Answer
$\frac{x^2}{1}+\frac{y^2}{16}=1$
See graph.
Work Step by Step
1. Based on the given conditions, we have $a=4, b=1$, center $(0,0)$ with a vertical major axis, thus $c=\sqrt {a^2-b^2}=\sqrt {16-1}=\sqrt {15}$, and $\frac{x^2}{1}+\frac{y^2}{16}=1$
2. See graph.