Answer
Vertices: $(0,\pm 3)$
Co-vertices: $(\pm 1,0)$
Foci: $(0,\pm 2\sqrt{2})$
Work Step by Step
Rewrite in the form of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$:
$9x^2+y^2=9$ (Divide by 9)
$\frac{x^2}{1}+\frac{y^2}{9}=1$
We obtain $a^2=1$ ($a=1$) and $b^2=9$ ($b=3$).
Since $b>a$, the major axis is vertical.
Find $c$:
$c=\sqrt{b^2-a^2}=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$
So, the ellipse has the vertices at $(0,\pm 3)$, the co-vertices at $(\pm 1,0)$, and the foci at $(0,\pm 2\sqrt{2})$.
