Answer
$\frac{(y-1)^{2}}{25}-\frac{(x+3)^{2}}{39}=1$
Work Step by Step
$\frac{(y-k)^{2}}{a^2}-\frac{(x-h)^{2}}{b^2}=1$
Here, $a=\frac{|-4-6|}{2}=5$ and $c=5$ from the given points
we can write
$\frac{(y-k)^{2}}{5^2}-\frac{(x+3)^{2}}{b^2}=1$
$2c=|-7-9|$
$c=8$
$c^2=a^2+b^2$
$8^2=5^2+b^2$
$b^2=39$
The equation of the hyperbola is
$\frac{(y-1)^{2}}{25}-\frac{(x+3)^{2}}{39}=1$
