Answer
$-\frac{x^2}{12}+\frac{y^2}{4}=1$ or $\frac{y^2}{4}-\frac{x^2}{12}=1$
Work Step by Step
Step 1. From the given conditions, we have $c=4, a=2$; thus $b=\sqrt {4^2-2^2}=2\sqrt 3$ with center at $(0,0)$ and foci on $x=0$
Step 2. We can write the equation as $-\frac{x^2}{12}+\frac{y^2}{4}=1$ or $\frac{y^2}{4}-\frac{x^2}{12}=1$
