Question

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

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

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

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