Check the picture below, so the hyperbola looks more or less like so, hmmm so is a vertical one, so is opening towards the y-axis, meaning the fraction that is positive is the one with the y-variable, its traverse axis length is 4 units, from +2 to -2, meaning that a = 2, and the distance either focus to the center is √5, so
[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)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} h=0\\ k=0\\ a=2\\ c=\sqrt{5} \end{cases}\qquad \cfrac{(y- 0)^2}{ 2^2}-\cfrac{(x- 0)^2}{ b^2}=1[/tex]
[tex]\sqrt{5}=\sqrt{2^2+b^2}\implies 5=4+b^2\implies 1=b^2\implies \sqrt{1}=b\implies 1=b \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(y- 0)^2}{ 2^2}-\cfrac{(x- 0)^2}{ 1^2}=1\implies \cfrac{y^2}{4}~~ - ~~x^2~~ = ~~1[/tex]
