## Calculus (3rd Edition)

$$\left(\frac{y}{6}\right)^{2}-\left(\frac{x}{12\sqrt 2}\right)^{2}=1.$$
Since the vertices are $(0,\pm 6)$ and $e=3$, then the axis of the hyperbola is the y-axis and then $b=6$, moreover, $e=\frac{c}{b}=\frac{c}{6}=3$, and hence $c=18$ and $a=\sqrt{c^2-b^2}=\sqrt{18^2-6^2}=12\sqrt{2}$. Hence the equation of the hyperbola is given by $$\left(\frac{y}{6}\right)^{2}-\left(\frac{x}{12\sqrt 2}\right)^{2}=1.$$