Question

The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$. If one of the directions 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 $4x-2y=1$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $e=\frac{1}{2}$

Directrix : $x=\frac{-a}{e}=-4\Rightarrow a=4e=4\left(\frac{1}{2}\right)=2$

$b^2=a^2(1-e^2)=4(1-\frac{1}{4})=4\left(\frac{3}{4}\right)=3$

The ellipse is : $\frac{x^2}{4}+\frac{y^2}{3}=1$

Differentiating w.r.t. $x$ on both sides, we get

$\frac{2x}{4}+\frac{2y{y}^{\prime }}{3}=0$

$y^{\prime }=\frac{-x/4}{y/3}=\frac{-3x}{4y}$

Slope of normal $=\frac{4y}{3x}$

Slope at $(1,\frac{3}{2})=\frac{4\left(3/2\right)}{3\left(1\right)}=2$

$\frac{y-3/2}{x-1}=2\Rightarrow 2y-3=4(x-1)$

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

$\Rightarrow 4x-2y-1=0$