Question

The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}.$ If one of its directrices is $x=-4,$ then the equation of the normal to it at $\left(1,\frac{3}{2}\right)$ is

• A. $2y-x=2$
• B. $4x-2y=1$
• C. $4x+2y=7$
• D. $x+2y=4$

Answer 1

The correct option is B $4x-2y=1$

Given: Eccentricity of ellipse$=\frac{1}{2}$

Now, $\frac{a}{e}=-4$

$\Rightarrow a=4\times \frac{1}{2}=2$

$\therefore b^2=a^2(1-e^2)$

$\Rightarrow a^2(1-\frac{1}{4})=3$

$\Rightarrow \frac{3}{4}{a}^2=3$

$\therefore a^2=4$

$\frac{x^2}{4}+\frac{y^2}{3}=1$

Differentiating w.r.t $x$ we get

$\frac{2x}{4}+\frac{2y}{3}\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=\frac{\frac{-2x}{4}}{\frac{2y}{3}}$

$\Rightarrow \frac{dy}{dx}=\frac{-3x}{4y}$

$\Rightarrow {\left[\frac{dy}{dx}\right]}_{(1,\frac{3}{2})}=\frac{-3}{4}\times \frac{2}{3}=\frac{-1}{2}$

Equation of normal at $(1,\frac{3}{2})$ is

$y-\frac{3}{2}=2(x-1)$

$\Rightarrow 2y-3=4x-4$

$\Rightarrow 4x-2y=1$ is the equation of the normal.