## Calculus (3rd Edition)

The vertices are $(\pm12,0)$. The foci are $(\pm \sqrt{468},0)$.
The equation $$\left(\frac{x}{2}\right)^{2}-\left(\frac{y}{3}\right)^{2}=36\Longrightarrow\left(\frac{x}{12}\right)^{2}-\left(\frac{y}{18}\right)^{2}=1$$ is hyperbola. We have $a=12$, $b=18$, $c=\sqrt{a^2+b^2}=\sqrt{468}$ and hence: The vertices are $(\pm12,0)$. The foci are $(\pm \sqrt{468},0)$.