Answer
$\frac{x^2}{4}+y^2=1$, foci $(\pm\sqrt 3,0)$
Work Step by Step
Step 1. From the given graph, we can identify $a=2, b=1$. Thus $c=\sqrt {a^2-b^2}=\sqrt {3}$. The ellipse is centered at $(0,0)$ with a horizontal major axis.
Step 2. We can write the equation as $\frac{x^2}{4}+y^2=1$ with foci at $(\pm\sqrt 3,0)$
