Answer
$\frac{(y-2)^2}{4}-\frac{(x-2)^2}{8}=1$
See graph.
Work Step by Step
1. Known a hyperbola with center $(2,2)$, vertex $(2,4)$ and point $(2+\sqrt {10},5)$ on the curve, we can find it has a vertical transverse axis with a general form $\frac{(y-2)^2}{a^2}-\frac{(x-2)^2}{b^2}=1$
2. With $a=4-2=2$, use the point $(2+\sqrt {10},5)$, we have $\frac{(5-2)^2}{4}-\frac{(\sqrt {10})^2}{b^2}=1$ which gives $b^2=8, c^2=12$, thus we have $\frac{(y-2)^2}{4}-\frac{(x-2)^2}{8}=1$
3. See graph.
