4 (B) 1. Find the equation of the tangent and normal to the ...

ELLIPSE

EXERCISE -- 4 (B)

1. Find the equation of the tangent and normal to the ellipse x2 + 8y2 = 33 at (–1, 2).

Sol. Given ellipse S = x2 + 8y2 = 33

Equation of the tangent is S1 =0

⇒1 1

2 2

xx yy1

a b+ =

x( 1) 8y(2) 33

x 16y 33

x 16y 33 0

− + =⇒ − + =⇒ − + =

Equation of the normal is 16x + y + k = 0

It passes through P(–1, 2) –16 + 2 + k = 0 ⇒ k = 14

Equation of the normal is 16x + y + 14 = 0.

2. Find the equation of the tangent and normal to the ellipse x2+2y2– 4x+12y + 14 = 0 at

(2, –1).

Sol. Given ellipse S =x2+2y2– 4x+12y + 14 = 0

Equation of the tangent is S1 =0

xx1 + 2yy1 – 2(x + x1) + 6(y + y1) + 14 = 0

⇒ 2x – 2y – 2(x + 2) + 6(y – 1) + 14 = 0

⇒ 4y + 4 = 0

y = –1 required equation of tangent.

Slope of tangent is 0

Equation of normal be 1

y 1 (x 2)0

−+ = −

x = 2 equation of normal.

3. Find the equation of the tangents to 9x2 + 16y2 = 144 which makes equal intercepts on coordinate axes.

Ans: x y 5 0± ± =

4. Find the coordinates of the points on the ellipse x2 + 3y2 = 37 at which the normal is parallel to the line 6x – 5y = 2.

Sol. Equation of the ellipse is x2 + 3y2 = 37 2 2x y

137373

⇒ + =

Slope of the normal = a sin 37 sin

3 tanb cos 37

cos3

θ θ= = θθ

θ

The normal is parallel to 6x – 5y = 2

∴ 6 6 2 3

3 tan tan5 55 3

θ = ⇒ θ = =

Case I :

The coordinates of P are (a cos θ, b sin θ)=5 37 2 3

37 , (5,2)37 3 37

⋅ =

Case II :

The coordinates of P are (a cos θ, b sin θ)=( 5) 37 2 3

37 , ( 5, 2)37 3 37

− −⋅ = − −

5. Find the value of k if 4x + y + k = 0 is a tangent to the ellipse x2 + 3y2 = 3.

Sol. Equation of the ellipse is x2 + 3y2 = 3

2 2x y

13 1

+ =

⇒ a2 = 3, b2 = 1

Equation of the line is 4x + y + k = 0

⇒y = –4x – k

⇒m = –4, c = –k

Above line is a tangent to the ellipse

⇒ c2 = a2 m2 + b2

(–k)2 = 3(–4)2 + 1

2 3

5 θ

37

2 3−

–5 θ

37

k2 = 48 + 1 = 49

k = ± 7.

6. Find the condition for the line xcosα + ysinα = p to be a tangent to the ellipse 2 2

2 2

x y1

a b+ = .

Sol. Equation of the ellipse is 2 2

2 2

x y1

a b+ = …(i)

Equation of the line is xcosα + ysinα = p

⇒ysinα = –xcosα + p

⇒cos p

y xsin sin

α= − +α α

∴ cos p

m ,csin sin

α= − =α α


⇒ c2 = a2m2 + b2

⇒

2 22 2

2 2

p cosa b

sin sin

α= +α α

⇒p2 = a2 cos2 α + b2 sin2 α.

II.

1. Find the equations of tangent and normal to the ellipse 2x2 + 3y2 = 11 at the point whose ordinate is 1.

Sol. Equation of the ellipse is S =2x2 + 3y2 = 11

Given y = 1 ( I,e., y coordinate is 1)

⇒2x2 + 3 = 11 ⇒ 2x2 = 8 ⇒ x = ±2

Points on the ellipse are P(2, 1) and Q(–2, 1)

Case I P(2, 1)

Equation of the tangent is S1 = 0

⇒ 2x 2 3y 1 11 4x 3y 11⋅ + ⋅ = ⇒ + =

The normal is perpendicular to the tangent.

Equation of the normal at P can be taken as 3x – 4y = k.

The normal passes through P(2, 1)

6 – 4 = k ⇒ k = 2

Equation of the normal at P is 3x – 4y = 2.

