Answer
$\frac{y^2}{16}-\frac{x^2}{20}=1$
See graph.
Work Step by Step
1. Given center $(0,0)$, focus $(0,-6)$, vertex $(0,4)$, we have a vertical transverse axis, $c=6,a=4$ and $b=\sqrt {c^2-a^2}=\sqrt {36-16}=\sqrt {20}=2\sqrt 5$,
2. thus the equation is $-\frac{x^2}{20}+\frac{y^2}{16}=1$ or $\frac{y^2}{16}-\frac{x^2}{20}=1$
3. See graph.
