Answer
$\frac{y^2}{16}-\frac{x^2}{16}=1$
Work Step by Step
Vertices: $(0,±a)=(0,±4)$
$a=4$
Foci: $(0,±c)=(0,±13)$
Hyperbola with vertical transverse axis:
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
The hyperbola passes through the point $(3,-5)$:
$\frac{(-5)^2}{4^2}-\frac{3^2}{b^2}=1$
$\frac{25}{16}-\frac{9}{b^2}=1$
$\frac{25}{16}-1=\frac{9}{b^2}$
$\frac{9}{16}=\frac{9}{b^2}$
$b^2=16$
$b=4$
Finally:
$\frac{y^2}{4^2}-\frac{x^2}{4^2}=1$
$\frac{y^2}{16}-\frac{x^2}{16}=1$
