Answer
$\frac{x^2}{16}+\frac{y^2}{7}=1$
Work Step by Step
The foci and the vertices lie in the x-axis. We have an ellipse with major axis horizontal. Also, its center is at the origin because the midpoint between the vertices is $\frac{(4,0)+(-4,0)}{2}=(0,0)$. So:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Vertices: $V(±a,0)=V(±4,0)$
$a=4$
Foci: $F(±c,0)=F(±3,0)$
$c=3$
$a^2=b^2+c^2$
$b^2=a^2-c^2=4^3-3^2=16-9=7$
$b=\sqrt 7$
We have:
$\frac{x^2}{4^2}+\frac{y^2}{(\sqrt 7)^2}=1$
$\frac{x^2}{16}+\frac{y^2}{7}=1$
