

 3k  1 4k  2 + –5=0  k 1 k 1 Required ratio is 1 : 2 extrenally.

k=–

1 2

Let the ratio is m : n 

(1 1  1 2  5) m 1 =– =– 1 ( 3)  1 4  5 n 2



ratio is 1 : 2 externally.

Self practice problem : (25)

If the line 2x – 3y +  = 0 divides the line joining the points A (– 1, 2) & B(– 3, – 3) internally in the ratio 2 : 3, find .

Answer :

18 5

Length of perpendicular from a point on a line :

The length of perpendicular from P(x 1, y1) on ax + by + c = 0 is

a x1  by1  c a 2  b2

.

Example # 25 : Find the distance between the line 12x – 5y + 9 = 0 and the point (2, 1) Solution :

The required distance =

12  2  5  1  9 12  ( 5) 2

2

=

| 24  5  9 | 28 = 13 13

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MATHS Example # 26 : Find all points on x + y = 4 that lie at a unit distance from the line 4x + 3y – 10 = 0.

Solution :

 

Note that the co-ordinates of an arbitrary point on x + y = 4 can be obtained by putting x = t (or y = t) and then obtaining y (or x) from the equation of the line, where t is a parameter.

Putting x = t in the equation x + y = 4 of the given line, we obtain y = 4 – t. So, co-ordinates of an arbitrary point on the given line are P(t, 4 – t). Let P(t, 4 – t) be the required point. Then, distance of P from the line 4x + 3y – 10 = 0 is unity i.e.

4t  3( 4  t )  10

=1 

42  32

|t + 2| = 5

t+2=±5  t = –7 or t = 3 Hence, required points are (–7, 11) and (3, 1)

Self practice problem : (26)

Find the length of the altitudes from the vertices of the triangle with vertices :(–1, 1), (5, 2) and (3, –1).

16 Answer :

13

8 ,

5

16 ,

37

Reflection of a point about a line : (i)

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

 ax  by1  c  x  x1 y  y1    1 2 2   a b a  b   (ii)

The image of a point (x 1, y1) about the line ax + by + c = 0 is

 ax  by1  c  x  x1 y  y1 .   2  1 2   a b a  b2   Example # 27 : Find the foot of perpendicular of the line drawn from P (– 3, 5) on the line x – y + 2 = 0.

Solution :

Slope of PM = – 1



Equation of PM is x+y–2=0 .........(i) solving equation (i) with x – y + 2 = 0, we get co-ordinates of M (0, 2)

Aliter Here,   

(1 ( 3)  ( 1)  5  2) x3 y 5 = =– (1)2  ( 1)2 1 1 x3 y 5 = =3 1 1 x+3=3  M is (0, 2)

x=0

and

y–5=–3



y=2

Example # 28 : Find the image of the point P(–1, 2) in the line mirror 2x – 3y + 4 = 0.

Solution : 

Let image of P is Q. PM = MQ & PQ  AB Let Q is (h, k)

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MATHS 

 h 1 k  2  ,  M is  2   2

It lies on 2x – 3y + 4 = 0. 

 h  1 k 2  –3   + 4 = 0.

2  2    2 

or

2h – 3k = 0

...........(i)

slope of PQ =

k2 h 1

PQ  AB 

2 k2 × = – 1.

3 h 1



3h + 2k – 1 = 0.

........(ii)

soving (i) & (ii), we get h =



Aliter

3 2 ,k= 13 13

 3 2  Image of P(– 1, 2) is Q  ,   13 13 

The image of P (– 1, 2) about the line 2x – 3y + 4 = 0 is

[2( 1)  3(2)  4] y2 x 1 = =–2 2 2  ( 3 ) 2 3 2

y2 8 x 1 = = 3 13 2



13x + 13 = 16





 3 2   image is  ,  13 13 

x=

3 & 13y – 26 = – 24 13



y=

2 13

Self practice problems : (27)

Find the foot of perpendicular of the line drawn from (– 2, – 3) on the line 3x – 2y – 1 = 0.

(28)

Find the image of the point (1, 2) in y-axis.

Answers

(27)

  23  41  ,    13 13 

(28)

(– 1, 2)

Bisectors of the angles between two lines: Equations of the bisectors of angles between the lines ax + by + c = 0 & ax + by + c  = 0 (ab

ab) are :

ax  by  c a2  b2

=±

ax  by  c  a  2  b 2

Note : Equation of straight lines passing through P(x 1, y1) & equally inclined with the lines a1x + b1y + c 1 = 0 & a2x + b2y + c 2 = 0 are those which are parallel to the bisectors between these two lines & passing through the point P. Example # 29 : Find the equations of the bisectors of the angle between the straight lines 3x – 4y + 7 = 0 and 12 x – 5y – 8 = 0.

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MATHS Solution :

The equations of the bisectors of the angles between 3x – 4y + 7 = 0 and 12 x – 5y – 8 = 0 are 3x  4y  7 3 2  ( 4 ) 2

or or

12 x  5 y  8

=±

12 2  ( 5) 2

3x  4y  7 12 x  5 y  8 =± 5 13

39x – 52y + 91 = ± (60 x – 25 y – 40) Taking the positive sign, we get 21 x + 27 y – 131 = 0 as one bisector Taking the negative sign, we get 99 x – 77 y + 51 = 0 as the other bisector.

Self practice problem : (29)

Find the equations of the bisectors of the angles between the following pairs of straight lines 3x + 4y + 13 = 0 and 12x – 5y + 32 = 0 Answer : 21x – 77y – 9 = 0 and 99x + 27y + 329 = 0

Methods to discriminate between the acute angle bisector & the obtuse angle bisector:  be the angle between one of the lines & one of the bisectors, find tan  If tan  < 1, then 2  < 90° so that this bisector is the acute angle bisector.

If tan > 1, then we get the bisector to be the obtuse angle bisector.

(i)

If

(ii)

Let L1 = 0 & L2 = 0 are the given lines & u1 = 0 and u2 = 0 are the bisectors between L1 = 0 & L2 = 0. Take a point P on any one of the lines L1 = 0 or L2 = 0 and drop perpendicular on u1 = 0 & u2 = 0 as shown in figure. If,

p < q  u1 is the acute angle bisector. p > q  u1 is the obtuse angle bisector. p = q  the lines L1 & L2 are perpendicular.

(iii)

If aa + bb < 0, then the equation of the bisector of this acute angle is

a x + by + c a 2  b2

=+

a  x + b  y + c a  2  b 2

If, however, aa+ bb > 0, the equation of the bisector of the obtuse angle is :

a x + by + c a b 2

2

=+

a  x + b  y + c a  2  b 2

Example # 30 : For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0, find the equation of the (i) bisector of the obtuse angle between them; (ii) bisector of the acute angle between them; Solution : (i) The equations of the given straight lines are 4x + 3y – 6 = 0 ........(1) 5x + 12y + 9 = 0 ........(2) The equation of the bisectors of the angles between lines (1) and (2) are

4x  3y  6 4 3 2

2

5 x  12y  9 =±

5  12 2

2

or

4x  3y  6 5 x  12 y  9 =± 5 13

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MATHS Taking the positive sign, we have or or

52x + 39y – 78 = 25x + 60y + 45 or 27x – 21y – 123 = 0 9x – 7y – 41 = 0 Taking the negative sign, we have

or

and

4x  3y  6 5 x  12 y  9 = 5 13

4x  3y  6 5 x  12 y  9 =– 5 13

52x + 39y – 78 = –25x – 60y – 45 or 77x + 99y – 33 = 0 or 7x + 9y – 3 = 0 Hence the equation of the bisectors are 9x – 7y – 41 = 0 ........(3) 7x + 9y – 3 = 0 ........(4) 4 9 Now slope of line (1) = – and slope of the bisector (3) = .

3 7 If  be the acute angle between the line (1) and the bisector (3), then

tan  =

9 4  7 3 9 4 1    7 3

27  28 21  36

=

=

55 11 = >1  15 3



 > 45º Hence 9x – 7y – 41 = 0 is the bisector of the obtuse angle between the given lines (1) and (2)

(ii)

Since 9x – 7y – 41 is the bisector of the obtuse angle between the given lines, therefore the other bisector 7x + 9y – 3 = 0 will be the bisector of the acute angle between the given lines.

2nd Method : Writing the equation of the lines so that constants become positive we have – 4x – 3y + 6 = 0 .......(1) and 5x + 12y + 9 = 0 .......(2) Here a1 = – 4, a2 = 5, b1 = –3, b2 = 12 Now a1a2 + b1b2 = – 20 – 36 = –56 < 0  origin does not lie in the obtuse angle between lines (1) and (2) and hence equation of the 4 x  3 y  6 5 x  12y  9 bisector of the obtuse angle between lines (1) and (2) will be 2 2 =– ( 4)  ( 3) 5 2  122 or 13(–4x – 3y + 6) = –5(5x + 12y + 9) or 27x – 21y – 123 = 0 or 9x – 7y – 41 = 0 and the equation of the bisector of the acute angle will be (origin lies in the acute angle) 4 x  3 y  6 ( 4)  ( 3) 2

or

2

5 x  12y  9 =

77x + 99y – 33 = 0

5 2  122 or

7x + 9y – 3 = 0

Self practice problem : (30)

Find the equations of the bisectors of the angles between the lines x + y – 3 = 0 and 7x – y + 5 = 0 and state which of them bisects the acute angle between the lines.

