Mathematics - Ellipse Question with Solution | TestHub

Given, $e=\frac{1}{2}$ and $\frac{a}{e}=4$

$\Rightarrow a=2,$

Also, $b^2=a^2\left(1-e^2\right)$

$\Rightarrow b^2=3$

 Equation of ellipse is  $\frac{x^2}{4}+\frac{y^2}{3}=1$
We know that, equation of normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $\left(x_1, y_1\right)$ is $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$
So, equation of normal at $\left(1, \frac{3}{2}\right)$ is  $\frac{4x}{1}-\frac{3y}{3/2}=4-3$ 
$\Rightarrow 4x-2y=1$