Answer
center $(0,2)$, foci $(-\sqrt 2,2),(\sqrt 2,2)$, vertices $(-\sqrt 3,2),(\sqrt 3,2)$
See graph.
Work Step by Step
1. $x^2+3y^2-12y+9=0\Longrightarrow (x)^2+3(y-2)^2=3$ or $\frac{(x)^2}{3}+\frac{(y-2)^2}{1}=1$,
2. we can find center $(0,2)$ with a horizontal major axis, $a=\sqrt 3, b=1$, $c=\sqrt {a^2-b^2}=\sqrt {3-1}=\sqrt 2$,
3. thus foci $(0\pm\sqrt 2,2)$ or $(-\sqrt 2,2),(\sqrt 2,2)$, vertices $(0\pm\sqrt 3,2)$ or $(-\sqrt 3,2),(\sqrt 3,2)$
4. See graph.