Case II : Q(–2, 1)

Equation of the tangent at Q is S2 = 0

⇒2x(–2) + 3y.1 = 11

⇒–4x + 3y = 11

4x – 3y + 11 = 0

Equation of the normal can be taken as 3x + 4y = k

The normal passes through Q(–2, 1)

–6 + 4 = k ⇒ k = –2

Equation of the normal at Q is 3x + 4y = –2

or 3x + 4y + 2 = 0.

2. Find the equations to the tangents to the ellipse, x2 + 2y2 = 3 drawn from the point (1, 2) and also find the angle between these tangents.

Sol. Equations of the ellipse is x2 + 2y2 = 3

⇒

2 2x y1

3 3/ 2+ =

⇒ a2 = 3, b2 = 3/2

Let m be the slope of the tangent which is passing through P(1, 2)

Equation of the tangent is

y – 2 = m(x – 1) = mx – m

y = mx + (2 – m)


⇒ c2 = a2m2 + b2

2 2

2 2

3(2 m) 3(m )

23

4 m 4m 3m2

⇒ − = +

⇒ + − = +

R

Q

P(1, 2)

2

2

52m 4m 0

2

4m 8m 5 0

(2m 1)(2m 5) 0

1 5m or

2 2

⇒ + − =

⇒ + − =⇒ − + =

= −

Case I : m = 1/2


⇒1 1 x 3

y x 22 2 2 2

= + − = +

⇒2y = x + 3

⇒x – 2y + 3 = 0

Case II : m = –5/2


5 5 5x 9

y x (2 )2 2 2 2

= − + + = − +

⇒2y = –5x + 9

⇒5x + 2y – 9 = 0

Angle between the tangents is given by

1 2

1 2

m mtan

1 m m

−θ =+

=

1 532 2 | 12 | 12

51 5 1142 2

+= = − =

−+ −

⇒θ = tan–1 12.

3. Find the equation of tangents to the ellipse 2x2 + y2 = 8 which are parallel to x – 2y + 4 = 0.

Sol. Equation of the ellipse is 2x2 + y2 = 8

2 2x y

14 8

⇒ + =

Equation of the tangent parallel to x – 2y + 4 = 0.

Is x – 2y + k = 0.

x k

2 2y⇒ = + ⇒ m =1/2 and c =k/2


⇒ c2 = a2m2 + b2 2

214. 8 36 6

4 4

kk k⇒ = + ⇒ = ⇒ = ±

Equation of tangents are

x – 2y ± 6 = 0

III.

1. Show that the feet of the perpendicular drawn from the centre on any tangent to the ellipse lies on the curve (x2 + y2)2 = a2x2 + b2y2.


2 2

x y1

a b+ =

Equation of the tangent at P(θ) is

x y

cos sin 1a b

θ + θ =

Slope of the tangent

PN =

cosbcosa

sin a sinb

θ − θ = −θ θ

Let N (x1, y1 ) be the foot of the perpendicular from C(0, 0) to any tangent.

Slope of CN = 1

1

y

x.

∴ Slope of PT × slope of CN = –1

1

1

b cos y1

a sin x

θ− ⋅ = −θ

2 2 2 21 1 1 1

cos sin 1k

ax by a x b y

θ θ= = =+

1 1x ycos sin 1

a bθ + θ =

1 1ax bycos , sin

k kθ = θ =

1 11 1

x yax by k

a b⋅ + ⋅ =

P

C

N(x1, y1)

2 21 1x y k+ =

N(x1, y1) is a point on x y

cos sin 1a b

θ + θ =

⇒1 1x y

cos sin 1a b

θ + θ =

⇒2 2 2 2 2 21 1 1 1x y a x b y+ = + (or)

( )22 2 2 2 2 21 1 1 1x y a x b y+ = +

Locus of N(x1, y1) is (x2 + y2)2 = a2x2 + b2y2

2. Show that the locus of the feet of the perpendiculars drawn from foci on any tangent of the ellipse is the auxiliary circle.


