## Precalculus (6th Edition) Blitzer

$\frac{y^2}{36}-\frac{(x-5)^2}{20}=1$
Step 1. With vertices at $(5,-6)$ and $(5,6)$, we have $a=\frac{6-(-6)}{2}=6$ and the hyperbola is centered at $(5,0)$ with a vertical transverse axis. Step 2. We can write the general form of the equation as $\frac{y^2}{36}-\frac{(x-5)^2}{b^2}=1$ Step 3. With point $(0,9)$ on the curve, we have $\frac{9^2}{36}-\frac{(0-5)^2}{b^2}=1$ which gives $b^2=20$ Step 4. The equation is $\frac{y^2}{36}-\frac{(x-5)^2}{20}=1$