Answer
Ellipse, center $C(2, 0)$, foci $F_1(-1,0)$ and $F_2(5,0)$, vertices $V(2\pm2\sqrt 3,0)$
See graph.
Work Step by Step
Step 1. Identify the type of the conics: rewrite the equation as $(x^2-4x+4)+4y^2=8+4$ or $(x-2)^2+4y^2=12$ which gives $\frac{(x-2)^2}{12}+\frac{y^2}{3}=1$, thus the equation is of the form of an ellipse.
Step 2. The center can be found at $C(2, 0)$
Step 3. With $a=2\sqrt 3, b=\sqrt 3, c=\sqrt {12-3}=3$, the foci can be found at $F(2\pm3, 0)$ or $F_1(-1,0)$ and $F_2(5,0)$
Step 4. The vertices can be found at $V(2\pm2\sqrt 3,0)$
Step 5. See graph.
