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
[tex]\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\\\\
-----------------------------\\\\[/tex]
[tex]\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}[/tex]
