Answer
$\frac{(y-1)^2}{9}-\frac{(x+2)^2}{16}=1$
Work Step by Step
Step 1. From the given conditions, the hyperbola has a center at $(-2,1)$ and a vertical transverse axis; thus the equation is in the form of
$\frac{(y-1)^2}{a^2}-\frac{(x+2)^2}{b^2}=1$
Step 2. From the given coordinates, we have
$a=4-1=3, c=6-1=5$
Thus
$b^2=c^2-a^2=16$
Step 3. The equation is
$\frac{(y-1)^2}{9}-\frac{(x+2)^2}{16}=1$
