Find the co-ordinates of the foci of the ellipse x 2 25 + y 2 4 = 1 .

Given ellipse be x 2 25 + y 2 4 = 1 . So on comparing with standard equation of horizontal ellipse, x 2 a 2 + y 2 b 2 = 1 . we have, a 2 = 25 , b 2 = 4 i.e., a = 5 ; b = 2 Now, b 2 = a 2 1 - e 2 ⇒ 4 = 25 1 - e 2 ⇒ 4 25 = 1 - e 2 ⇒ e 2 = 1 - 4 25 ⇒ e 2 = 21 25 ⇒ e = 21 5 Thus, foci are ± a e , 0 . ⇒ ± 21 5 × 5 , 0 ⇒ ± 21 , 0

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

Embibe Experts Solutions for Chapter: Ellipse, Exercise 1: Exercise-1

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Ellipse, Exercise 1: Exercise-1

Attempt the free practice questions on Chapter 18: Ellipse, Exercise 1: Exercise-1 with hints and solutions to strengthen your understanding. Alpha Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Ellipse, Exercise 1: Exercise-1 with Hints & Solutions

EASY

JEE Main/Advance

IMPORTANT

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

Find the co-ordinates of the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1$ .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 24

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

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 25

Find the equation of common tangent to circle $x^{2} + y^{2} = 5$ and ellipse $x^{2} + 9 y^{2} = 9$

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 26

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

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 32

The position of the point $(1, 3)$ with respect to the ellipse $4 x^{2} + 9 y^{2} - 16 x - 54 y + 61 = 0$

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 18 - Ellipse>Exercise-1>Q 36

Identify the following statements for true/false (T/F) in order
$S 1 :$ A latus rectum of an ellipse is a line passing through a focus
$S 2 :$ A latus rectum of an ellipse is a line through the centre
$S 3 :$ A latus rectum of an ellipse is a line perpendicular to the major axis
$S 4 :$ A latus rectum of an ellipse is a line parallel to the minor axis