MATH SOLVE

4 months ago

Q:
# Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y = ± 3/4x

Accepted Solution

A:

Check the picture below.so the hyperbola looks more or less like so, with a = 6, and its center at the origin.[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} a=6\\ h=0\\ k=0\\ \stackrel{asymptotes}{y=\pm\frac{3}{4}x} \end{cases}\implies \stackrel{\textit{using the positive asymptote}}{0+\cfrac{6}{b}(x-0)=\cfrac{3}{4}x}\implies \cfrac{6x}{b}=\cfrac{3x}{4}\implies 24x=3xb \\\\\\ \cfrac{24x}{3x}=b\implies 8=b \\\\[-0.35em] ~\dotfill\\\\ \cfrac{(y- 0)^2}{ 6^2}-\cfrac{(x- 0)^2}{ 8^2}=1\implies \cfrac{y^2}{36}-\cfrac{x^2}{64}=1[/tex]