Question

Let $x=4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$. If $P(1,\beta ),\beta >0$ is a point on this ellipse, then the equation of the normal to it at $P$ is :-

• A. $7x-4y=1$
• B. $4x-2y=1$
• C. $4x-3y=2$
• D. $8x-2y=5$

Answer 1

Answer: (B)

Ellipse : $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\text{ directrix }:x=\frac{a}{e}=4\&e=\frac{1}{2}$
$\Rightarrow a=2\&b^2=a^2\left(1-e^2\right)=3$
$\Rightarrow$ Ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$
$P$ is $\left(1,\frac{3}{2}\right)$
Normal is: $\frac{4x}{1}-\frac{3y}{3/2}=4-3$
$\Rightarrow 4x-2y=1$