Find the equation of the ellipse whose eccentricity is 2 3 , focus 3 , 4 and directrix is 3 x + 4 y = 5 .

Let P x , y be the point on the ellipse and F 3 , 4 is the focus. According to the definition of an ellipse, the distance of a point P from focus is e times its distance from directrix, where e is eccentricity. Given directrix is 3 x + 4 y = 5 . If M be the foot of the perpendicular from P on directrix, then P F = e P M ⇒ P F 2 = e 2 P M 2 ⇒ x - 3 2 + y - 4 2 = 4 9 × 3 x + 4 y - 5 3 2 + 4 2 2 ⇒ x 2 + 9 - 6 x + y 2 + 16 - 8 y = 4 9 × 1 25 ( 9 x 2 + 16 y 2 + 25 + 24 x y - 40 y - 30 x ) ⇒ 225 x 2 + 2025 - 1350 x + 225 y 2 + 3600 - 1800 y = 36 x 2 + 64 y 2 + 100 + 96 x y - 160 y - 120 x ⇒ 189 x 2 + 161 y 2 - 96 x y - 1230 x - 1640 y + 5525 = 0

Find the equation of the hyperbola, whose foci are ± 5 , 0 and length of transverse axis is 8 .

Given foci = ± 5 , 0 and length of transverse axis is 8 . Since foci lie on the x - axis then the equation of the hyperbola is of the form x 2 a 2 - y 2 b 2 = 1 . W know that length of transverse axis is 2 a = 8 ∴ a = 4 We have foci = ± a e , 0 = ± 5 , 0 ∴ a e = 5 ⇒ e = 5 4 We know that e = a 2 + b 2 a ⇒ 5 4 = 16 + b 2 4 ⇒ 25 = 16 + b 2 ⇒ b 2 = 9 ∴ Required equation of the hyperbola is x 2 16 - y 2 9 = 1 .

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Conic Sections>Uttar Pradesh Board-2018

Embibe Experts Solutions for Chapter: Conic Sections, Exercise 1: Uttar Pradesh Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Conic Sections, Exercise 1: Uttar Pradesh Board-2018

Attempt the free practice questions on Chapter 14: Conic Sections, Exercise 1: Uttar Pradesh Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.