2 2

x y1

a b+ =

Equation of the tangent to the ellipse is

2 2 2y mx a m b= ± +

⇒ 2 2 2y mx a m b− = ± + …(1)

Equation to the perpendicular from either focus (±ae, 0) on this tangent is

1

y (x ae) my (x ae)m

= − ± ⇒ = − ±

⇒ my x ae+ = ± …(2)

Squaring and adding (1) and (2)

(y – mx)2 + (my + x)2 = a2m2 + b2 + a2e2

⇒y2 + m2x2 – 2mxy + m2y2 + x2 + 2mxy= a2m2 + a2 – a2e2 + a2e2

⇒ (x2 + y2)(1 + m2) = a2(1 + m2)

⇒ x2 + y2 = a2

The locus is the auxiliary circle concentric with the ellipse.

S S′

L L′

3. The tangent and normal to the ellipse x2 + 4y2 = 4 at a point P(θ) meets the major axis at Q and R respectively. If 0 < θ < π/2 and QR = 2, then show that θ = cos–1(2/3).

Sol.

Equation of the ellipse is x2 + 4y2 = 4

2 2x y

14 1

+ =

Equation of the tangent at P(θ) is

x y

cos sin 12 1

⋅ θ + θ =

Equation of x-axis ( i.e., major axis) is y = 0

x 2

cos 1 x2 cos

⋅ θ = ⇒ =θ

Coordinates of Q are 2

,0cos

θ

Equation of the normal at P(θ) is

2 2ax by 2x ya b 3

cos sin cos sin− = − ⇒ − =

θ θ θ θ

Substituting y = 0 we get 2x

3cos

=θ

⇒3

x cos2

= θ

Coordinates of R are 3

cos ,02

θ

23 2 3cos 4QR cos

2 cos 2cos

− θ + = − θ + = θ θ

Given QR = 2 2

2

3cos 42

2cos

3cos 4 4cos

− θ +⇒ =

θ

⇒ − θ + = θ

⇒3cos2 θ + 4 cos θ – 4 = 0

⇒ (3 cos θ – 2)(cosθ + 2) = 0

Q R

P(θ)

⇒3 cosθ – 2 = 0 ⇒ cosθ = 2/3

⇒ cosθ + 2 = 0 ⇒ cosθ = –2

⇒ cosθ = 2/3 or –2

⇒2

cos3

θ =

i.e. 1 2cos

3− θ =

.

THEOREM

The equation of the polar of the point P(x1, y1) with respect to the ellipse S = 0 is S1 = 0.

Note : If P is an external point of the ellipse S=0, then the polar of P meets the ellipse in two points and the polar becomes the chord of contact of P.

Note : If P lies on the ellipse S = 0, then the polar of P becomes the tangents at P to the ellipse S = 0.

Note 3 : If P is an internal point of the ellipse S=0, then the polar of P does not meet the ellipse S = 0.

THEOREM

The pole of the line lx + my + n = 0 (n ≠ 0) with respect to the ellipse 2 2

2 2

x yS 1 0

a b≡ + − = is

2 2a l b m,

n n

− −

.

Proof :

Equation of the ellipse 2 2

2 2

x yS 1 0

a b≡ + − =

Let P(x1, y1) be the pole of the line :

lx + my + n = 0 …(1)

The polar of P with respect to the ellipse is S1=0

⇒1 1

2 2

xx yy1 0

a b+ − = …(2)

Now (1) and (2) represent the same line.

∴ 2 2

1 11 12 2

x y 1 a l b mx , y

n n na l b m

− − −= = ⇒ = =

∴ Pole P = 2 2a l b m

,n n

− −

.

CONJUGATE POINTS

Two points P and Q are said to be conjugate with respect to the ellipse S = 0 if the polar of P with respect to S = 0 passes through Q.

Note : The condition for the points P(x1, y1),Q(x2, y2) to be conjugate with respect to the ellipse S = 0 is S12 = 0.

THEOREM

The condition for the lines l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0 to be conjugate with

respect to the ellipse 2 2

2 2

x y1

a b+ = is 2 2

1 2 1 2 1 2a l l b m m n n+ = .

Proof :

Pole of l1x + m1y + n1 = 0 with respect to 2 2

2 2

x y1

a b+ = is

2 21 1

1 1

a l b mP ,

n n

− −

.

Given lines are conjugate ⇒

P lies on l2x + m2y + n2 = 0 2 2

1 12 2 2

1 1

a l b ml m n 0

n n

− −⇒ + + =

2 21 2 1 2 1 2

2 21 2 1 2 1 2

a l l b m m n n 0

a l l b m m n n

⇒ − − + =

⇒ + =

THEOREM

The equation to the pair of tangents to the ellipse S = 0 from P(x1, y1) is 21 11S S S= .

