Answer
$\dfrac{x^2}{(64/3)}+\dfrac{y^2}{(16/3)}=1$
Work Step by Step
The eccentricity of the ellipse $\dfrac{x^2}{p^2}+\dfrac{y^2}{q^2}=1$ when $p \gt q$ is given by:
$e=\dfrac{\sqrt {p^2-q^2}}{p}$
The foci of the ellipse are: $(\pm pe,0)$ and the directrices are given as: $x=\pm \dfrac{p}{e}$
We are given that $e=0.24$ ; focus: $(4,0)$
The directrix is $x=\dfrac{16}{3} \implies e=\dfrac{\sqrt 3}{2}$
Thus, the equation of the ellipse becomes:
$[(x-4)^2+(y-0)^2]^{1/2}=\dfrac{3}{4} (x-\dfrac{16}{3})^2 \implies \dfrac{x^2}{(64/3)}+\dfrac{y^2}{(16/3)}=1$
