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 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.

x - 2 y + 4 = 0 is a common tangent to y 2 = 4 x & x 2 4 + y 2 b 2 = 1 . Then find the value of ' b ' and the other common tangent.

Given, equation of ellipse is x 2 4 + y 2 b 2 = 1 ⇒ a 2 = 4 Therefore, Equation of tangent in slope form to the ellipse is y = m x ± a 2 m 2 + b 2 … … … 1 And Given equation of tangent is x - 2 y + 4 = 0 Or, y = x 2 + 2 … … 2 Equation (1) and (2) represent same tangent therefore on comparing them we get, m = 1 2 & 2 = a 2 m 2 + b 2 ⇒ 4 = 4 × 1 4 + b 2 ⇒ b = ± 3 Now, Equation of parabola is y 2 = 4 x Equation of tangent in slope form will be y = m x + 1 m … … 3 On comparing equation (1) and (3) we get, 1 m = a 2 m 2 + b 2 ⇒ 1 m 2 = 4 m 2 + 3 ⇒ 4 m 4 + 3 m 2 - 1 = 0 ⇒ 4 m 4 + 4 m 2 - m 2 - 1 = 0 4 m 2 m 2 + 1 - 1 m 2 + 1 = 0 ⇒ m = ± 1 2 For m = - 1 2 ,Substituting in equation (3) x + 2 y + 4 = 0 is the another common tangent.

If latus rectum of ellipse x 2 25 + y 2 16 = 1 is double ordinate of parabola y 2 = 4 a x , then find the value of a

Given, equation of ellipse x 2 25 + y 2 16 = 1 ⇒ a 2 = 25 & b 2 = 16 Length of latus rectum of ellipse = 2 b 2 a = 2 × 16 5 = 32 5 Now, for parabola y 2 = 4 a x let the double ordinate be latus rectum of parabola Therefore, Length of latus rectum = 4 a ⇒ 32 5 = 4 a ⇒ a = 8 5

E M B I B E

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

HARD

JEE Main/Advance

IMPORTANT

Earn 100

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

Important Questions on Ellipse

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 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