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 :
Correct Option: , 4
Solution:
$\frac{a}{e}=4 \Rightarrow a=4 \times \frac{1}{2}=2$
Now, $b^{2}=a^{2}\left(1-e^{2}\right)$
$\Rightarrow b^{2}=4\left(1-\frac{1}{4}\right)=4 \times \frac{3}{4}=3$
So, equation $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$
$\Rightarrow 3 x^{2}+4 y^{2}=12$ .....(i)
Now, $P(1, \beta)$ lies on it
$\Rightarrow 3+4 \beta^{2}=12 \Rightarrow \beta=\frac{3}{2}$
So, equation of normal at $P\left(1, \frac{3}{2}\right)$
$\Rightarrow \frac{a^{2} x}{1}-\frac{b^{2} y}{3 / 2}=a^{2}-b^{2} \Rightarrow 4 x-2 y=1$