Answer.

x – 3y + 10 = 0 (bisector of the obtuse angle); 6x + 2y – 5 = 0 (bisector of the acute angle)

To discriminate between the bisector of the angle containing a point : To discriminate between the bisector of the angle containing the origin & that of the angle not containing the origin. Rewrite the equations, ax + by + c = 0 & ax + by + c = 0 such that the constant terms c, c  are positive. Then ;

a x + by + c a 2  b2

= +

a  x + b  y + c a  2  b 2

gives the equation of the bisector of the

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MATHS angle containing the origin &

a x + by + c a b 2

2

=

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 ,

a x  by  c 2

a b

2

=

a  x  b y  c 2

a   b

2

or

a x  by  c 2

a b

2

=



a  x  b y  c 2

a   b

2

according as

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

Example # 31 : For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0, find the equation of the bisector of the angle which contains the origin.

Solution :

For point O(0, 0), 4x + 3y – 6 = –6 < 0 and 5x + 12y + 9 = 9 > 0 Hence for point O(0, 0) 4x + 3y – 6 and 5x + 12y + 9 are of opposite signs.

Hence equation of the bisector of the angle between the given lines containing the origin will be 4x  3y  6 ( 4 )  (3 ) 2

2

5 x  12y  9 =–

5 2  122

or

4x  3y  6 5 x  12 y  9 =– 5 13

or

52x + 39y – 78 = –25x – 60y – 45.

or

77x + 99y – 33 = 0

or

7x + 9y – 3 = 0

Self practice problem : (31)

Find the equation of the bisector of the angle between the lines x + 2y – 11 = 0 and 3x – 6y – 5 = 0 which contains the point (1, – 3).

Answer : 3x – 19 = 0

Condition of Concurrency : Three lines a1x + b1y + c 1 = 0, a2x + b2y + c 2 = 0 & a3x + b3y + c 3 = 0 are concurrent if

a1 a2

b1 b2

a3

b3

c1 c2 = 0. c3

Alternatively : If three constants A, B & C (not all zero) can be found such that A(a1x + b1y + c 1) + B(a2x + b2y + c 2) + C(a3x + b3y + c 3)  0, then the three straight lines are concurrent.

Example # 32 : Prove that the straight lines 4x + 7y = 9, 5x – 8y + 15 = 0 and 9x – y + 6 = 0 are concurrent.

Solution :

and

Given lines are 4x + 7y – 9 = 0 5x – 8y + 15 = 0 9x – y + 6 = 0

=

4 7 9 5  8 15 9

1

........(1) ........(2) ........(3)

= 4(–48 + 15) – 7 (30 – 135) – 9 (– 5 + 72) = –132 + 735 – 603 = 0

6

Hence lines (1), (2) and (3) are concurrent.

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MATHS Self practice problem : (32)

Find the value of m so that the lines 3x + y + 2 = 0, 2x – y + 3 = 0 and x + my – 3 = 0 may be concurrent.

Answer : 4

Family of Straight Lines : The equation of a family of straight lines passing through the point of intersection of the lines, L1  a1x + b1y + c 1 = 0 & L2  a2x + b2y + c 2 = 0 is given by L1 + k L2 = 0 i.e. (a1x + b1y + c 1) + k(a2x + b2y + c 2) = 0, where k is an arbitrary real number.

Note :(i)

(ii)

If u1 = ax + by + c, u2 = ax + by + d, u3 = ax + by + c, u4 = ax + by + d then u1 = 0;u2 = 0; u3 = 0 ; u4 = 0 form a parallelogram.

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

The diagonal AC is also given by u1 +  u4 = 0 and

u2 + u3 = 0, if the two equations are identical for some real [For getting the values of

 and   & compare the coefficients of x, y & the constant terms].

Example # 33 : Find the equation of the straight line which passes through the point (2, –3) and the point of intersection of the lines x + y + 4 = 0 and 3x – y – 8 = 0.

Solution :

or

Any line through the intersection of the lines x + y + 4 = 0 and 3x – y – 8 =0 has the equation (x + y + 4) +  (3x – y – 8) = 0 .........(i) This will pass through (2, –3) if (2 – 3 + 4) +  (6 + 3 – 8) = 0 or 3 +  = 0   = – 3.

Putting the value of  in (i), the required line is (x + y + 4) + (–3) (3x – y – 8) = 0 – 8x + 4y + 28 = 0 or 2x – y – 7 = 0

Aliter Solving the equations x + y + 4 = 0 and 3x – y – 8 = 0 by cross-multiplication, we get x = 1, y = –5 So the two lines intersect at the point (1, –5). Hence the required line passes through (2, –3) and (1, –5) and so its equation is y + 3 =

–5  3 (x – 2) or 2x – y – 7 = 0 1 2

Example # 34 : Obtain the equations of the lines passing through the intersection of lines 4x – 3y – 1 = 0 and 2x – 5y + 3 = 0 and equally inclined to the axes.

Solution : or

The equation of any line through the intersection of the given lines is (4x – 3y – 1) +  (2x – 5y + 3) = 0 x (2  + 4) – y (5 + 3) + 3 – 1 = 0 .......(i) Let m be the slope of this line. Then m =

2  4 5  3

As the line is equally inclined with the axes, therefore m = tan 45º or m = tan 135º



m = ±1,

2  4 =±1 5  3

1 , putting the values of  in (i), we get 2x + 2y – 4 = 0 and 14x – 14y = 0 3



 = –1 or

i.e.

x + y – 2 = 0 and x = y as the equations of the required lines.

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MATHS Self practice problem : (33)

Find the equation of the lines through the point of intersection of the lines x – 3y + 1 = 0 and 2x + 5y – 9 = 0 and whose distance from the origin is Answer :

5

2x + y – 5 = 0

A Pair of straight lines through origin: (i)

A homogeneous equation of degree two, "ax² + 2hxy + by² = 0" always represents a pair of straight lines passing through the origin if : (a) (b) (c)

 h² = ab  h² < ab  h² > ab

lines are real & distinct . lines are coincident . lines are imaginary with real point of intersection i.e. (0, 0)

This equation is obtained by multiplying the two equations of lines (m 1x – y) (m 2x – y) = 0  m 1m 2x 2 – (m 1 + m 2) xy + y2 = 0 (ii)

If y = m 1x & y = m 2x be the two equations represented by ax² + 2hxy + by² = 0, then; m1 + m2 = 

(iii)

a 2h & m1 m2 = . b b

If  is the acute angle between the pair of straight lines represented by,

ax² + 2hxy + by² = 0, then tan (iv)

 =

2 h 2  ab .

ab

The condition that these lines are : (a) (b) (c)

at right angles to each other is a + b = 0. i.e. coefficient of x² + coefficient of y² = 0. coincident is h² = ab . equally inclined to the axis of x is h = 0.i.e. coeff. of xy = 0 .

Note that a homogeneous equation of degree n represents n straight lines passing through origin. (v)

The equation to the pair of straight lines bisecting the angles between the straight lines ax² + 2hxy + by² = 0 is

xy x2  y2 = . h ab

Example # 35 : Show that the equation 6x 2 – 5xy + y2 = 0 represents a pair of distinct straight lines, each passing through the origin. Find the separate equations of these lines.

Solution :



The given equation is a homogeneous equation of second degree. So, it represents a pair of straight lines passing through the origin. Comparing the given equation with ax 2 + 2hxy + by2 = 0, we obtain a = 6, b = 1 and 2h = – 5.

25 1 –6= > 0  h2 > ab 4 4 Hence, the given equation represents a pair of distinct lines passing through the origin. h2 – ab =

2

2

2

Now, 6x – 5xy + y = 0



y y   –5   +6=0 x x

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MATHS 2

 

y y y   –3   –2   +6=0 x x x

y  y    3    2 = 0 x  x 

y y – 3 = 0 or – 2 = 0  y – 3x = 0 or y – 2x = 0 x x So the given equation represents the straight lines y – 3x = 0 and y – 2x = 0.

Example # 36 : Find the equations to the pair of lines through the origin which are perpendicular to the lines represented by 2x 2 – 7xy + 3y2 = 0.

Solution :  



We have 2x 2 – 7xy + 3y2 = 0.

2x 2 – 6xy – xy + 3y2 = 0  2x(x – 3y) – y (x – 3y) = 0 (x – 3y) (2x – y) = 0  x – 3y = 0 or 2x – y = 0 Thus the given equation represents the lines x – 3y = 0 and 2x – y = 0. The equations of the lines passing through the origin and perpendicular to the given lines are y – 0 = –3 (x – 0) 1 and y – 0 = – (x – 0) [  (Slope of x – 3 y = 0) is 1/3 and (Slope of 2x – y = 0) is 2] 2 y + 3x = 0 and 2y + x = 0

Example # 37 : Find the angle between the pair of straight lines 4x 2 + 24xy + 11y2 = 0 Solution :

Given equation is 4x 2 + 24xy + 11y2 = 0 Here a = coeff. of x 2 = 4, b = coeff. of y2 = 11 and 2h = coeff. of xy = 24  h = 12 Now tan  =

2 h 2  ab ab

=

2 144  44 4  11

=

4 3

Where  is the acute angle between the lines. 

4 4 acute angle between the lines is tan–1   and obtuse angle between them is  – tan–1   3 3

Example # 38 : Find the equation of the bisectors of the angle between the lines represented by 3x 2 – 5xy + 4y2 = 0 Solution :

Given equation is 3x 2 – 5xy + 4y2 = 0 comparing it with the equation ax 2 + 2hxy + by2 = 0 we have a = 3, 2h = –5; and b = 4

.......(1) .......(2)

Now the equation of the bisectors of the angle between the pair of lines (1) is

or

or

x2  y2 xy = ; 5 34  2

or

x2  y2 xy = ab h

2xy x2  y2 = 5 1

5x 2 – 2xy – 5y2 = 0

Self practice problems : (34)

Find the area of the triangle formed by the lines y2 – 9xy + 18x 2 = 0 and y = 9.

