Answer
$\dfrac{x^2}{4}-\dfrac{y^2}{12}=1$
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 vertices are: $(\pm 2,0)$ and $e=2$
so, $e=\dfrac{c}{a}=2\implies c=2p=(2)(2)=4$
Now, $q^2=c^2-p^2=16-4=12$
Thus, the equation of the ellipse becomes:
$\dfrac{x^2}{4}-\dfrac{y^2}{12}=1$
