Answer
$\dfrac{x^2}{64}+\dfrac{y^2}{48}=1$
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 $x=-16 \implies e=\dfrac{1}{2}$
Thus, the equation of the ellipse becomes:
$[(x+4)^2+(y-0)^2]^{1/2}=\dfrac{1}{2} (x+16)^2 \implies \dfrac{x^2}{64}+\dfrac{y^2}{48}=1$