(35)

If the pairs of straight lines x 2 – 2pxy – y2 = 0 and x 2 – 2qxy – y2 = 0 be such that each pair bisects the angle between the other pair, prove that pq = –1.

Answers :

(34)

27 sq. units 4

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MATHS General equation of second degree representing a pair of Straight lines : (i)

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

a h g abc + 2fgh

 af²  bg²  ch² = 0, i.e. if h b f = 0. g f

c

Such an equation is obtained again by multiplying the two equation of lines (a1x + b1y + c 1) (a2x + b2y + c 2) = 0 (ii)

The angle

 between the two lines representing by a general equation is the same as that

between the two lines represented by its homogeneous part only.

Example # 39 : Prove that the equation 2x 2 + 5xy + 3y2 + 6x + 7y + 4 = 0 represents a pair of straight lines.

Find the co-ordinates of their point of intersection.

Solution :



Given equation is 2x 2 + 5xy + 3y2 + 6x + 7y + 4 = 0 Writing the equation (1) as a quadratic equation in x we have 2x 2 + (5y + 6) x + 3y2 + 7y + 4 = 0 x=

=

=

 (5 y  6)  (5y  6)2  4.2(3y 2  7 y  4) 4

 (5 y  6)  25y 2  60y  36  24y 2  56 y  32 4  (5 y  6)  y 2  4 y  4 4

=

(5 y  6)  ( y  2) 4

5 y  6  y  2 5 y  6  y  2 , 4 4



x=

or or

4x + 4y + 4 = 0and 4x + 6y + 8 = 0 x + y + 1 = 0 and 2x + 3y + 4 = 0 Hence equation (1) represents a pair of straight lines whose equation are x+y+1=0 .....(1) 2x + 3y + 4 = 0 .....(2) Solving these two equations, the required point of intersection is (1, – 2).

and

Self practice problem : (36)

Find the combined equation of the straight lines passing through the point (1, 1) and parallel to the lines represented by the equation x 2 – 5xy + 4y2 + x + 2y – 2 = 0 and find the angle between them.

Answer :

3 x 2 – 5xy + 4y2 + 3x – 3y = 0, tan–1   5

Homogenization : This method is used to write the joint equation of two lines connecting origin to the points of intersection of a given line and a given second degree curve.

The equation of a pair of straight lines joining origin to the points of intersection of the line

 x + my + n = 0 and a second degree curve S  ax² + 2hxy + by² + 2gx + 2fy + c = 0 L

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MATHS 2

 x  my   x  my   x  my    2fy    c   = 0. is ax² + 2hxy + by² + 2gx   n   n   n  The equation is obtained by homogenizing the equation of curve with the help of equation of line.

Notes : (i)

Here we have written 1 as

x  my and converted all terms of the curve to second degree n

expressions (ii)

Equation of any curve passing through the points of intersection of two curves C 1 = 0 and C2 = 0 is given by  C1 +  C2 = 0, where

 &  are parameters.

Example # 40 : Prove that the angle between the lines joining the origin to the points of intersection of the straight line y = 3x + 2 with the curve x 2 + 2xy + 3y2 + 4x + 8y – 11 = 0 is tan–1 Solution :

2 2 .

3

Equation of the given curve is x 2 + 2xy + 3y2 + 4x + 8y – 11 = 0 and equation of the given straight line is y – 3x = 2;



y  3x =1 2

Making equation (1) homogeneous equation of the second degree in x any y with the help  y  3x   y  3x   y  3x   + 8y   – 11  of (1), we have x 2 + 2xy + 3y2 + 4x  1   2   2   2 

2

=0

1 11 2 (4xy + 8y2 – 12x 2 – 24 xy) – (y – 6xy + 9x 2) = 0 2 4

or

x 2 + 2xy + 3y2 +

or or or

4x 2 + 8xy + 12y2 + 2(8y2 – 12x 2 – 20xy) – 11 (y2 – 6xy + 9x 2) = 0 –119x 2 + 34xy + 17y2 = 0 or 119x 2 – 34xy – 17y2 = 0 7x 2 – 2xy – y2 = 0 This is the equation of the lines joining the origin to the points of intersection of (1) and (2).

Comparing equation (3) with the equation ax 2 + 2hxy + by2 = 0 we have a = 7, b = –1 and 2h = –2 i.e. h = –1 If  be the acute angle between pair of lines (3), then

tan  =

2 h 2  ab ab

=

2 1 7 7 1

=

2 8 2 2 = 6 3



 = tan–1

2 2 .

3

Self practice problems : (37)

Find the equation of the straight lines joining the origin to the points of intersection of the line 3x + 4y – 5 = 0 and the curve 2x 2 + 3y2 = 5.

(38)

Find the equation of the straight lines joining the origin to the points of intersection of the line lx + my + n = 0 and the curve y2 = 4ax. Also, find the condition of their perpendicularity.

Answers : (37) x 2 – y2 – 24xy = 0 (38) 4alx 2 + 4amxy + ny2 = 0; 4al + n = 0

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MATHS

Ellipse In this chapter we are going to discuss in detail the nature of path in which on planets move around the sun.

They follow on elliptical path with the sun at one of its foci. Let us look at the definition of ellipse.

Definitions : It is locus of a point which moves in such a way that the ratio of its distance from a fixed point and a fixed line (not passes through fixed point and all points and line lies in same plane) is constant (e), which is less than one.

The fixed point is called - focus The fixed line is called -directrix.

The constant ratio is called - eccentricity, it is denoted by 'e'.

Example # 1 : Find the equation to the ellipse whose focus is the point (– 1, 1), whose directrix is the straight line x – y + 3 = 0 and eccentricity is Solution :

1 .

2

Let P  (h, k) be moving point,

e=

PS 1 = PM 2 1 4

hk 3   2  

2



(h + 1)2 + (k – 1)2 =



locus of P(h, k) is 8 {x 2 + y2 + 2x – 2y + 2} = (x 2 + y2 – 2xy + 6x – 6y + 9) 7x 2 + 7y2 + 2xy + 10x – 10 y + 7 = 0.

Note : The general equation of a conic with focus (p, q) & directrix x + my + n = 0 is: (2 + m 2) [(x  p)2 + (y  q)2] = e2 (x + my + n)2  ax 2 + 2hxy + by2 + 2gx + 2fy + c = 0 represent ellipse if 0 < e < 1;   0, h² < ab Self Practice Problem :

1 (1)

Find the equation to the ellipse whose focus is (0, 0) directrix is x + y – 1 = 0 and e =

2

.

3x 2 + 3y2 – 2xy + 2x + 2y – 1 = 0.

Answer :

Standard Equation Standard equation of an ellipse referred to its principal axes along the coordinate axes is

x2 a2



y2 b2

= 1,

where a > b & b² = a² (1  e²).

b2 Eccentricity: e = 1  2 , (0 < e < 1) a

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MATHS Focii : S  (a e, 0) & S  ( a e, 0). a a &x= . e e

Equations of Directrices : x =

Major Axis : The line segment AA in which the focii S & S lie is of length 2a & 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).

Minor Axis : The yaxis intersects the ellipse in the points B  (0,  b) & B  (0, b). The line segment BB 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 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.

2b 2 minor axis   2a 1 e 2 a major axis



2

Length of latus rectum (LL) =



= 2 e (distance from focus to the corresponding directrix) Centre : The point which bisects every chord of the conic drawn through it, is called the centre of the 2 y2 conic. C  (0, 0) the origin is the centre of the ellipse x 2  2 = 1.

a

Note : (i) (ii)

b

2 y2 If the equation of the ellipse is given as x 2  2 = 1 and nothing is mentioned, then the rule is

a

b

to assume that a > b.

If b > a is given, then the yaxis will become major axis and x-axis will become the minor axis and all other points and lines will change accordingly.

x2 y2 + =1 a2 b2

Equation :

Foci

(0, be)

a2 = b2 (1 – e2), a < b.

(0, b) ;

Vertices  (L·R.) =

y = 

b e



e=

1–

L.R.

y = be

Directrices :

2a 2 , b

a2 b2

centre : (0, 0)

Example # 2 : Find the equation to the ellipse whose centre is origin, axes are the axes of co-ordinate and passes through the points (2, 2) and (3, 1).

Solution :

Let the equation to the ellipse is

x2 a

2

+

y2 b2

=1

Since it passes through the points (2, 2) and (3, 1) 

4 a2

+

4 b2

=1

..........(i)

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MATHS 9

and

a

2

1

+

=1

b2

.........(ii)

from (i) – 4 (ii), we get

4  36 a

= 1– 4 

2

a2 =

32 3

from (i), we get

1 b

2

b2 = 

=

3 83 1 – = 32 32 4

32 5

Ellipse is 3x 2 + 5y2 = 32 1 3 Since both focus lies on x-axis, therefore x-axis is major axis and mid point of focii is origin which is centre and a line perpendicular to major axis and passes through centre is minor axis which is y-axis.

Example # 3 : Find the equation of the ellipse whose focii are (4, 0) and (– 4, 0) and eccentricity is Solution :

Let equation of ellipse is

x2 a2



ae = 4

and



a = 12

and



1  b2 = 144 1   9 

+

y2 b2

=1

1 (Given) 3 2 2 b = a (1 – e2)

e=

b2 = 16 × 8 b=8 2 Equation of ellipse is

y2 x2 + =1 128 144

Example # 4 : If minor-axis of ellipse subtend a right angle at its focus then find the eccentricity of ellipse.

Solution :

Let the equation of ellipse is BSB =

and

OB = OB



BSO =



OS = OB



e2 =

a

2

a2

+

y2 b2

=1

(a > b)

 2



b2

x2

 4

= 1 – e2



ae = b



e=

1 2

Example # 5 : From a point Q on the circle x 2 + y2 = a2, perpendicular QM are drawn to x-axis, find the locus of point 'P' dividing QM in ratio 2 : 1.

