College Algebra, Chapter 8, 8.4, Section 8.4, Problem 16

Find the center, foci, vertices and asymptotes of the hyperbola $\displaystyle \frac{(y - 1)^2}{25} - (x + 3)^2 = 1$. Sketch its graph.

The shifted hyperbola has center at $(-3, 1)$ and a vertical transverse axis. It is derived from the hyperbola $\displaystyle \frac{y^2}{25} - x^2 = 1$ with center at the origin. Since $a^2 = 25$ and $b^2 = 1$, we have $a = 5, b = 1$ and $c = \sqrt{25 + 1} = \sqrt{26}$. Thus, the foci lie $\sqrt{26}$ units above and below the center. Consequently, the vertices of the hyperbola lies $5$ units above and below the center. By applying transformations, we get

Foci

$\displaystyle (-3, 1) \to (-3, 1 + \sqrt{26})$

$\displaystyle (-3, 1) \to (-3, 1 - \sqrt{26})$

Vertices

$\displaystyle (-3, 1) \to (-3, 1 + 5) = (-3, 6)$

$\displaystyle (-3, 1) \to (-3, 1 - 5) = (-3, -4)$

The asymptotes of the unshifted hyperbola are $\displaystyle y = \pm \frac{a}{b}x = \pm 5x$, so the asymptotes of the shifted hyperbola are


$
\begin{equation}
\begin{aligned}

y - 1 =& \pm (5x + 3)
\\
\\
y - 1 =& \pm 5x \pm 15
\\
\\
y =& 5x + 16 \text{ and } y = -5x - 14



\end{aligned}
\end{equation}
$



Therefore, the graph is