College Algebra, Chapter 8, 8.3, Section 8.3, Problem 36

Find an equation for the hyperbola with vertices $(0, \pm 6)$ and asymptotes $\displaystyle y = \pm \frac{1}{3}x$.
The parabola $\displaystyle \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ has vertices on $(0,\pm a)$ its asymptotes are defined as $\displaystyle
y = \pm \frac{a}{b} x$. Thus we have $a = 6$ and

$
\begin{equation}
\begin{aligned}
\frac{a}{b} &= \frac{1}{3} && \text{Substitute the value of } a\\
\\
\frac{6}{b} &= \frac{1}{3} && \text{Apply cross multiplication}\\
\\
b &= 18
\end{aligned}
\end{equation}
$

Therefore, the equation is
$\displaystyle \frac{y^2}{6^2} - \frac{x^2}{18^2} = 1 \qquad \text{or} \qquad \frac{y^2}{36} - \frac{x^2}{324} = 1$