Answer
$\frac{x^2}{4^2}+\frac{y^2}{7}=1$
Work Step by Step
The graph of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a\gt b\gt 0$ is an ellipse that has vertices $(\pm a,0)$ and foci $(\pm c,0)$, where $c=\sqrt{a^2-b^2}$.
Hence the equation is of an ellipse that has vertices $(\pm 4,0)$ and foci $(\pm 3,0)$, where $9=\sqrt{4^2-b^2}$ is $\frac{x^2}{4^2}+\frac{y^2}{7}=1$
