College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 20

Determine the center, vertices, foci and asymptotres of the hyperbola $\displaystyle x^2 - 4y^2 + 16 = 0$. Then, sketch its graph

$
\begin{equation}
\begin{aligned}
x^2 - 4y^2 &= -16 && \text{Subtract 16}\\
\\
\frac{y^2}{4} - \frac{x^2}{16} &= 1 && \text{Divide both sides by } -16
\end{aligned}
\end{equation}
$


The hyperbola has the form $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with center at origin and vertical transverse axis since the denominator
of $y^2$ is positive. This gives $a^2 = 4$ and $b^2 = 16$, so $a = 2, b = 4$ and $c = \sqrt{a^2 + b^2} = \sqrt{4+16} = 2 \sqrt{5}$
Then, the following are determined as

$
\begin{equation}
\begin{aligned}
\text{center } (h,k) && \rightarrow && (0,0)\\
\\
\text{vertices } (0, \pm 2)&& \rightarrow && (0, \pm 2)\\
\\
\text{foci } (\pm c, 0) && \rightarrow && (0, \pm 2 \sqrt{5})\\
\\
\text{asymptote } y = \pm \frac{b}{a}x && \rightarrow && y = \pm \frac{1}{2}x
\end{aligned}
\end{equation}
$

Therefore, the graph is