Answer
$\frac{x^2}{9}-\frac{y^2}{16}=1$
See graph.
Work Step by Step
1. Given foci $(-5,0)$ and $(5,0)$, vertex $(3,0)$, we have a horizontal transverse axis, center $(0,0)$ (midpoint of foci), $c=5,a=3$ and $b=\sqrt {c^2-a^2}=\sqrt {25-9}=4$,
2. thus the equation is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ or $\frac{x^2}{9}-\frac{y^2}{16}=1$
3. See graph.
