Answer
$\frac{x^2}{4}-\frac{y^2}{12}=1$
Work Step by Step
Vertices: $(±a,0)=(±2,0)$
$a=2$
Foci: $(±c,0)=(±4,0)$
$c^2=a^2+b^2$
$b^2=c^2-a^2=4^2-2^2=16-4=12$
$b=2\sqrt 3$
Hyperbola with horizontal transverse axis:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{x^2}{2^2}-\frac{y^2}{(2\sqrt 3)^2}=1$
$\frac{x^2}{4}-\frac{y^2}{12}=1$
