Answer
Ellipse, center $C(3, -3)$, foci $F(3, -3\pm\frac{\sqrt 2}{2})$, vertices $V_1(3,-4)$ and $V_2(3,-2)$
See graph.
Work Step by Step
Step 1. Identify the type of the conics: rewrite the equation as $2(x^2-6x+9)+(y^2+6y+9)=18+9-26$ or $2(x-3)^2+(y+3)^2=1$, thus the equation is of the form of an ellipse.
Step 2. The center can be found at $C(3, -3)$
Step 3. With $a=1, b=\frac{\sqrt 2}{2}, c=\sqrt {1-\frac{1}{2}}=\frac{\sqrt 2}{2}$, the foci can be found at $F(3, -3\pm\frac{\sqrt 2}{2})$
Step 4. The vertices can be found at $V(3, -3\pm1)$ or $V_1(3,-4)$ and $V_2(3,-2)$
Step 5. See graph.
