something noteworthy is that in the equation the variable with the positive sign is the "y", meaning is a vertical hyperbola that has a vertical traverse axis.
[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)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf 9y^2-16x^2=144\implies \cfrac{9y^2-16x^2}{144}=1\implies \cfrac{9y^2}{144}-\cfrac{16x^2}{144}=1 \\\\\\ \cfrac{y^2}{16}-\cfrac{x^2}{9}=1\implies \cfrac{(y-0)^2}{4^2}-\cfrac{(x-0)^2}{3^2}=1\qquad \begin{cases} h = 0\\ k = 0\\ a = 4\\ b = 3\\ c = 5 \end{cases}[/tex]
Check the picture below.
