if you notice, the positive fraction, is the one with the "x" variable in it, that simply means, the hyperbola is opening over the x-axis, so is a horizontal one, therefore
[tex]\bf \textit{hyperbolas, horizontal traverse axis }
\\\\
\cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1
\qquad
\begin{cases}
center\ ( h, k)\\
vertices\ ( h\pm a, k)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}
\end{cases}\\\\
-------------------------------[/tex]
[tex]\bf \cfrac{(x-5)^2}{81}-\cfrac{(y-1)^2}{144}=1\implies \cfrac{(x-5)^2}{9^2}-\cfrac{(y-1)^2}{12^2}=1~~
\begin{cases}
a=9\\
b=12
\end{cases}
\\\\\\
c=\sqrt{9^2+12^2}\implies c=15\\\\
-------------------------------\\\\
vertices~~(5\pm 9,1)\implies
\begin{cases}
(14,1)\\
(-4,1)
\end{cases}~~foci~~(5\pm 15,1)\implies
\begin{cases}
(20,1)\\
(-10,1)
\end{cases}[/tex]
