## Precalculus (6th Edition) Blitzer

See graph; foci $(\pm\sqrt {41},0)$; asymptotes $y=\pm\frac{5}{4}x$
Step 1. From the given equation $\frac{x^2}{16}-\frac{y^2}{25}=1$, we have $a=4, b=5, c=\sqrt {a^2+b^2}=\sqrt {41}$ centered at $(0,0)$ with a horizontal transverse axis. Step 2. We can find the vertices as $(\pm4,0)$ and asymptotes as $y=\pm\frac{5}{4}x$ Step 3. We can graph the equation as shown in the figure with foci at $(\pm\sqrt {41},0)$