College Algebra, Chapter 8, 8.3, Section 8.3, Problem 38

Find an equation for the hyperbola with vertices $(0, \pm 6)$ and passes through $(-5,9)$.
The hyperbola $\displaystyle \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ has vertices on $(0,\pm a)$. So, the value of
$a = 6$. Also, if the hyperbola passes through the point $(-5,9)$. Then the point is a solution for the equation. Thus,

$
\begin{equation}
\begin{aligned}
\frac{(a)^2}{6^2} - \frac{(-5)^2}{b^2} &= 1 && \text{Substitute the given}\\
\\
\frac{81}{36} - \frac{25}{b^2} &= 1 && \text{Add } \frac{25}{b^2}\\
\\
\frac{81}{36} &= \frac{25}{b^2} && \text{Apply cross multiplication}\\
\\
81b^2 &= 900 && \text{Divide both sides by 81}\\
\\
b^2 &= \frac{100}{9} && \text{Take the square root}\\
\\
b &= \frac{10}{3}
\end{aligned}
\end{equation}
$

Therefore, the equation is
$\displaystyle \frac{y^2}{36} - \frac{x^2}{\frac{100}{9}} = 1 \qquad \text{or} \qquad \frac{y^2}{36} - \frac{9x^2}{100} = 1$