Answer
$\frac{y^2}{4}-\frac{x^2}{12}=1$. See graph.
Work Step by Step
1. Based on the given conditions, the hyperbola has a vertical transverse axis with center $(0,0)$, thus we have a general form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
2. With $a=2, c=4$, we have $b^2=c^2-a^2=12$, thus the equation $\frac{y^2}{4}-\frac{x^2}{12}=1$ with asymptotes $y=\pm\frac{a}{b}x=\pm\frac{\sqrt 3}{3}x$
3. See graph.
