Answer
Ellipse, center $C(0, 6)$, foci $F(0, 6\pm\sqrt {33})$, vertices $V_1(0,0)$ and $V_2(0,12)$
See graph.
Work Step by Step
Step 1. Identify the type of the conics: rewrite the equation as $12x^2+y^2-12y=0$ or $12x^2+(y-6)^2=36$ which gives $\frac{x^2}{3}+\frac{(y-6)^2}{36}=1$, thus the equation is of the form of an ellipse.
Step 2. The center can be found at $C(0, 6)$
Step 3. With $a=6, b=\sqrt 3, c=\sqrt {36-3}=\sqrt {33}$, the foci can be found at $F(0, 6\pm\sqrt {33})$
Step 4. The vertices can be found at $V(0, 6\pm6)$ or $V_1(0,0)$ and $V_2(0,12)$
Step 5. See graph.
