Answer
See graph and explanations.
Work Step by Step
Step 1. Given $\frac{x^2}{16}+\frac{y^2}{4}=1$, we can identify $a^2=16, b^2=4$; 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 graph the equation as shown in the figure, and the foci can be located at $(\pm2\sqrt 3,0)$
