## Precalculus (6th Edition) Blitzer

$\frac{(x-7)^2}{4}+\frac{(y-6)^2}{9}=1$
Step 1. Based on the given conditions, we have $a=\frac{9-3}{2}=3, b=\frac{9-5}{2}=2$ with a vertical major axis. The ellipse is centered at $(\frac{5+9}{2},\frac{9+3}{2})$ or $(7,6)$. Step 2. We can write the equation as $\frac{(x-7)^2}{4}+\frac{(y-6)^2}{9}=1$