The eccentricity of an ellipse whose center 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

 

As $x=-\frac{a}{e}=-4$

We have, $a=4e=4\cdot \frac{1}{2}=2$

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

Thus the equation to ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$

Differentiating $w.r.t.x$, we get

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

At $\left(1,\frac{3}{2}\right),\frac{dy}{dx}=-\frac{3}{4}\cdot \frac{1\cdot 2}{3}=-\frac{1}{2}$

So the slope of normal is $2$. The equation is

$y-\frac{3}{2}=2(x-1)\text{ i.e., }4x-2y-1=0$