The eccentricity of an ellipse whose centre is at the origin is...

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

$⇒ a = 2,$

Also, ${b}^{2} = {a}^{2}\left(1 - {e}^{2}\right)<br /><br />$⇒ {b}^{2} = 3$<br /><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/210efedd-9922-4eb2-883c-4b98159607fd-image.png" style="width: 330px; height: 202px;" />$

Equation of ellipse is $\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{3} = 1$
We know that, equation of normal to the ellipse $\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ at \left({x}_{1}, {y}_{1}\right)$is$\dfrac{{a}^{2}x}{{x}_{1}} - \dfrac{{b}^{2}y}{{y}_{1}} = {a}^{2} - {b}^{2}$<br />So, equation of normal at \left(1, \dfrac{3}{2}\right)$ is $\dfrac{4x}{1}$-$\dfrac{3y}{3 / 2}=4 - 3$
$⇒ 4x - 2y = 1$