Answer
$\frac{(y-1)^{2}}{25}-\frac{(x+3)^{2}}{39}=1$, equation of a hyperbola.
Work Step by Step
$\frac{(y-k)^{2}}{a^2}-\frac{(x-h)^{2}}{b^2}=1$ ... (1)
From the given points, we have $a=\frac{|-4-6|}{2}=5$
Equation (1) can be written as:$\frac{(y-1)^{2}}{5^2}-\frac{(x+3)^{2}}{b^2}=1$
Also, $2c=|-7-9|$
and $c=8$
Now, $c^2=a^2+b^2$
$5^2+b^2=8^2$
or, $b^2=39$
Hence, $\frac{(y-1)^{2}}{25}-\frac{(x+3)^{2}}{39}=1$, equation of a hyperbola.