THEOREM

The equation of the tangent at P(θ) on the ellipse S = 0 is x y

cos sin 1a b

θ + θ = .

THEOREM

The equation of the normal at P(θ) on the ellipse S = 0 is 2 2ax bya b

cos sin− = −

θ θ.

THEOREM Four normals can be drawn from any point to the ellipse and the sum of the eccentric angles of their feet is an odd multiple of π.

EXERCISE – 4(C)

1. Find the pole of the line 21x – 16y – 12 = 0 w.r.t to the ellipse 3x2 + 4y2 = 12.

Sol. Equation of the ellipse is 3x2 + 4y2 = 12

S = 2 2x y

14 3

+ =

Pole of the line 21x – 16y – 12 = 0 is 2 2a b m

,n n

− −

�

( )4 21 3( 16), 7, 4

12 12

− ⋅ − − = = − − −

Pole = (7, –4)

2. Show that the focus of an ellipse is the pole of the corresponding directrix.


2 2

x y1

a b+ =

Equation of one of the the directrix is x = a/e

⇒a

x 0e

− =

Pole = 2 2a b m

,n n

− −

�

( )

2 2a 1 b 0, ae,0

( a / e) ( a / e)

− ⋅ − ⋅= = − −

Pole = (ae, 0) is the corresponding focus.

Similarly we can prove the result for the other directrix and the corresponding focus.

3. Find the pole of the line 5x + 7y + 8 = 0 w.r.to 5x2 + 7y2 = 8.

Ans: (–1, –1)

4. Show that (2, –3), (18, 4) are conjugate points w.r.to 2 2x y

19 4

+ = .

Sol. Equation of the ellipse is S = 2 2x y

19 4

+ =

P(2, –3), Q(18, 4) are the given points.

12

2.18 3.44 4 0

9 4S

−⇒ = + = − =

Therefore, given points are conjugate points.

II METHOD.

Polar of P w.r.to the ellipse is 2x 3y

19 4

− =

2 18 34 4 3 1

9 4

⋅ − ⋅ = − =

Polar of P passes through Q

∴ P and Q are conjugate points w.r.to the ellipse.

5. Find the value of k if the lines 2x + 3y + 1= 0, x + y + k = 0 are conjugate w.r.to the ellipse 3x2 + 4y2 = 12.


⇒

2 2x y1

4 3+ =

⇒ a2 = 4, b2 = 3

Equation of the first line is 2x + 3y + 1 = 0

Pole is 2 2a b m

,n n

− −

� ( )4 2 3 3, 8, 9

1 1

− ⋅ − ⋅ = = − −

Since given lines are conjugate, this point lies on x + y + k = 0

–8 – 9 + k = 0 ⇒ k = 17.

6. Show that the conjugate lines through focus of a ellipse S = 0 are at right angles.


2 2

x y1

a b+ =

Let l1x + m1y + n1 = 0, l2x + m2y + n2 = 0 be conjugate lines,

∴ 2 21 2 1 2 1 2a l l b m m n n+ = …(1)

The given lines pass through S(ae, 0)

1 1 1 1

2 2 2 2

l ae 0 n 0 l ae n

l ae 0 n 0 l ae n

+ + = ⇒ = −

+ + = ⇒ = −

2 21 2 1 2l l a e n n⋅ = …(2)

From (1), (2), we get

2 2 2 21 2 1 2 1 2

2 2 2 21 2 1 2

2 2 2 2 21 2 1 2

a l l b m m l l a e

l l (a a e ) b m m 0

b l l b m m 0 [a (1 e ) b ]

+ =

⇒ − + =

⇒ + = − =

l1l2 + m1m2 = 0 ⇒ 1 2 1 2

1 2 1 2

l l l l1 1

m m m m

− −= − ⇒ = −

∴ The given conjugate lines are at right angles.

7. Find the value of k, if (1, 2)(k, –1) are conjugate points with respect to 2x2 + 3y2 = 6.


⇒

2 2x y1

3 2+ =

Given (1, 2), (k, –1) are conjugate points ,

⇒S12 = 0 ⇒ 1 2 1 22 2

x x y y1

a b+ =

1 k 2( 1)1

3 2

⋅ −⇒ + =

k1 1 2

3⇒ = + =

⇒ k = 6

II.

