## Calculus (3rd Edition)

$$\left(\frac{x}{3}\right)^{2}-\left(\frac{y}{4}\right)^{2}=1.$$
Since the vertices are $(\pm 3,0)$ and the foci are $(\pm 5,0)$, then we have $a=3, c=5$, and hence $b=\sqrt{c^2-a^2}=\sqrt{25-9}=4$. Hence the equation of the hyperbola is given by $$\left(\frac{x}{3}\right)^{2}-\left(\frac{y}{4}\right)^{2}=1.$$