Question

What are the vertices of the hyperbola with equation 4y^2 - 25x^2 = 100?

Answer 1

bearing in mind that the variable with the positive sign is the "y", meaning the hyperbola  is opening vertically, and thus its vertices/foci will be along the center's y-coordinate.

$\bf \textit{hyperbolas, vertical traverse axis } \\\\ \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf 4y^2-25x^2=100\implies \cfrac{4y^2-25x^2}{100}=1\implies \cfrac{4y^2}{100}-\cfrac{25x^2}{100}=1 \\\\\\ \cfrac{y^2}{25}-\cfrac{x^2}{4}=1\implies \cfrac{(y-0)^2}{5^2}-\cfrac{(x-0)^2}{2^2}=1\qquad \begin{cases} h=0\\ k=0\\ a=5\\ b=2 \end{cases} \\\\\\ \stackrel{\textit{vertices}}{(0, 0\pm 5)}\implies \begin{cases} (0,5)\\ (0,-5) \end{cases}$