Solution : Let Q  (a cos, a sin)

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MATHS M  (a cos, 0) Let P  (h, k) a sin  3



h = a cos, k =



 3k  h   +   =1  a  a



Locus of P is

2

2

x2 a2

+

y2

=1

(a / 3 ) 2

Example # 6 : Find the equation of axes, directrix, co-ordinate of focii, centre, vertices, length of latus rectum and eccentricity of an ellipse

Solution :

( x  3) 2 ( y  2) 2 + = 1.

25 16

Let x – 3 = X, y – 2 = Y, so equation of ellipse becomes as   

equation of major axis is Y = 0 equation of minor axis is X = 0 centre (X = 0, Y = 0)

X2 5

2

+

Y2 42

= 1.

y = 2. x = 3. x = 3, y = 2 C  (3, 2)

Length of semi-major axis a = 5 Length of major axis 2a = 10 Length of semi-minor axis b = 4 Length of minor axis = 2b = 8.

Let 'e' be eccentricity  b2 = a2 (1 – e2) a2  b2

e=

a2

=

25  16 3 = .

25 5

Length of latus rectum = LL =

2  16 32 2b 2 = = 5 5 a

Co-ordinates focii are X = ± ae, Y = 0  S  (X = 3, Y = 0) &  S  (6, 2) &

S (X = –3, Y = 0) S (0, 2)

Co-ordinate of vertices Extremities of major axis A  (X = a, Y = 0)  A  (x = 8, y = 2) A  (8, 2) Extremities of minor axis B (X = 0, Y = b) B (x = 3, y = 6) B  (3, 6) a Equation of directrix X = ± e x–3=±

25 3



x=

& & & & & &

34 3

A (X = – a, Y = 0) A = (x = – 2, 2) A (– 2, 2) B  (X = 0, Y = – b) B (x = 3, y = – 2) B (3, – 2)

&

x=–

16 3

Self Practice Problem (2)

Find the equation to the ellipse whose axes are of lengths 6 and 2 6 and their equations are x – 3y + 3 = 0 and 3x + y – 1 = 0 respectively.

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MATHS (3)

Find the eccentricity of ellipse whose minor axis is double the latus rectum.

(4)

Find the co-ordinates of the focii of the ellipse 4x 2 + 9y2 = 1.

(5)

Find the standard ellipse

(6)

A point moves so that the sum of the squares of its distances from two intersecting lines is constant (given that the lines are neither perpendicular nor they make complimentry angle).

Prove that its locus is an ellipse.

Hint. : Assume the lines to be y = mx and y = – mx.

Answers :

(2)

x2 a

2

+

y2 b

2

= 1 passing through (2, 1) and having eccentricity

1 .

2

3(x – 3y + 3)2 + 2(3x+ y – 1)2 = 180, 21x 2 – 6xy + 29y2 + 6x – 58y – 151 = 0.

(3)

3 2

(4)

    5 , 0  6   

3x 2 + 4y2 = 16

(5)

Auxiliary Circle / Eccentric Angle : 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 & the auxiliary circle respectively. ‘’ is called the Eccentric Angle of the point P on the ellipse ( < ). Q  (a cos , a sin) P  (a cos , b sin) Note that :

(PN) b Semi minor axis   (QN) a Semi major axis NOTE : 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.

Example # 7 : Find the focal distance of a point P() on the ellipse Solution :

x2 a

2

+

y2 b2

=1

(a > b)

Let 'e' be the eccentricity of ellipse.  PS = e . PM

and

a  = e   a cos   e  PS = (a – a e cos) PS = e. PM

a  = e  a cos    e  PS = a + ae cos  focal distance are (a ± ae cos) Note : PS + PS = 2a PS + PS = AA

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MATHS Eample # 8 :

Find the distance from centre of the point P on the ellipse

Solution :

angle  with x-axis.

Let P  (a cos, b sin)  m (op) =

b tan = tan a

OP =



a 2 cos 2   b 2 sin 2  =

a2  b2  =

a 2  b 2 tan 2 

OP=

1  tan 2 

=

1

tan =

x2 a2

+

y2 b2

= 1 whose radius makes

a tan  b

a 2  b 2 tan 2  sec 2 

a2 tan2  2 b

a2 tan 2  2 b

ab a sin   b 2 cos 2  2

2

Self Practice Problem (7)

Find the distance from centre of the point P on the ellipse

x2 a2

+

y2 b2

= 1 whose eccentric

angle is  (8)

Find the eccentric angle of a point on the ellipse

x2 y2 + = 1 whose distance from the 6 2

centre is 2.

(9)

Show that the area of triangle inscribed in an ellipse bears a constant ratio to the area of the triangle formed by joining points on the auxiliary circle corresponding to the vertices of the first triangle.

Answers :

(7)

r  a 2 cos 2   b 2 sin 2 

(8)

±

 3 ,± 4 4

Parametric Representation : 2 y2 The equations x = a cos  & y = b sin  together represent the ellipse x 2  2 = 1.

a

b

Where  is a parameter. Note that if P()  (a cos  b sin ) is on the ellipse then; Q()  (a cos  a sin ) is on the auxiliary circle.

The equation to the chord of the ellipse joining two points with eccentric angles  &  is given by x  y    cos  sin  cos . a 2 b 2 2

Example # 9 : Write the equation of chord of an ellipse

Solution :

Equation of chord is

  5  x2 y2 . + =1 joining two points P   and Q  4  4  25 16

  5    5    5           x y 4 4  4 4  4 4    cos + . sin = cos 5 2 4 2 2

 3   3  x y  +  =0 . cos  . sin  5  4   4  4

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MATHS –

x y + =0 5 4



4x = 5y

Example # 10 : If P() and P() are extremities of a focal chord of ellipse then prove that its eccentricity

e=

Solution :

  cos    2  .  cos   2 

Let the equation of ellipse is



equation of chord is

x2 a2

+

y2 b2

=1

      x y  +  = cos   cos  sin  2 2 a b      2 

Since above chord is focal chord,  it passes through focus (ae, 0) or (– ae, 0) 



Note : 

     = cos   ± e cos   2   2 

  cos    2  e=  cos   2    cos 2 ±e=  cos 2

Ans.

  . tan 2 2 ±e=   1  tan . tan 2 2 1  tan



2 1 e =    e 1 2 tan . tan 2 2 1 e e 1   tan tan = or e 1 1 e 2 2

Applying componendo and dividendo

Example # 11 : Find the angle between two diameters of the ellipse eccentricity angle  and  =  + Solution :



Let ellipse is

x2 a2

y2

+

b2

Slope of OQ = m 2 =

b sin  b =– cot a cos  a

=

y2 b2

= 1. Whose extremities have

=1

b sin  b = tan a cos  a

tan =

a2

+

 .

2

Slope of OP = m 1 =

m1  m 2 1  m1m 2

x2

b (tan   cot  ) a b2 1 2 a

given  =  +

=

 2

2ab (a  b 2 ) sin 2 2

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MATHS Self Practice Problem : (10)

Find the sum of squares of two diameters of the ellipse

eccentric angles differ by

(11)

a2

+

y2 b2

= 1 whose extremities have

 and show that it is constant.

2

Show that the sum of squares of reciprocals of two perpendicular diameters of the ellipse x2 a

(12)

x2

2

+

y2 b2

= 1 is constant. Find the constant also.

Find the locus of the foot of the perpendicular from the centre of the ellipse chord joining two points whose eccentric angles differ by

Answers :

(10)

4(a2 + b2)

1 1 1   2  2 4a b 

(11)

x2 a2

+

y2 b2

= 1 on the

 .

2 2(x 2 + y2)2 = a2 x 2 + b2 y2.

(12)

Position of a Point w.r.t. an Ellipse : The point P(x 1, y1) lies outside, inside or on the ellipse according as ;

x12 a2



y12 b2

Example # 12 : Check whether the point P(3, 2) lies inside or outside of the ellipse

Solution :

S1  

 1 > < or = 0.

x2 y2 + = 1.

25 16

9 4 9 1 + –1= + –1<0 25 16 25 4

Point P  (3, 2) lies inside the ellipse.

Example # 13 : Find the set of value(s) of '' for which the point P(, – ) lies inside the ellipse Solution :

If P(, – ) lies inside the ellipse  S1 < 0 

2 2 + –1<0 16 9



 12 12  .    ,  5 5 

25 . 2 < 1 144





2 <

x2 y2 + = 1.

16 9

144 25

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

x2 a

+

2

y2 b2

= 1 in two points real, coincident or imaginary

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

2 y2 Hence y = mx + c is tangent to the ellipse x 2  2 = 1 if c² = a²m² + b².

a

b

Example # 14 : Find the set of value(s) of '' for which the line 3x – 4y +  = 0 intersect the ellipse

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8

MATHS x2 y2 + = 1 at two distinct points.

16 9 Solution :

Solving given line with ellipse, we get

( 4 y   )2 y2 + =1 9  16 9

y 2y 2 2 – + –1 = 0 18 9 144

Since, line intersect the parabola at two distinct points,  roots of above equation are real & distinct  D0

2



(18)2

–

 2  8   .  144  1 > 0 9  



–12 2 <  < 12 2

Self Practice Problem (13)

Find the value of '' for which 2x – y +  = 0 touches the ellipse Answer :

=±

x2 y2 + =1 25 9

109

Tangents : (a)

Slope form: y = mx ±

(b)

Point form :

(c)

Parametric form:

x x1 a2



y y1 b2

2

2

a m b

2

is tangent to the ellipse

 1 is tangent to the ellipse

x2 a2

+

x2 a2

y2 b2

+

y2 b2

= 1 for all values of m.

