Answer
Ellipse, center $C(1, 4)$, foci $F(1, 4\pm\sqrt {15})$, vertices $V(1, 4\pm2\sqrt 5)$
See graph.
Work Step by Step
Step 1. Identify the type of the conics: rewrite the equation as $4(x^2-2x+1)+(y^2-8y+16)=4+16$ or $4(x-1)^2+(y-4)^2=20$ which gives $\frac{(x-1)^2}{5}+\frac{(y-4)^2}{20}=1$, thus the equation is of the form of an ellipse.
Step 2. The center can be found at $C(1, 4)$
Step 3. With $a=2\sqrt 5, b=\sqrt 5, c=\sqrt {20-5}=\sqrt {15}$, the foci can be found at $F(1, 4\pm\sqrt {15})$
Step 4. The vertices can be found at $V(1, 4\pm2\sqrt 5)$
Step 5. See graph.
