College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 40

Identify the type of curve which is represented by the equation $\displaystyle 36x^2 - 4y^2 - 36x - 8y = 31 $
Find the foci and vertices(if any), and sketch the graph

$
\begin{equation}
\begin{aligned}
36 (x^2 - x + \quad) - 4(y^2 + 2y + \quad)&= 31 && \text{Group terms and factor}\\
\\
36 \left( x^2 - x + \frac{1}{4} \right) - 4 (y^2 + 2y + 1) &= 31 + 9 - 4 && \text{Complete the square; Add } \left( \frac{2}{2} \right)^2 =1
\text{ on the left and subtract. Then, add 9 on the right side and subtract 4}\\
\\
36 \left( x - \frac{1}{2} \right) - 4 (y + 1)^2 &= 36 && \text{Perfect square}\\
\\
\left( x - \frac{1}{2} \right)^2 - \frac{(y+1)^2}{9} &= 1 && \text{Divide by 36}
\end{aligned}
\end{equation}
$


The equation is hyperbola that has the form $\displaystyle \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ with center at $(h,k)$ and horizontal transverse axis.
Since the $x^2$-term is positive. The graph of the shifted hyperbola is obtained by shifting the graph of $\displaystyle x^2 - \frac{y^2}{9} = 1$, by
$\displaystyle \frac{1}{2}$ units to the right and 1 unit downward. This gives us $a^2 = 1$ and $b^2 = 9$, so $a = 1, b =3$ and $c = \sqrt{a^2+b^2} = \sqrt{1+9} = \sqrt{10}$.
Thus, by applying transformations, we have

$
\begin{equation}
\begin{aligned}
\text{center } & (h,k) && \rightarrow && \left( \frac{1}{2}, -1 \right)\\
\\
\text{vertices } & (a,0)&& \rightarrow && (1,0) && \rightarrow && \left( 1 + \frac{1}{2}, 0 - 1 \right) && = && \left( \frac{3}{2}, -1 \right)\\
\\
& (-a,0)&& \rightarrow && (-1,0) && \rightarrow && \left( -1 + \frac{1}{2}, 0 - 1 \right) && = && \left( - \frac{1}{2}, -1 \right)\\
\\
\text{foci } & (c,0)&& \rightarrow && (\sqrt{10},0) && \rightarrow && \left( \sqrt{10} + \frac{1}{2}, 0 - 1 \right) && = && \left( \sqrt{10} + \frac{1}{2}, - 1 \right)\\
\\
& (-c,0)&& \rightarrow && (-\sqrt{10},0) && \rightarrow && \left( -\sqrt{10} + \frac{1}{2}, 0 - 1 \right) && = && \left( -\sqrt{10} + \frac{1}{2}, - 1 \right)
\end{aligned}
\end{equation}
$

Therefore, the graph is