A focus of an ellipse is at origin.

Directrix is $x=4$ and eccentricity is $e=\frac{1}{2}$

Major axis is along $x$-axis ,

$\Rightarrow \frac{a}{e}-ae=4$

$\Rightarrow \frac{a}{{\displaystyle \frac{1}{2}}}-\frac{a}{2}=4$

$\Rightarrow a=\frac{8}{3}$

By the help of options we get the idea that this case is of $\left(a>b\right)$

Then we can say that $e=\sqrt{1-\frac{b^2}{a^2}}$

$$\begin{gathered} \Rightarrow \frac{1}{4}=1-\frac{9b^2}{64} \\ \Rightarrow b^2=\frac{16}{3} \end{gathered}$$

By this time we can say that either the correct option is $\left(1\right)$ or $\left(4\right)$

The standard form of equation of an ellipse is given by $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

where centre is $\left(h,k\right)$

The foci are $\left(h-\sqrt{a^2-b^2},k\right) \& \left(h+\sqrt{a^2-b^2},k\right)$

By this we can say that $h+\sqrt{a^2+b^2}=0$

Now putting all values, we get

$$\begin{gathered} h=-\sqrt{\frac{64}{9}-\frac{16}{3}} \\ h=\frac{-4}{3} \end{gathered}$$

And $k=0$

So by this we can say that the required equation of ellipse is ${\left(3x+4\right)}^2+12y^2=64$