Answer
$\frac{y^2}{4}-\frac{x^2}{32}=1$
See graph.
Work Step by Step
1. Given focus $(0,6)$, vertices $(0,-2)$ and $(0,2)$, we have a vertical transverse axis, center $(0,0)$ (midpoint of vertices), $c=6,a=2$ and $b=\sqrt {c^2-a^2}=\sqrt {36-4}=4\sqrt 2$,
2. thus the equation is $-\frac{x^2}{32}+\frac{y^2}{4}=1$ or $\frac{y^2}{4}-\frac{x^2}{32}=1$
3. See graph.
