check the picture below, so the hyperbola looks more or less like so.
so, we know its center is at the origin and it's "a" component is 4 units.
[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}\\
asymptotes\quad y= k\pm \cfrac{a}{b}(x- h)
\end{cases}\\\\
-------------------------------\\\\
\begin{cases}
h=0\\
k=0\\
a=4
\end{cases}\implies \cfrac{(y- 0)^2}{ 4^2}-\cfrac{(x- 0)^2}{ b^2}=1[/tex]
[tex]\bf \begin{cases}
y= k+ \cfrac{a}{b}(x- h)\implies &y=0+\cfrac{4}{b}(x-0)\\\\
&y=\cfrac{4x}{b}\\\\
y=\cfrac{1}{3}x\implies &y=\cfrac{x}{3}
\end{cases}
\\\\\\
\cfrac{x}{3}=\cfrac{4x}{b}\implies \cfrac{b}{3}=4\implies \boxed{b=12}
\\\\\\
\cfrac{(y- 0)^2}{ 4^2}-\cfrac{(x- 0)^2}{ 12^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{144}=1[/tex]
