Respuesta :
so, since we know the vertices are at (0, 8) and (0, -8), that means is a vertical hyperbola, or one with a vertical traverse axis, check the picture below.
which means, its center is at (0,0), and the traverse axis is 8 units long.
because the hyperbola is opening over the y-axis, that means for the equation, the fraction with the "y" is the positive one, so
[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]\bf \begin{cases} h=0\\ k=0\\ a=8 \end{cases}\implies \cfrac{(y-0)^2}{8^2}-\cfrac{(x-0)^2}{b}=1 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{we know the asymptote is}}{\pm \cfrac{1}{2}x~~=~~0\pm\cfrac{8}{b}(x-0)}\implies \stackrel{\textit{using only the positive asymptote}}{\cfrac{x}{2}=\cfrac{8x}{b}\implies bx=16x}\\\\\\ b=\cfrac{16x}{x}\implies b=16 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(y-0)^2}{8^2}-\cfrac{(x-0)^2}{16^2}=1\implies \cfrac{y^2}{64}-\cfrac{x^2}{256}=1[/tex]
