Question

The equation of a hyperbola is given by (y-2)^2/81-(x+10)^2/9=1.

Identify the coordinates of the center of the hyperbola.
Find the length of the transverse and conjugate axes.
Find the slopes of the asymptotes.
Find the coordinates of the foci.
Graph the hyperbola. Label the center, midpoints of the associated rectangle, and foci.

Answer 1

so... for a hyperbola, the fraction with the positive sign, in this case the fraction with the "y" variable, is where the traverse axis is, so in this case is a vertical traverse axis, so the hyperbola looks more or less like the one in the picture below


`\bf \cfrac{(y-2)^2}{81}-\cfrac{(x+10)^2}{9}=1\implies \cfrac{(y-2)^2}{9^2}-\cfrac{[x-(-10)]^2}{3^2}=1\\\\ -----------------------------\\\\`


`\bf \textit{hyperbolas, vertical traverse axis }\\\\ \cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1 \qquad \begin{cases} center\ ({{ h}},{{ k}})\\ vertices\ ({{ h}}, {{ k}}\pm a)\\ asymptotes\quad y={{ k}}\pm \cfrac{a}{b}(x-{{ h}})\\ \textit{slope of the asymptote}\\ \qquad \qquad \pm\cfrac{a}{b}\\\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{{{ a }}^2+{{ b }}^2}\\\\ foci=k, k\pm c \end{cases}`