1. Find the equation of a straight line through the point (2, 1) and conjugate to the straight line 9x + 2y = 1 with respect to the ellipse 3x2 + 2y2 = 1.

Sol. Equation of ellipse is 3x2 + 2y2 = 1

2 2x y 1

1/ 3 1/ 2⇒ + =

Equation of the given line

9x + 2y = 1 i.e., 9x + 2y – 1=0

Pole of the line is 2 2a b m

,n n

− −

�

( )1 1

9 23 2, 3,1

1 1

= − − = − −

Required line is passing through (3, 1) and also through the point ( 2, 1 ).

Equation of the line is 1 1

y 1 (x 2)3 2

−− = −−

y – 1 = 0

2. An ellipse has OB as minor axis, S, S′ are its foci and the angle. SBS′ is a right angle, then find the eccentricity of the ellipse.


2 2

x y1

a b+ =

The foci are S(ae, 0) and S′ (–ae, 0)

Coordinates of B are (0, b)

Slope of SB = b

ae−, slope of S′B =

b

ae

SBS′ = 90° ⇒ Slope of SB × slope of S′B = –1

2 2 2

b b1

ae ae

b a e

− ⋅ = −

=

2 2 2 2a (1 e ) a e− =

2 2 21 e e 2e 1− = ⇒ =

2 1 1e e

2 2= ⇒ =

III.

1. Show that the poles of tangents to the ellipse 2 2

2 2

x y1

a b+ = w.r.to the circle x2 + y2 = a2

lies on the curve a2x2 + b2y2 = a4.

Sol.

Equation of the circle is x2 + y2 = a2

Let P(x1, y1) the ploe.

The polar of P with respect to the circle x2 + y2 = a2 is S1 =0

⇒xx1 + yy1 – a2 = 0 -----(i)

(i) is a tangent to the ellipse 2 2

2 2

x y1

a b+ =

c2 = a2m2 + b2

O S (ae,c)

S′ (–ae,c)

B(0, b)

⇒ 2 22

2 21

1 1

a xa b

y y

− −= ⋅ +

Places on the curve a4 = a2x2 + b2y2.

2. Show that the poles of the tangents to the circle x2 + y2 = a2 + b2 w.r.to the ellipse 2 2

2 2

x y1

a b+ = lies on

2 2

4 4 2 2

x y 1

a b a b+ =

+.

Sol. Let P(x1, y1) be the pole.

Equation of the ellipse is 2 2

2 2

x y1

a b+ =

Equation of the polar is S1 =0⇒ 1 12 2

xx yy1 0

a b+ − = ---(i)

Equation of the circle is x2 + y2 = a2 + b2

Centre = (0, 0), 2 2r a b= +

(i) is a tangent to the circle ⇒ radius = perpendicular distance from centre to the lilne

⇒2 2

2 21 14 4

| 0 0 1|a b

x y

a b

+ − = +

+

2 2

2 2 1 12 2 4 4 2 21 14 4

x y1 1a b

x y a b a b

a b

⇒ = + ⇒ + =++

Locus of P(x1, y1) is 2 2

4 4 2 2

x y 1

a b a b+ =

+.

PROLEMS FOR PRACTICE

1. Find the eccentricity, coordinates of foci. Length of latus rectum and equations of directrices of the following ellipses.

i) 9x2 + 16y2 – 36x + 32y – 92 = 0

ii) 3x2 + y2 – 6x – 2y – 5 = 0

2. Find the equation of the ellipse referred to its major and minor axes as the coordinate axes x, y respectively with latus rectum of length 4 and the distance between foci 4 2 .

Ans. x2 + 2y2 = 16

3. If the length of the latus rectum is equal to half of its minor axis of an ellipse in the standard form, then find the eccentricity of the ellipse.

Ans. e = 3 / 2

4. If θ1, θ2 are the eccentric angles of the extremities of a focal chord of the ellipse 2 2

2 2

x y1

a b+ = (a > b) and e its eccentricity, then show that

i) 1 2 1 2ecos cos2 2

θ + θ θ − θ=

ii) 1 2e 1cot cot

e 1 2 2

+ θ θ = ⋅ −

5. C is the center, AA′ and BB′ are major and minor axis of the ellipse 2 2

2 2

x y1

a b+ = . If PN

is the ordinate of a point P on the ellipse then show that 2 2

2

(PN) (BC)

(A N)(AN) (CA)=

′.

