## Precalculus (6th Edition) Blitzer

$\frac{x^2}{25}+\frac{y^2}{9}=1$
Step 1. With the given conditions, we have $c=4, a=5$; thus $b^2=5^2-4^2=9$ Step 2. The center is at $(0,0)$, so we can write the equation as $\frac{x^2}{25}+\frac{y^2}{9}=1$