Answer
$\frac{x^2}{9}-\frac{4y^2}{9}=1$
Work Step by Step
Vertices: $(±a,0)=(±3,0)$
$a=3$
Asymptotes:
$y=±\frac{b}{a}x$
$y=±\frac{1}{2}x$
$\frac{b}{a}=\frac{1}{2}$
$\frac{b}{3}=\frac{1}{2}$
$b=\frac{3}{2}$
Hyperbola with horizontal transverse axis:
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{x^2}{3^2}-\frac{y^2}{(\frac{3}{2})^2}=1$
$\frac{x^2}{9}-\frac{y^2}{\frac{9}{4}}=1$
$\frac{x^2}{9}-\frac{4y^2}{9}=1$