6. If S and T are the foci of an ellipse and B is one end of the minor axis. If STB is an equilateral triangle, then find the eccentricity of the ellipse.

Sol.

Equation of the ellipse is 2 2

2 2

x y1

a b+ =

Foci are S(ae, 0), T(–ae, 0)

B(0, b) is the end of the minor axis

STB is an equilateral triangle

SB = ST ⇒ SB2 = ST2

⇒ a2e2 + b2 = 4a2e2

⇒b2 = 3a2e2

⇒ a2(1 – e2) = 3a2e2

1 – e2 = 3e2

4e2 = 1 ⇒ e2 = 1

4

Eccentricity of the ellipse : 1

e2

= .

B(0, b)

T S

7. Find the equation of the tangent and normal to the ellipse 9x2 + 16y2 = 144 at the end of the latus rectum in the first quadrant.

Sol. Given ellipse is 9x2 + 16y2 = 144

⇒

2 2x y1

16 9+ =

⇒2 2

2

a b 16 9 7e

16 4a

− −= = =

End of the latus rectum in first Quadrant

2b 9P ae, 7,

a 4

=

Equation of the tangent at P is S1 =0⇒ 1 12 2

xx yy1

a b+ =

7 y 9x 1

16 9 4

7x y1

16 4

⋅ + =

+ =

7x 4y 16+ =

Equation of the normal at P is

⇒

2 22 2

1 1

a x b ya b

x y− = −

⇒16x 9y

16 9(9 / 4)7

− = −

16x

4y 7 16x 4 7y 7 77

− = ⇒ − =

8. If a tangent to the ellipse 2 2

2 2

x y1

a b+ = (a > b) meets its major axis and minor axis in M

and N respectively, then prove that 2 2

2 2

a b1

(CM) (CN)+ = . Where e is the centre of the

ellipse.

9. Find the condition for the line

i) lx + my + n = 0 to be a tangent to the ellipse 2 2

2 2

x y1

a b+ = .

ii) lx + my + n = 0 to be a normal to the ellipse 2 2

2 2

x y1

a b+ = .

Sol. i) Equation of the ellipse is 2 2

2 2

x y1

a b+ =

Equation of the tangent at P(θ) is x y

cos sin 1a b

θ + θ = …(1)

Equation of the given line is lx + my = –n …(2)

(1) and (2) are representing the same line. Therefore,

cos sin 1

al bm n

θ θ= =−

⇒cos sin 1

al bm n

θ θ− =−

⇒al bm

cos sinn n

θ = − θ = −

SINCE 2 2cos sin 1θ + θ =

⇒

2 2 2 2

2 2

a l b m1

n n+ =

⇒ a2l2 + b2m2 = n2 is the required condition.

ii) Let lx + my + n = 0 be normal at P(a)Equation of the normal at P(a) is

2 2ax bya b

cos sin− = −

θ θ …(1)

BUT equation of the normal is Lx + my = –n …(2)

Comparing (1) and (2)

2 2

l m na b a b

cos sin

= =− −

θ θ

2 2

l cos msin n

a b a b

θ − θ −= =−

2 2 2 2

an bncos , sin

l(a b ) m(a b )

−θ = θ =− −

2 2cos sin 1θ + θ =

2 2 2 2

2 2 2 2 2 2 2 2

a n b n1

l (a b ) m (a b )+ =

− −

2 2 2 2 2

2 2 2

a b (a b )

l m n

−+ = is the required condition.

10. If the normal at one end of a latus rectum of the ellipse 2 2

2 2

x y1

a b+ = passes through one

end of the minor axis, then show that e4 + e2 = 1.


2 2

x y1

a b+ =

One end of the latusrectum is L(ae, b2/a)

Equation of the normal at L(ae, b2/a) is

2 2

2 22

a x b ya b

ae (b / a)− = −

2 22 2

1 1

a x b ya b

x y

− = −

∵

2 2axay a e

e− =

This normal passes through B′(0, –b)

ab = a2e2

⇒ b = ae2

⇒b2 = a2e4

⇒ a2(1 – e2) = a2e4

⇒ e4 + e2 = 1.

