Answer
center $(0,0)$, transverse axis $y$-axis, vertices $(0,-3),(0,3)$, foci $(0,-\sqrt {10}),(0,\sqrt {10})$, and asymptotes $y =\pm3x$.
See graph.
Work Step by Step
1. Given $y^2-9x^2=9\Longrightarrow \frac{y^2}{9}-x^2=1$, we can find the center $(0,0)$, transverse axis $y$-axis, $a=3,b=1$, vertices $(0,-3),(0,3)$, $c=\sqrt {a^2+b^2}=\sqrt {9+1}=\sqrt {10}$, foci $(0,-\sqrt {10}),(0,\sqrt {10})$, and asymptotes $y=\pm\frac{a}{b}x=\pm3x$.
2. See graph.
