Calculus: Early Transcendentals 8th Edition

$\frac{(y-1)^{2}}{25}-\frac{(x+3)^{2}}{39}=1$
$\frac{(y-k)^{2}}{a^2}-\frac{(x-h)^{2}}{b^2}=1$ 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: $\frac{(y-1)^{2}}{25}-\frac{(x+3)^{2}}{39}=1$