11. If PN is the ordinate of a point P on the ellipse 2 2

2 2

x y1

a b+ = and the tangent at P meets

X - axis at T, then show that (CN)(CT) = a2 where C is the centre of the ellipse.

12. Show that the point of intersection of perpendicular tangents to an ellipse lie on a circle.


2 2

x y1

a b+ =

Let P(x1, y1) be the point of intersection of the tangents.


X

Y

C

L(ae,b2/a)

B′(0, –b)

2 2 2y mx a m b= ± +

This tangent is passing through P(x1, y1)

2 2 21 1y mx a m b= ± +

⇒2 2 2

1 1y mx a m b− = ± +

⇒2 2 2 2

1 1(y mx ) a m b− = +

⇒2 2 2 2 2 2

1 1 1 1m x y 2mx y a m b x+ − − −

⇒2 2 2 2 2

1 1 1 1m (x a ) 2mx y (y b ) 0− − + − =

This is a quadratic equation in m giving two values for m say m1 and m2. These are the slopes of the tangents passing through (x1,y1).

The tangents are perpendicular ⇒ m1m2 = –1

2 2

2 2 2 2 2 2 2 211 1 1 12 2

1

y b1 y b x a x y a b

x a

− = − ⇒ − = − + ⇒ + = +−

Locus of P(x1, y1) is x2 + y2 = a2 + b2 which is a circle.

This circle is called Director circle of the Ellipse.

13. Find the pole of the line 3x – 5y – 9 = 0 w.r. to the ellipse 4x2 + 8y2 – 16x + 15 = 0.

Ans. Pole = 9 5

,4 24

−

14. Show that the poles of the tangents to the auxiliary circle w.r. to the ellipse lie on the

curve 2 2

4 4 2

x y 1

a b a+ = .

Sol. Let P(x1, y1) be the pole.

Polar of P w.r. to the ellipse 2 2

2 2

x y1

a b+ = is

R

Q

P(x1, y1)

1 12 2

xx yy1

a b+ = …(1)

Equation of the auxiliary circle is x2 + y2 = a2

Polar (1) is a tangent to the auxiliary circle.

2 21 14 4

| 0 0 1|a

x y

a b

+ − =

+

2 21 14 4 2

x y 1

a b a+ =


4 4 2

x y 1

a b a+ = .

15. A chord PQ of an ellipse S = 0 subtends a right angle at the centre of the ellipse. Show that the point of intersection of tangents at P and Q lies on another ellipse

2 2

4 4 2 2

x y 1 1

a b a b+ = + .

16. Show that the poles of the tangents of y2 = 4kx (k > 0) w.r. to the ellipse 2 2

2 2

x y1

a b+ = lies

on a parabola.

Sol. Let P(x1, y1) be the pole .

Pplar of P w.r.to ellipse 2 2

2 2

x y1 0

a b+ − =

is 1 12 2

xx yy1 0

a b+ − =

2 2

12

11

b x by x

ya y

−= ⋅ +

This is a tangent to y2 = 4kx ⇒ a

cm

=

A

R Q

P(x1, y1) A′

221

21 11

21

421 12

kyb k

y bxb x

a y

by x

a k

−= =− −

−⇒ =

Locus of P(x1, y1) is the parabola 4

22

by x

a k

= −

.

17. Show that the poles of normal chords of the ellipse 2 2

2 2

x y1

a b+ = lie on the curve

6 62 2 2

2 2

a b(a b )

x y+ = − .

Sol.

Equation of the ellipse si 2 2

2 2

x y1

a b+ =

Equation of the normal at P(θ) is 2 2 2ax by(a b )

cos sin− = −

θ θ … (1).

Let P(x1, y1) be the pole of (1)

Polar of P is 1 12 2

xx yy1

a b+ = …(2)

Comparing (1) and (2)

1 12 2

2 2

x y1a b

a b a bcos sin

= = −− θ θ

1 13 3 2 2

x cos y sin 1

a b a b

θ θ= =− −

3 32 2 2 2

1 1

a b(a b ) cos , (a b )sin

x y

−− θ = − θ =

6 62 2 2 2 2

2 21 1

a b(a b ) (cos sin )

x y− θ + θ = +


2 2 22 2

a b(a b )

x y+ = − .

4 (B) 1. Find the equation of the tangent and normal to the ...

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