Answer
$\dfrac{x^2}{4}-\dfrac{y^2}{12}=1$
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: Vertices: $(\pm 2,0); e=2$
so, $e=\dfrac{c}{a}=2$ or, $c=2(m)=(2)(2)=4$
Also, $n^2=c^2-m^2=16-4=12$
so, the equation of the ellipse is:
$\dfrac{x^2}{4}-\dfrac{y^2}{12}=1$
