JEE Main Solved Paper 2017 Question 27
Question: The eccentricity of an ellipse whose centre is at the origin is $ \frac{1}{2}. $ If one of its directices is $ x=-4, $ then the equation of the normal to it at $ ( 1,\frac{3}{2} ) $ is:- JEE Main Solved Paper-2017
Options:
A) $ x+2y=4 $
B) $ 2y-x=2 $
C) $ 4x-2y=1 $
D) $ 4x+2y=7 $
Show Answer
Answer:
Correct Answer: C
Solution:
- Eccentricity of ellipse $ =\frac{1}{2} $ Now, $ -\frac{a}{e}=-4\Rightarrow a=4\times \frac{1}{2}=2 $
$ \therefore $ $ b^{2}=a^{2}(1-e^{2})=a^{2}( 1-\frac{1}{4} )=3 $
$ \therefore $ Equation of ellipse $ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1 $
$ \Rightarrow $ $ \frac{x}{2}+\frac{2y}{3}\times y’=0,\Rightarrow y’=-\frac{3x}{4y} $ $ y’{| _{(1,3/2)}}=-\frac{3}{4}\times \frac{2}{3}=-\frac{1}{2} $
$ \therefore $ Equation of normal at $ ( 1,\frac{3}{2} ) $ $ y-\frac{3}{2}=2(x-1)\Rightarrow 2y-3=4x-4 $
$ \therefore $ $ 4x-2y=1 $
