Answer
$\frac{y^2}{9}-\frac{x^2}{16}=1$
See graph
Work Step by Step
1. Given center $(0,0)$, focus $(0,5)$, vertex $(0,3)$, we have a vertical transverse axis, $c=5,a=3$ and $b=\sqrt {c^2-a^2}=\sqrt {25-9}=4$,
2. thus the equation is $-\frac{x^2}{16}+\frac{y^2}{9}=1$ or $\frac{y^2}{9}-\frac{x^2}{16}=1$
3. See graph.