= 1 at (x 1, y1).

x2 y2 xcos ysin   1 is tangent to the ellipse 2 + 2 = 1 at the point a b a b

(a cos , b sin ).

Note : (i)

There are two tangents to the ellipse having the same m, i.e. there are two tangents parallel to any given direction.These tangents touches the ellipse at extremities of a diameter.

(ii)

 cos   sin  2   2  a , b Point of intersection of the tangents at the point  &  is,  cos   cos  2   2 

(iii)

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

Example # 15 : Find the equations of the tangents to the ellipse 3x 2 + 4y2 = 12 which are perpendicular to the line y + 2x = 4.

1 Solution : Slope of tangent = m = 2 y2 x2 Given ellipse is + =1 3 4 Equation of tangent whose slope is 'm' is y = mx ± 4m 2  3 

1 2 2y = x ± 4 m=



y=

1 x± 2

1 3

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MATHS Example # 16 : A tangent to the ellipse

Solution :

x2

y2

+

= 1 touches at the point P on it in the first quadrant and a2 b2 meets the co-ordinate axes in A and B respectively. If P divides AB in the ratio 3 : 1, find the equation of the tangent.

Let P  ( a cos, b sin)  equation of tangent is x y cos + sin = 1 a b

A  (a sec, 0) B  (0, b cosec)  P divide AB internally in the ratio 3 : 1 

a cos =

a sec  4



cos 2 =

1 4

and

b sin  =

3b cos ec 4



sin =

3 2



tangent is



bx +

x 3y + =1 2a 2b



cos =

1 2

3 ay = 2ab

Example # 17 : Prove that the locus of the point of intersection of tangents to an ellipse at two points whose eccentric angle differ by a constant  is an ellipse.

Solution : Let P (h, k) be the point of intersection of tangents at A() and B() to the ellipse.



  b sin  a cos     2   2  h= &k=   cos  cos     2   2 



 h k     +   = sec 2  a b  2     

2

2

but given that  –  =  

x2

locus is

 a 2 sec 2   2

+

y2  b 2 sec 2   2

=1

Example # 18 : Find the locus of foot of perpendicular drawn from centre to any tangent to the ellipse x2

Solution :

+

y2

= 1. a2 b2 Let P(h, k) be the foot of perpendicular to a tangent y = mx + from centre



k h .m=–1  m=– h k P(h, k) lies on tangent



k = mh +



a 2m 2  b 2

a 2m 2  b 2

.......(i)

.......(ii)

.......(iii)

from equation (ii) & (iii), we get 2

2  k  h  k 

2 2   = a h + b2  k2 



locus is (x 2 + y2)2 = a2x 2 + b2y2

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MATHS Self Practice Problem (14)

Show that the locus of the point of intersection of the tangents at the extremities of any focal chord of an ellipse is the directrix corresponding to the focus.

(15)

Show that the locus of the foot of the perpendicular on a varying tangent to an ellipse from either of its foci is a concentric circle.

(16)

Prove that the portion of the tangent to an ellipse intercepted between the ellipse and the directrix subtends a right angle at the corresponding focus.

(17)

Find the area of parallelogram formed by tangents at the extremities of latera recta of the ellipse

(18)

x2 a2



y2 b2

 1.

If y1 is ordinate of a point P on the ellipse then show that the angle between its focal radius and tangent at it, is tan

(19)

–1

 b2     aey  .

1  x2

Find the eccentric angle of the point P on the ellipse

y2

+

a2

b2

= 1 tangent at which, is

equally inclined to the axes.

Answers :

(17)

2a 3 a2  b2

(19)

b b b = ± tan–1   ,  – tan–1   , –  + tan–1   a a a

Normals : x2

= 1 is

a2 x b2 y  = a²  b². x1 y1

Equation of the normal at (x 1, y1) to the ellipse

(ii)

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

a2

+

y2

(i)

b2

ax. sec   by. cosec  = (a²  b²).

Equation of a normal in terms of its slope 'm' is y = mx 

Example # 19 : P and Q are corresponding points on the ellipse

Solution :

x2 a

2

+

a2

a  b m 2

(iii)

x2

b2

y2 b2

= 1 is;

2

a 2  b 2m 2

y2

+

.

= 1 and the auxiliary circles

respectively. The normal at P to the ellipse meets CQ in R, where C is the centre of the ellipse.

Prove that CR = a + b Let P  (acos , b sin)  Q  (a cos, a sin) Equation of normal at P is (a sec) x – (b cosec ) y = a2 – b2 ..........(i) equation of CQ is y = tan . x .........(ii) Solving equation (i) & (ii), we get (a – b) x = (a2 – b2) cos x = (a + b) cos, & y = (a + b) sin  R  ((a + b) cos, (a + b) sin)  CR = a + b

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MATHS Example # 20 : Find the shortest distance between the line x + y = 10 and the ellipse Solution :

x2 y2  =1 16 9

Shortest distance occurs between two non-intersecting curve always along common normal.

Let 'P' be a point on ellipse and Q is a point on given line for which PQ is common normal.  Tangent at 'P' is parallel to given line 

Equation of tangent parallel to given line is (y = mx ±

a 2m 2  b 2 )

y=–x±5  x + y + 5 = 0 or x+y–5=0 minimum distance = distance between x + y – 10 = 0 & x + y – 5 = 0



| 10  5 |



shortest distance =

1 1

5 =

2

Example # 21 : Prove that, in an ellipse, the distance between the centre and any normal does not exceed the difference between the semi-axes of the ellipse. x2

y2

1 a2 b2 Equation of normal at P () is (a sec)x – (bcosec )y – a2 + b2 = 0 distance of normal from centre

Solution :

Let the equation of ellipse is



| a2  b2 |

= OR =

a 2  b 2  (a tan )2  (b cot )2 | a2  b2 |

=

(a  b)2  (a tan   b cot )2

(a + b)2 + (a tan – b cot)2  (a + b)2



or



|OR| (a – b)

| a2  b2 | (a  b ) 2

Self Practice Problem (20)

Find the value(s) of 'k' for which the line x + y = k is a normal to the ellipse

(21)

If the normal at the point P() to the ellipse

x2 a2



y2 b2

1

x2 y2  = 1 intersects it again at the point Q(2) 14 5

then cos = (A) –

2 3

Answers :

(20)

(B) k= 

a

2

2 3

 b2

(C) –



6 7

(D)

6 7

2

a2  b2

(21)

A

Pair of Tangents : The equation to the pair of tangents which can be drawn from any point (x 1, y1) to the ellipse x2 a

2



y2 b2

= 1 is given by: SS1 = T² where :

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MATHS S

x2 a

2



y2 b

2

2

–1

;

x1

S1 =

a

2

2



y1 b

2

–1;

T

xx1 a

2

+

yy1 b2

– 1.

Example # 22 : How many real tangents can be drawn from the point (4, 3) to the ellipse Solution :

x2 y2 + = 1. Find 16 9

the equation of these tangents & angle between them.

Given point P  (4, 3) ellipse S 



S1 

x2 y2 + –1=0 16 9 16 9 + –1=1>0 16 9

 Point P  (4, 3) lies outside the ellipse.  Two tangents can be drawn from the point P(4, 3).

Equation of pair of tangents is SS1 = T 2 

2  x2 y2     1 . 1 =  4 x  3 y  1  16 9   16 9   



xy 2y x2 y2 x2 y2 x + –1= + +1+ – – 6 3 16 9 16 9 2

 

– xy + 3x + 4y – 12 = 0 x=4&y=3

and angle between them =



(4 – x) (y – 3) = 0

 2

Alternative By direct observation x = 4, y = 3 are tangents.

Example # 23 : Find the locus of point of intersection of perpendicular tangents to the ellipse Solution :

x2 a2



y2 b2

=1

Let P(h, k) be the point of intersection of two perpendicular tangents equation of pair of tangents is SS1 = T 2 

2  x2 y2   h2 k 2   hx ky        1   1  a2 b2   a2 b2  =  2  2  1 b a     



2 x 2  k  1 y2  2  + 2 a2  b b 

 h2    1 + ........ = 0  a2   

.........(i)

Since equation (i) represents two perpendicular lines  

1 a2

 k2  1    1  b2  + b2  

k 2 – b2 + h2 – a2 = 0

 h2     1  a2  =0   

locus is x 2 + y2 = a2 + b2

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MATHS Self Practice Problem (22)

Find the locus of point of intersection of the tangents drawn at the extremities of a focal chord of the ellipse

x2 a

2

Answer :

y2

+

b2

x=±

= 1.

a e

Director Circle : 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 & minor axes.

Example # 24 : An ellipse slides between two perpendicular lines. Show that the locus of its centre is a circle.

Solution : Let length of semi-major and semi-minor axis are 'a' and 'b' and centre is C  (h, k) Since ellipse slides between two perpendicular lines, therefore point of intersection of two perpendicular tangents lies on director circle.

Let us consider two perpendicular lines as x & y axes  point of intersection is origin O  (0, 0)  OC = radius of director circle   

2 2 h2  k 2 = a  b locus of C  (h, k) is x 2 + y2 = a2 + b2

which is a circle

Self Practice Problem (23)

A tangent to the ellipse x 2 + 4y2 = 4 meets the ellipse x 2 + 2y2 = 6 at P and Q. Prove that the tangents at P and Q of the ellipse x 2 + 2y2 = 6 are at right angles.

