Answer
$\frac{x^2}{16}+\frac{y^2}{4}=1$, foci $(\pm2\sqrt 3,0)$
Work Step by Step
Step 1. From the given graph, we can identify $a=4, b=2$; thus $c=\sqrt {a^2-b^2}=2\sqrt {3}$. The ellipse is centered at $(0,0)$ with a horizontal major axis.
Step 2. We can write the equation as $\frac{x^2}{16}+\frac{y^2}{4}=1$ with foci at $(\pm2\sqrt 3,0)$
