Question

PLEASE HELP WITH PRE-CALC

Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ±1 divided by 3 times x

Answer 1

check the picture below, so the hyperbola looks more or less like so.

so, we know its center is at the origin and it's "a" component is 4 units.


`\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 √( a ^2 + b ^2)\\ asymptotes\quad y= k\pm \cfrac{a}{b}(x- h) \end{cases}\\\\ -------------------------------\\\\ \begin{cases} h=0\\ k=0\\ a=4 \end{cases}\implies \cfrac{(y- 0)^2}{ 4^2}-\cfrac{(x- 0)^2}{ b^2}=1`


`\bf \begin{cases} y= k+ \cfrac{a}{b}(x- h)\implies &y=0+\cfrac{4}{b}(x-0)\\\\ &y=\cfrac{4x}{b}\\\\ y=\cfrac{1}{3}x\implies &y=\cfrac{x}{3} \end{cases} \\\\\\ \cfrac{x}{3}=\cfrac{4x}{b}\implies \cfrac{b}{3}=4\implies \boxed{b=12} \\\\\\ \cfrac{(y- 0)^2}{ 4^2}-\cfrac{(x- 0)^2}{ 12^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{144}=1`