Chord of Contact : Equation to the chord of contact of tangents drawn from a point P(x 1, y1) to the ellipse T = 0, where

T=

xx1 a

2

+

yy1 b2

x2 a

2

+

y2 b2

= 1 is

–1

Example # 25 : If tangents to the parabola y2 = 4ax intersect the ellipse

x2 a2

+

y2 b2

= 1 at A and B, then find the

locus of point of intersection of tangents at A and B. Let P  (h, k) be the point of intersection of tangents at A & B xh yk  equation of chord of contact AB is =1 .............(i) 2 + a b2 which touches the parabola equation of tangent to parabola y2 = 4ax a y = mx + m a  mx – y = – .............(ii) m equation (i) & (ii) as must be same

Solution :

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MATHS 

a  1 m = = m  k  1  h   2  2 b   a 



m=–



–

2 ak h b 2 & m = b2 k a

hb 2

ak

=

ka 2



b2

locus of P is y2 = –

b4 a3

.x

Self Practice Problem (24)

Find the locus of point of intersection of tangents at the extremities of normal chords of the ellipse

(25)

x2

y2

+

a2

= 1.

b2

Find the locus of point of intersection of tangents at the extremities of the chords of the ellipse

x2 a2

Answers :

y2

+

= 1 subtending a right angle at its centre.

b2

a6

(24)

x

2

b6 +

y

2

= (a2 – b2)2 (25)

x2 a

4

+

y2 b

4

1

=

a

2

+

1 b2

Chord with a given middle point : 2 y2 Equation of the chord of the ellipse x 2  2 = 1 whose middle point is (x 1, y1) is T = S1,

a

2

where S1 =

x1

a2

2



y1

b2

–1;

T

b

xx1 a

2

+

yy1 b2

– 1.

Example # 26 : Find the locus of the mid - point of focal chords of the ellipse Solution :

Let P  (h, k) be the mid-point 

equation of chord whose mid-point is given

xh a

2

+

yk b

2

x2 a2

+

–1=

y2 b2 h2 a2

= 1.

+

k2 b2

–1

since it is a focal chord,  it passes through focus, either (ae, 0) or (–ae, 0) If it passes through (ae, 0) 

locus is

x2 y2 ex = 2 + 2 a a b

If it passes through (–ae, 0) 

locus is –

ex x2 y2 = 2 + 2 a a b

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MATHS Example # 27 : Find the condition on 'a' and 'b' for which two distinct chords of the ellipse

x2 2a 2

+

y2 2b 2

=1

passing through (a, – b) are bisected by the line x + y = b.

Solution :

Let the line x + y = b bisect the chord at P(, b – )  equation of chord whose mid-point is P(, b – )

x 2a

y(b   )

+

2

2b

2

=

2 2a

2

+

(b   ) 2 2b 2

Since it passes through (a, –b) 

2 (b   ) 2  (b   ) – = + 2a 2b 2a 2 2b 2



1   1 1  1 2     – 1 = 2  2  2  – +1 a b b a b    



1   1  3 1 2  2  2  –     + 2 = 0 a b   b a

since line bisect two chord  above quadratic equation in  must have two distinct real roots 2

1   1 3 1    –4  2  2 .2>0 b a a b    



9

  

b

2

+

2

1 a

2

+

8 8 6 – 2 – 2 >0 ab a b

1



b

2

–

7 a

2

+

6 >0 ab

2

a – 7b + 6ab > 0 a2 > 7b2 – 6ab which is the required condition.

Self Practice Problem

x2 y2 + = 1 which is bisected at (2, 1).

36 9

(26)

Find the equation of the chord

(27)

Find the locus of the mid-points of normal chords of the ellipse

(28)

Find the length of the chord of the ellipse

Answers :

(26)

x + 2y = 4

(28)

7 5

(27)

x2 a2

+

y2 b2

= 1.

 1 2 x2 y2 + = 1 whose middle point is  ,  2 5 25 16

 x2 y2      a2 b2   

2

 a6 b6     2 2 2  x 2 y 2  = (a – b )  

41

Important Highlights : Refering to the ellipse (1)

x2 a2



y2 b2

=1

If P be any point on the ellipse with S & S as its foci then

 (SP) +  (SP) = 2a.

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MATHS

(2)

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 well known 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.

(3)

The product of the length’s of the perpendicular segments from the foci on any tangent to the ellipse is b² and the feet of these perpendiculars lie on its 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 similiar ellipse as that of the original one.

(4)

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

(5)

If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively, & 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. [where S and S are the focii of the ellipse and T is the point where tangent at P meet the major axis]

(6)

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.

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MATHS ď&#x20AC; (7)

If the tangent at the point P of a standard ellipse meets the axis in T and t and CY is the perpendicular on it from the centre then, (i) T t. PY = a2 ď&#x20AC;­ b2 and (ii) least value of T t is a + b.

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18

MATHS

Parabola This chapter focusses on parabolic curves, which constitutes one category of various curves obtained by slicing a cone by a plane, called conic sections. A cone (not necessarily right circular) can be out in various ways by a plane, and thus different types of conic sections are obtained.

Let us start with the definition of a conic section and then we will see how are they obtained by slicing a right circular cone.

Conic Sections: A conic section, or conic is the locus of a point which moves in a plane so that its distance from a fixed point is in a constant ratio to its perpendicular distance from a fixed straight line.

•

The fixed point is called the Focus.

•

The fixed straight line is called the Directrix.

•

The constant ratio is called the Eccentricity denoted by e.

PS e PM •

The line passing through the focus & perpendicular to the directrix is called the Axis.

•

A point of intersection of a conic with its axis is called a Vertex.

If S is (p, q) & directrix is x + my + n = 0 then

PS =

( x –  )2  ( y – )2 & PM =

| x  my  n | 2  m2

PS =e  (2 + m2) [(x – p)2 + (y – q)2] = e2 (x + my + n)2 PM Which is of the form ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

Section of right circular cone by different planes A right circular cone is as shown in the figure – 1

A Ge ne rat or axis

Vertex

B Circular Base Figure 1

(i)

Section of a right circular cone by a plane passing through its vertex is a pair of straight lines passing through the vertex as shown in the figure - 2.

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1

MATHS V

P

Q Figure –2

(ii)

Section of a right circular cone by a plane parallel to its base is a circle as shown in the figure  3.

(iii)

Section of a right circular cone by a plane parallel to a generator of the cone is a parabola as shown in the figure–4.

Figure-4 (iv)

Section of a right circular cone by a plane neither parallel to any generator of the cone nor perpendicular or parallel to the axis of the cone is an ellipse or hyperbola as shown in the figure  5 & 6.

Figure -5

Figure -6

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MATHS 3D View :

Distinguishing various conics : The nature of the conic section depends upon the position of the focus S w.r.t. the directrix & also upon the value of the eccentricity e. Two different cases arise.

Case (I) When The Focus Lies On The Directrix.

In this case   abc + 2fgh  af 2  bg2  ch2 = 0 & the general equation of a conic represents a pair of straight lines if: e > 1  h2 > ab the lines will be real & distinct intersecting at S. e = 1  h2 > ab the lines will coincident. e < 1  h2 < ab the lines will be imaginary.

Case (II) When The Focus Does Not Lie On Directrix. a parabola

an ellipse

a hyperbola

rectangular hyperbola

e = 1;   0,

0 < e < 1;   0;

e > 1;   0;

e > 1;   0

h² = ab

h² < ab

h² > ab

h² > ab; a + b = 0

PARABOLA Definition and terminology 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).

Four standard forms of the parabola are y² = 4ax; y² =  4ax; x² = 4ay; x² =  4ay For parabola y2 = 4ax: (i) Vertex is (0, 0) (ii) focus is (a, 0) (iii) Axis is y = 0 (iv) Directrix is x + a = 0 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.

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.

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MATHS Example # 1: Find the equation of the parabola whose focus is at (– 1, – 2) and the directrix is x – 2y + 3 = 0.

Solution :

    

Let P(x, y) be any point on the parabola whose focus is S(– 1, – 2) and the directrix x – 2y + 3 = 0. Draw PM perpendicular to directrix x – 2y + 3 = 0. Then by definition, SP = PM SP2 = PM2

 x  2y  3   (x + 1) + (y + 2) =  1 4   2

2

2

5 [(x + 1)2 + (y + 2)2] = (x – 2y + 3)2 5(x 2 + y2 + 2x + 4y + 5) = (x 2 + 4y2 + 9 – 4xy + 6x – 12y) 4x 2 + y2 + 4xy + 4x + 32y + 16 = 0 This is the equation of the required parabola.

Example # 2 : Find the vertex, axis, focus, directrix, latusrectum of the parabola, also draw their rough sketches. 4y2 + 12x – 20y + 67 = 0 Solution : The given equation is 67 4y2 + 12x – 20y + 67 = 0  y2 + 3x – 5y + =0 4 2

67 4



y2 – 5y = – 3x –



5  42  y   = – 3x – 2 4 



5 5 67 y2 – 5y +   = – 3x – +   4 2 2



7  5  y   = – 3 x   2 2  

2

2

2

....(i)

7 5 ,y=Y+ ....(ii) 2 2 Using these relations, equation (i) reduces to Y2 = – 3X ....(iii) This is of the form Y2 = – 4aX. On comparing, we get 4a = 3  a = 3/4.

Vertex - The coordinates of the vertex are (X = 0, Y = 0) So, the coordinates of the vertex are Let

x=X–

 7 5  ,   2 2

[Putting X = 0, Y = 0 in (ii)]

Axis The equation of the axis of the parabola is Y = 0.

So, the equation of the axis is y=

5 2

[Putting Y = 0 in (ii)]

FocusThe coordinates of the focus are (X = –a, Y = 0) i.e.

(X = – 3/4, Y = 0).

So, the coordinates of the focus are (–17/4, 5/2) [Putting X = 3/4 in (ii)] Directrix The equation of the directrix is X = a i.e. X =

3 .

4

So, the equation of the directrix is

11 [Putting X = 3/4 in (ii)] 4 Latusrectum The length of the latusrectum of the given parabola is 4a = 3. x=–

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MATHS Self Practice Problems (1)

Find the equation of the parabola whose focus is the point (0, 0)and whose directrix is the straight line 3x – 4y + 2 = 0.

