question_answer

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 \[\left( 1,\frac{3}{2} \right)\]is:-                      JEE Main Solved Paper-2017

A) \[x+2y=4\]                         

B) \[2y-x=2\]

C) \[4x-2y=1\]                        

D) \[4x+2y=7\]

Correct Answer: C

Solution :

 Eccentricity of ellipse \[=\frac{1}{2}\]                 Now,     \[-\frac{a}{e}=-4\Rightarrow a=4\times \frac{1}{2}=2\] \[\therefore \]  \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})={{a}^{2}}\left( 1-\frac{1}{4} \right)=3\] \[\therefore \]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 \[\left( 1,\frac{3}{2} \right)\]                 \[y-\frac{3}{2}=2(x-1)\Rightarrow 2y-3=4x-4\]                 \[\therefore \]  \[4x-2y=1\]