Answer
vertices $(0,-4),(0,4)$, foci $(0,-2\sqrt {3}),(0,2\sqrt {3})$.
See graph.
Work Step by Step
1. Based on the given equation, we have $\frac{x^2}{4}+\frac{y^2}{16}=1$, with $a=4, b=2$, center $(0,0)$ and a vertical major axis, thus $c=\sqrt {a^2-b^2}=\sqrt {16-4}=2\sqrt {3}$, vertices $(0,-4),(0,4)$, foci $(0,-2\sqrt {3}),(0,2\sqrt {3})$.
2. See graph.
