College Algebra, Chapter 8, 8.3, Section 8.3, Problem 40

Find an equation for the hyperbola with foci $(\pm 3, 0)$ and passes through $(4,1)$.
The hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has foci $(\pm c,0)$. So, the value of
$c$ is $3$. Thus, the first equation gives us $a^2 + b^2 = 9$. Also, if the hyperbola passes through the given point, then the point is a
solution to the equation,

$
\begin{equation}
\begin{aligned}
\frac{(4)^2}{a^2} - \frac{(1)^2}{b^2} &= 1 && \text{Substitute the given}\\
\\
\frac{16}{a^2} - \frac{1}{b^2} &= 1 && \text{Evaluate}\\
\\
\frac{16}{a^2} &= \frac{1}{b^2} + 1 && \text{Add } \frac{1}{b^2}\\
\\
16b^2 &= a^2 + a^2 b^2 && \text{Multiply } a^2 b^2\\
\\
a^2 &= \frac{16b^2}{1 + b^2} && \text{Simplify, thus gives us the second equation}
\end{aligned}
\end{equation}
$

By substituting the second equation to the first equation, we get

$
\begin{equation}
\begin{aligned}
\frac{16b^2}{1+b^2} + b^2 &=9 && \text{Multiply } (1 + b^2)\\
\\
16b^2 + b^2 + b^4 &= 9 + 9b^2 && \text{Simplify and combine like terms}\\
\\
b^4 + 8b^2 &= 9 && \text{Subtract 9}\\
\\
b^4 + 8b^2 - 9 &= 0 && \text{Factor}\\
\\
(b^2+9)(b^2-1) &= 0 && \text{Zero Product Property}\\
\\
b^2 + 9 &= 0 \text{ and } b^2 - 1 = 0 && \text{Choose the value of $b$ that will give real roots}\\
\\
b^2 -1 &= 0 && \text{Solve for } b^2\\
\\
b^2 &= 1
\end{aligned}
\end{equation}
$

We back substitute $b^2$to the Equation, we have

$
\begin{equation}
\begin{aligned}
a^2 + 1 &= 9 && \text{Solve for } a^2\\
\\
a^2 &= 8
\end{aligned}
\end{equation}
$

Therefore, the equation is
$\displaystyle \frac{x^2}{8} - y^2 = 1$