Answer
$\dfrac{x^2}{64}+\dfrac{y^2}{48}=1$
Work Step by Step
The eccentricity of the ellipse $\dfrac{x^2}{m^2}+\dfrac{y^2}{n^2}=1$ when $m \gt n$ is given by:
$e=\dfrac{\sqrt {m^2-n^2}}{m}$
The foci of the ellipse are: $(\pm me,0)$ and the directrices are given as: $x=\pm \dfrac{m}{e}$
Given: $e=0.24$ ; focus: $(-4,0)$
the directrix is $x=-16 $ or, $ e=\dfrac{1}{2}$
So, the equation of the ellipse is:
$[(x+4)^2+(y-0)^2]^{1/2}=(\dfrac{1}{2}) (x+16)^2$
or, $\dfrac{x^2}{64}+\dfrac{y^2}{48}=1$