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