(2)

Find the extremities of latus rectum of the parabola y = x2 – 2x + 3.

(3)

Find the latus rectum & equation of parabola whose vertex is origin & directrix is x + y = 2.

(4)

Find the vertex, axis, focus, directrix, latusrectum of the parabola y2 – 8y – x + 19 = 0. Also draw their roguht sketches.

(5)

Find the equation of the parabola whose focus is (1, – 1) and whose vertex is (2, 1). Also find its axis and latusrectum.

Answers :

(1)

16x 2 + 9y2 + 24xy – 12x + 16y – 4 = 0

(3)

4 2 , x2 + y2 – 2xy + 8x + 8y = 0

(2)

1 9 3 9  ,   ,  2 4 2 4

(4)

(2x – y – 3)2 = – 20 (x + 2y – 4), Axis 2x – y – 3 = 0. LL = 4 5 .

(5)

Parametric representation: The simplest & the best form of representing the coordinates of a point on the parabola is (at², 2at) i.e. the equations x = at² & y = 2at together represents the parabola y² = 4ax, t being the parameter.

Parametric form for : y2 = – 4ax (–at2, 2at) 2 x = 4ay (2at , at2) 2 x = – 4ay (2at , – at2) Example # 3 : Find the parametric equation of the parabola (x – 1)2 = –12 (y – 2) Solution :

 4a = – 12 x – 1 = 2 at 

 a = – 3, y – 2 = at2 x = 1 – 6t, y = 2 – 3t2

Self Practice Problems (6)

Find the parametric equation of the parabola x2 = 4ay x = 2at, y = at2.

Answer :

Position of a point relative to a parabola: The point (x 1 , y1) lies outside, on or inside the parabola y² = 4ax according as the expression y1²  4ax 1 is positive, zero or negative.

Outside Inside P(x1, y1)

S1 : y12 – 4ax1

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MATHS S1 < 0  Inside S1 > 0  Outside

Example # 4 : Check whether the point (3, 4) lies inside or outside the paabola y2 = 4x.

Solution :  

y2 – 4x = 0 S1 y12 – 4x 1 = 16 – 12 = 4 > 0 (3, 4) lies outside the parabola.

Self Practice Problems Find the set of value's of for which (, – 2 – ) lies inside the parabola y2 + 4x = 0.

(7)

  (– 4 – 2 3 , – 4 + 2 3 )

Answer :

Line & a parabola: The line y = mx + c meets the parabola y² = 4ax in two points real, coincident or imaginary according Tangent as a  cm  condition of tangency is, c = a/m.



Secant

Length of the chord intercepted by the parabola on the line y = m x + c is :  4   2 m 

A

a(1  m2 )(a  mc ) .

B

NOTE : 1.

The equation of a chord joining t1 & t2 is 2x  (t1 + t2) y + 2 at1 t2 = 0.

2.

If t1 & t2 are the ends of a focal chord of the parabola y² = 4ax then t1t2 = 1. Hence the coordinates at the extremities of a focal chord can be taken as (at², 2at) &

 a ,  2 a   t2 t 

Focal chord A S (focus) B

3.

Length of the focal chord making an angle  with the x axis is 4acosec² 

Example # 5 : Discuss the position of line y = x + 1 with respect to parabola y2 = 4x.

Solution : Solving we get (x + 1) 2 = 4x  (x – 1)2 = 0 so y = x + 1 is tangent to the parabola.

Example # 6 : Prove that focal distance of a point P(at2, 2at) on parabola y2 = 4ax (a > 0) is a(1 + t2).

Solution :



PS = PM = a + at2

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MATHS PS = a (1 + t2).

Example # 7 : If t1, t2 are end points of a focal chord then show that t1 t2 = –1.

Solution : Let parabola is y2 = 4ax 2

P (at1 , 2at1)



since P, S & Q are collinear mPQ = mPS



2t 1 2 = 2 t1  t 2 t1  1

 

t12 – 1 = t12 + t1t2 t1t2 = – 1

S(a, 0) 2

Q (at2 , 2at2)

Example # 8 : If the endpoint t1, t2 of a chord satisfy the relation t1 t2 = k (const.) then prove that the chord always passes through a fixed point. Find the point?

Solution : Equation of chord joining (at12, 2at1) and (at22, 2at2) is

2 y – 2at1 = t  t (x – at12) 1 2 (t1 + t2) y – 2at12 – 2at1t2 = 2x – 2at12

2 y = t  t (x + ak) 1 2 

(

t1t2 = k)

This line passes through a fixed point (– ak, 0).

Self Practice Problems (8)

If the line y = 3x + intersect the parabola y2 = 4x at two distinct point's then set of value's of '' is

(9)

Find the midpoint of the chord x + y = 2 of the parabola y2 = 4x.

(10)

If one end of focal chord of parabola y2 = 16x is (16, 16) then coordinate of other end is.

(11)

If PSQ is focal chord of parabola y2 = 4ax (a > 0), where S is focus then prove that 1 1 1 + = .

PS SQ a

(12)

Find the length of focal chord whose one end point is ‘t’.

Answers :

(8)

(– , 1/3)

(12)

 1 a t   t 

(9)

(4, – 2)

(10)

(1, – 4)

2

Tangents to the parabola y² = 4ax : Equation of tangent at a point on the parabola can be obtained by replacement method or using derivatives.

In replacement method, following changes are made to the second degree equation to obtain T. x2  x x1, y2  y y1, 2xy  xy1 + x1y, 2x  x + x1, 2y  y + y1 So, it follows that the targents are : (i)

y y1 = 2 a (x + x 1) at the point (x 1, y1) ;

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MATHS a  a 2 a  (m  0) at  2 , m m m

(ii)

y = mx +

(iii)

t y = x + a t² at (at², 2at).

(iv)

Point of intersection of the tangents at the point t1 & t2 is { at1 t2 , a(t1 + t2) }.

Example # 9 : Prove that the straight line y = mx + c touches the parabola y2 = 4a (x + a) if c = ma +

a m

Equation of tangent of slope ‘m’ to the parabola y2 = 4a(x + a) is

Solution :

y = m(x + a) +

a m



1  y = mx + a  m   m 

but the given tangent is y = mx +c 

c = am +

a m

Example # 10 : A tangent to the parabola y2 = 8x makes an angle of 45° with the straight line y = 3x + 5. Find its equation and its point of contact.

Solution :

Slope of required tangent’s are m = m 1 = – 2,

m2 =

3 1 1 3

1 2



Equation of tangent of slope m to the parabola y2 = 4ax is y = mx +



1  tangent’s y = – 2x – 1 at  ,  2  2  

y=

a . m

1 x + 4 at (8, 8) 2

Example # 11 : Find the equation to the tangents to the parabola y2 = 9x which goes through the point (4, 10).

Solution :

Equation of tangent to parabola y2 = 9x is y = mx +

9 4m

Since it passes through (4, 10) 

10 = 4m + m=



9 4m



16 m 2 – 40 m + 9 = 0

1 9 , 4 4

equation of tangent’s are

y=

x +9 4

&

y=

9 x + 1.

4

Example # 12 : Find the equations to the common tangents of the parabolas y2 = 4ax and x 2 = 4by.

Solution : Equation of tangent to y2 = 4ax is

a m Equation of tangent to x 2 = 4by is y = mx +

........(i)

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MATHS x = m 1y +

b m1



1 b x– m1 (m1 ) 2

y=

........(ii)

for common tangent, (i) & (ii) must represent same line. 

1 m1 = m

&

a b = – 2 m m1



a = – bm 2 m



 a m =    b



 a equation of common tangent is y =     b

1/ 3

1/ 3

 b x + a    a

1/ 3

.

Self Practice Problems (13)

Find equation tangent to parabola y2 = 4x whose intercept on y–axis is 2.

(14)

Prove that perpendicular drawn from focus upon any tangent of a parabola lies on the tangent at the vertex.

(15)

Prove that image of focus in any tangent to parabola lies on its directrix.

(16)

Prove that the area of triangle formed by three tangents to the parabola y2 = 4ax is half the area of triangle formed by their points of contacts.

Answers :

(13)

y

x 2 2

Normals to the parabola y² = 4ax : P

Normal is obtained using the slope of tangent.

Normal

2a Slope of tangent at (x1 , y1) = y 1 y1 2a



Slope of normal = –

(i)

y  y1 = –

(ii) (iii)

y = mx  2am  am 3 at (am 2,  2am) y + tx = 2at + at3 at (at2, 2at).

y1 (x  x 1) at (x 1, y1) ; 2a

NOTE : (i)

Point of intersection of normals at t1 & t2 is (a (t 12 + t 22 + t1t2 + 2),  a t1 t2 (t1 + t2)).

(ii)

If the normals to the parabola y² = 4ax at the point t1, meets the parabola again at the point

 2 t2, then t2 = –  t1   . t1  

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MATHS P(t1)

Q(t2)

(iii)

If the normals to the parabola y² = 4ax at the points t1 & t2 intersect again on the parabola at the point 't3' then t1 t2 = 2; t3 =  (t1 + t2) and the line joining t1 & t2 passes through a fixed point (2a, 0).

