let's notice something on this hyperbola, the fraction that is positive, is the fraction with the "y" variable, that simply means that the hyperbola is opening vertically, namely runs over the y-axis or it has a vertical traverse axis, which means, that, the foci will be a certain "c" distance from the center over the y-axis, well, with that mouthful, let's proceed.
[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} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\cfrac{(y-3)^2}{1}-\cfrac{(x+2)^2}{4}=1\implies \cfrac{[y-3]^2}{1^2}-\cfrac{[x-(-2)]^2}{2^2}=1~~ \begin{cases} h=-2\\ k=3\\ a=1\\ b=2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ c=\sqrt{a^2+b^2}\implies c=\sqrt{1+4}\implies c=\sqrt{5} \\\\\\ \stackrel{\textit{so then the foci are at}}{(-2~~,~~3\pm \sqrt{5})}\qquad \qquad \qquad \stackrel{\textit{and its vertices are at }}{(-2~~,~~3\pm 1)}\implies \begin{cases} (-2,4)\\ (-2,2) \end{cases}[/tex]
now let's check for the asymptotes.
[tex]\bf y=3\pm \cfrac{1}{2}[x-(-2)]\implies y=3\pm \cfrac{1}{2}(x+2) \\\\[-0.35em] ~\dotfill\\\\ y=3+ \cfrac{1}{2}(x+2)\implies y=3+\cfrac{x+2}{2}\implies y=\cfrac{6+x+2}{2} \\\\\\ y=\cfrac{x+8}{2}\implies y=\cfrac{1}{2}x+4 \\\\[-0.35em] ~\dotfill\\\\ y=3- \cfrac{1}{2}(x+2)\implies y=3-\cfrac{(x+2)}{2}\implies y=\cfrac{6-(x+2)}{2} \\\\\\ y=\cfrac{6-x-2}{2}\implies y=\cfrac{-x+4}{2}\implies y=-\cfrac{1}{2}x+2[/tex]
