Question

Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ±2/3x.

Answer 1

Check the picture below.

so the hyperbola looks more or less like so, is a hyperbola with a vertical traverse axis, with a = 4 and h = 0, k = 0.

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

$\bf \stackrel{\textit{the asymptotes are}}{\pm\cfrac{2}{3}x}\implies \pm\cfrac{2}{3}x=k\pm \cfrac{a}{b}(x- h)~~ \begin{cases} h=0\\ k=0\\ a=4 \end{cases} \\\\\\ +\cfrac{2}{3}x=0+\cfrac{4}{b}(x-0)\implies \cfrac{2x}{3}=\cfrac{4x}{b}\implies 2bx=12x \\\\\\ b=\cfrac{12x}{2x}\implies b=6 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(y-0)^2}{4^2}-\cfrac{(x-0)^2}{6^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{36}=1$