Answer
$\dfrac{x^2}{1600}+\dfrac{y^2}{1536}=1$
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 $e=0.2$ ; foci: $(\pm 8,0)$
Now, $p=\dfrac{c}{e}=\dfrac{8}{0.2}=40$ and $q^2=p^2-c^2=1600-64=1536$
Thus, the equation of the ellipse becomes:
$\dfrac{x^2}{1600}+\dfrac{y^2}{1536}=1$
