Check the picture below, so the hyperbola looks more or less like so, then
[tex]\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) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} h=5\\ k=3\\ a=6\\ b=9 \end{cases}\implies \cfrac{(y-3)^2}{6^2}~~ - ~~\cfrac{(x-5)^2}{9^2}=1\implies \cfrac{(y-3)^2}{36}~~ - ~~\cfrac{(x-5)^2}{81}=1[/tex]
