If ' P ' be a moving point on the ellipse x 2 25 + y 2 16 = 1 in such a way that tangent at ' P ' intersect x = 25 3 at Q then circle on P Q as diameter passes through a fixed point. Find that fixed point.

Equation of ellipse is x 2 25 + y 2 16 = 1 , ⇒ a 2 = 25 , b 2 = 16 Therefore, b 2 = a 2 1 - e 2 ⇒ 16 = 25 1 - e 2 ⇒ e 2 = 1 - 16 25 ⇒ e = 3 5 Therefore, Focus of ellipse is S a e , 0 = 3 , 0 and S - a e , 0 = - 3 , 0 . Equation of directrix is x = ± a e = ± 25 3 . Therefore, the line at which Q lies on directrix of the ellipse. Let P 5 cos θ , 4 sin θ By definition of ellipse P S = e P Q . P S and P Q are perpendicular to each other. And P Q is diameter of circle therefore circle passes through focus of ellipse. Therefore, the fixed point is 3 , 0 .

Find the equation of tangents to the ellipse x 2 50 + y 2 32 = 1 , which passes through a point ( 15 , - 4 ) .

Given the equation of ellipse x 2 50 + y 2 32 = 1 . The equation of tangent passing through 15 , - 4 . From ellipse, we have a 2 = 50 , b 2 = 32 . We know the equation of tangent to the ellipse x 2 a 2 + y 2 b 2 = 1 is y = m x ± a 2 m 2 + b 2 . . . 1 So, y - m x = ± a 2 m 2 + b 2 ( passing through 15 , - 4 ) - 4 - m × 15 = ± 50 m 2 + 32 Squaring both sides, 16 + 225 m 2 + 120 m = 50 m 2 + 32 175 m 2 + 120 m - 16 = 0 = m + 4 5 m - 4 35 So, m = - 4 5 , 4 35 The equation of the tangent to the ellipse x 2 a 2 + y 2 b 2 = 1 is y = m x ± a 2 m 2 + b 2 Case 1 for m = - 4 5 y = - 4 5 x + 50 × 16 25 + 32 ⇒ 5 y + 4 x = 1600 ⇒ 4 x + 5 y = 40 Case 2 for m = 4 35 y = 4 35 x + 4 35 2 × 50 + 32 ⇒ 35 y = 4 x + 4000 ⇒ 4 x - 35 y = 200

If 3 x + 4 y = 12 intersect the ellipse x 2 25 + y 2 16 = 1 at P and Q , then find the point of intersection of tangents at P and Q .

Given equation of ellipse x 2 25 + y 2 16 = 1 . . . 1 So, a 2 = 25 & b 2 = 16 . And also equation of line 3 x + 4 y = 1 . . . 1 Let h , k are the point of intersection of tangent. Then equation of chord of contact h x 25 + k y 16 = 1 . . . 2 ∴ Equations 1 & 2 represent the same line. So, 3 h 25 = 4 k 16 = 12 1 So, 3 = 12 × h 25 ⇒ h = 3 × 25 12 ⇒ h = 25 4 And k 16 × 12 = 4 ⇒ k = 4 × 16 12 ⇒ k = 16 3 So, the point is 25 4 , 16 3 .

Find the equation of chord of ellipse x 2 25 + y 2 16 = 1 whose mid-point is ( 3 , 1 ) .

Given equation of ellipse = x 2 25 + y 2 16 = 1 . Equation of chord having mid-point 3 , 1 . We know the equation of chord having mid-point x , y is T = S 1 Where T = x x 1 a 2 + y y 1 b 2 - 1 S = x 1 2 a 2 + y 1 2 b 2 - 1 So, by the condition, we have x 50 + y 40 - 1 = 1 25 × 4 + 4 25 × 16 - 1 ⇒ 4 x + 5 y - 200 200 = 2 + 2 - 200 200 ⇒ 4 x + 5 y = 4

Find the condition so that the line p x + q y = r intersects the ellipse x 2 a 2 + y 2 b 2 = 1 in points whose eccentric angles differ by π 4 .

Let the points are A = a cos θ , b sin θ & B = A cos φ , B sin φ Given, θ - φ = π 4 The equation of the chord joining these two points is x a cos θ + φ 2 + y b sin θ + φ 2 = cos θ - φ 2 ⇒ x a cos θ + φ 2 + y b sin θ + φ 2 = cos π 8 It is same as p x + q y = r cos θ + φ 2 a · p = sin θ + φ 2 b · q = cos π 8 r We know, cos 2 θ + φ 2 + sin 2 θ + φ 2 = 1 ⇒ a 2 p 2 cos 2 π 8 r 2 + b 2 q 2 cos 2 π 8 r 2 = 1 ⇒ a 2 p 2 + b 2 q 2 = r 2 sec 2 π 8 This is the condition so that the line intersect the ellipse.

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 15

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Any tangent to an ellipse is cut by the tangents at the ends of major axis in the points $T$ and $T^{'}$ . Prove that the circle, whose diameter is $T T^{'}$ will pass through the foci of the ellipse.

Important Questions on Ellipse

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 16

If

' P'

be a moving point on the ellipse

\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1

in such a way that tangent at

' P'

intersect

x = \frac{25}{3}

at

Q

then circle on

P Q

as diameter passes through a fixed point. Find that fixed point.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 17

Find the equation of tangents to the ellipse

\frac{x^{2}}{50} + \frac{y^{2}}{32} = 1

, which passes through a point

(15, - 4)

.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 18

If

3 x + 4 y = 12

intersect the ellipse

\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1

at

P

and

Q,

then find the point of intersection of tangents at

P

and

Q

.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 19

Find the equation of chord of ellipse

\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1

whose mid-point is

(3, 1)

.

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 20

Find the condition so that the line

p x + q y = r

intersects the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

in points whose eccentric angles differ by

\frac{π}{4}

.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 21

If the normal at an end of a latus-rectum of an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

passes through one extremity of the minor axis, show that the eccentricity of the ellipse is given by

e^{4} + e^{2} - 1 = 0

.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 22

A ray emanating from the point

(- 4, 0)

is incident on the ellipse

9 x^{2} + 25 y^{2} = 225

at the point

P

with abscissa

3 .

Find the equation of the reflected ray after first reflection.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 23

An ellipse and a hyperbola have the same centre origin, the same foci and the minor-axis of the one is the same as the conjugate axis of the other. If