Example # 13 : If the normal at point ‘t1’ intersects the parabola again at ‘t2’ then show that t2 = –t1 –

Solution :

Slope of normal at P = – t1 and slope of chord PQ =



– t1 =

2 t1

2 t1  t 2

2 t1  t 2

t1 + t2 = –

2 t1



t2 = – t1 –

2 . t1

Example # 14 : If the normals at points t1, t2 meet at the point t3 on the parabola then prove that (i) t1 t2 = 2 (ii) t1 + t2 + t3 = 0 Solution : Since normal at t1 & t2 meet the curve at t3 

t3 = – t1 –

t3 = – t2 –  

2 t1 2 t2

.....(i)

.....(ii)

(t12 + 2) t2 = t1 (t22 + 2) t1t2 (t1 – t2) + 2 (t2 – t1) = 0 t1  t2 , t1t2 = 2 ......(iii) Hence (i) t1 t2 = 2 from equation (i) & (iii), we get t3 = – t1 – t2 Hence (ii) t1 + t2 + t3 = 0

Example # 15 : Find the locus of the point N from which 3 normals are drawn to the parabola y2 = 4ax are such that (i) Two of them are equally inclined to x-axis (ii) Two of them are perpendicular to each other Solution : Equation of normal to y2 = 4ax is y = mx – 2am – am3 Let the normal passes through N(h, k)  k = mh – 2am – am3  am3 + (2a – h) m + k = 0 For given value’s of (h, k) it is cubic in ‘m’.

Let m1, m2 & m3 are root’s of above equation  m1 + m2 + m3 = 0 ......(i)

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MATHS m 1m 2 + m 2m 3 + m 3m 1 = m 1m 2m 3 = – (i) (ii)

2a  h a

......(ii)

k a

......(iii)

If two normal are equally inclined to x-axis, then m1 + m2 = 0  m3 = 0  y=0 If two normal’s are perpendicular  m1 m2 = – 1 from (3)

m3 =

k a

from (2)

–1+

k 2a  h (m1 + m2) = a a

from (1)

m1 + m2 = –

.....(iv)

k a

.....(v)

.....(vi)

from (5) & (6), we get –1–

h k2 =2– a a

y2 = a(x – 3a) Self Practice Problems (17)

Find the points of the parabola y2 = 4ax at which the normal is inclined at 30° to the axis.

(18)

If the normal at point P(1, 2) on the parabola y2 = 4x cuts it again at point Q then Q = ?

(19)

Find the length of normal chord at point ‘t’ to the parabola y2 = 4ax.

(20)

If normal chord at a point 't' on the parabola y2 = 4ax subtends a right angle at the vertex then prove that t2 = 2

(21)

Prove that the chord of the parabola y2 = 4ax, whose equation is y – x 2 + 4a 2 = 0, is a normal to the curve and that its length is 6 3a .

(22)

If the normals at 3 points P, Q & R are concurrent, then show that (i) The sum of slopes of normals is zero, (ii) Sum of ordinates of points P, Q, R is zero (iii) The centroid of PQR lies on the axis of parabola.

Answers :

(17)

a 2a   a 2a   , ,  ,  3    3  3 3  

3

(18)

(9, – 6)

(19)



4a( t 2  1) 2 t2

Pair of tangents: The equation to the pair of tangents which can be drawn from any point (x 1, y1) to the parabola y² = 4ax is given by: SS1 = T² where : S  y²  4ax ; S1 = y1²  4ax 1 ; T  y y1  2a(x + x 1).

A (x1, y1) P B

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MATHS Example # 16 : Write the equation of pair of tangents to the parabola y2 = 4x drawn from a point P(–1, 2) Solution :

We know the equation of pair of tangents are given by SS1 = T²  (y2 – 4x) (4 + 4) = (2y – 2 (x – 1))2  8y2 – 32x = 4y2 + 4x2 + 4 – 8xy + 8y – 8x  y2 – x2 + 2xy – 6x – 2y = 1

Example # 17 :Find the focus of the point P from which tangents are drawn to parabola y2 = 4ax having slopes m1, m2 such that (i) m1 + m2 = m0 (const) (ii) 1 + 2 = 0 (const) 2 Solution : Equation of tangent to y = 4ax, is

a m Let it passes through P(h, k)  m2h – mk + a = 0 y = mx +

k h



(i)

m1 + m2 = m0 =

(ii)

m1  m 2 k /h tan0 = 1  m m = 1  a/h 1 2



y = (x – a) tan0

y = m 0x

Self Practice Problem (23)

If two tangents to the parabola y2 = 4ax from a point P make angles 1 and 2 with the axis of the parabola, then find the locus of P in each of the following cases. (i) tan21 + tan22 =  (a constant) (ii) cos 1 cos 2 =  (a constant) Answer :

(i)

y2 – 2ax = x2 , (ii)

x2 = 2 {(x – a)2 + y2}

Director circle: Locus of the point of intersection of the perpendicular tangents to a curve is called the Director Circle.

For parabola y2 = 4ax it’s equation is x + a = 0 which is parabola’s own directrix.

Chord of contact: Equation to the chord of contact of tangents drawn from a point P(x 1, y1) is yy1 = 2a (x + x 1).

A P (x1, y1)

Chord of contact B

NOTE : The area of the triangle formed by the tangents from the point (x 1, y1) & the chord of contact is 1 (y ²  4ax 1)3/2 2a 1

Example # 18 : Find the length of chord of contact of the tangents drawn from point (x1, y1) to the parabola y2 = 4ax.

Solution : Let tangent at P(t1) & Q(t2) meet at (x1, y1)  at1t2 = x1 & a(t1 + t2) = y1 

PQ =

(at12  at 22 )2  (2a( t1  t 2 ))2

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MATHS =a

(( t1  t 2 )2  4t 1t 2 )((t1  t 2 )2  4) =

( y12  4ax 1 )( y12  4a 2 ) a2

Example # 19 : If the line x – y – 1 = 0 intersect the parabola y2 = 8x at P & Q, then find the point of intersection of tangents at P & Q. Solution :

Let (h, k) be point of intersection of tangents then chord of contact is yk = 4(x + h) 4x – yk + 4h = 0 .....(i) But given is x – y – 1 = 0   

4 k 4h = = 1 1 1 h = – 1, k = 4 point (–1, 4)

Example # 20 : Find the locus of point whose chord of contact w.r.t to the parabola y2 = 4bx is the tangents of the parabola y2 = 4ax.

Solution :

a ......(i) m Let it is chord of contact for parabola y2 = 4bx w.r.t. the point P(h, k)  Equation of chord of contact is yk = 2b(x + h) Equation of tangent to y2 = 4ax is y = mx +

2b 2bh x+ k k From (i) & (ii) y=

m=

.....(ii)

2b a 2bh , = k m k

locus of P is y2 =



a=

4b 2h k2

4b 2 x. a

Self Practice Problems (24)

Prove that locus of a point whose chord of contact w.r.t. parabola passes through focus is directrix

(25)

If from a variable point ‘P’ on the line x – 2y + 1 = 0 pair of tangent’s are drawn to the parabola y2 = 8x then prove that chord of contact passes through a fixed point, also find that point.

Answers :

(25)

(1, 8)

Chord with a given middle point: Equation of the chord of the parabola y² = 4ax whose middle point is (x 1, y1) is y  y1 =

2a (x  x 1)  T = S1 y1

M(x,y)

Example # 21 : Find the locus of middle point of the chord of the parabola y2 = 4ax which pass through a given point (p, q).

Solution : Let P(h, k) be the mid point of chord of parabola y2 = 4ax, so equation of chord is yk – 2a(x + h) = k2 – 4ah.

Since it passes through (p, q)  qk – 2a (p + h) = k2 – 4ah

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MATHS 

Required locus is y2 – 2ax – qy + 2ap = 0.

Example # 22 : Find the locus of middle point of the chord of the parabola y2 = 4ax whose slope is ‘m’.

Solution :

Let P(h, k) be the mid point of chord of parabola y2 = 4ax, so equation of chord is yk – 2a(x + h) = k2 – 4ah.

2a but slope = =m k 2a  locus is y = m

Self Practice Problems (26)

Find the equation of chord of parabola y2 = 4x whose mid point is (4, 2).

(27)

Find the locus of mid - point of chord of parabola y2 = 4ax which touches the parabola x2 = 4by.

Answers :

(26)

x–y–2=0

(27)

y (2ax – y2) = 4a2b

Important Highlights: (i)

If the tangent & normal at any point ‘P’ of the parabola intersect the axis at T & 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 & the perpendicular from P on the directrix. From this we conclude that all rays emanating from S will become parallel to the axis of theparabola after reflection.

(ii)

The portion of a tangent to a parabola cut off between the directrix & the curve subtends a right angle at the focus.

See figure above.

(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 2 at the vertex and intercepts a chord of length a 1  t 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.

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MATHS

(v)

If the tangents at P and Q meet in T, then:  TP and TQ subtend equal angles at the focus S.  ST 2 = SP. SQ &  The triangles SPT and STQ are similar.

P T S

TSP = TSQ

Q

(vi)

Semi latus rectum of the parabola y² = 4ax, is the harmonic mean between segments of any focal chord of the parabola.

P

2(PS )(SQ ) = 2a PS  SQ

S Q

(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  am 3 i.e. am 3 + m(2a  h) + k = 0. m1 + m2 + m3 = 0 ;

m 1m 2 + m 2m 3 + m 3m 1 =

2a  h k ; m 1 m2 m3 =  . a a

Where m 1, m 2, & m 3 are the slopes of the three concurrent normals. Note that A P(h, k)

A, B, C  Conormal points B C



algebraic sum of the slopes of the three concurrent normals is zero.



algebraic sum of the ordinates of the three conormal 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 froma point P (h, k) is h > 2a & k 2 <

4 3.

27a (h – 2a)

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MATHS (ix)

Length of subtangent at any point P(x, y) on the parabola y² = 4ax equals twice the abscissa of the point P. Note that the subtangent is bisected at the vertex. y P

T

(x)

O

D N

x

TD = 2(OD) , DN = 2a

Length of subnormal is constant for all points on the parabola & is equal to the semi latus rectum.

See figure above.

Note : Students must try to proof all the above properties.

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