Answer
$ \dfrac{x^2}{15}+\dfrac{y^2}{16}=1$
Work Step by Step
The standard form of an ellipse is: $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1, \text{ where } b \gt a.$
Let us consider an ellipse with $x$-intercepts at $\pm \sqrt {15}$ and $y$-intercepts at $\pm 4$.
(Due to the intercepts, the center will be at origin.)
Note that the $x$-intercepts are $\sqrt{15}$ units away from the center while the $y$-intercepts are $4$ units away. Since $\sqrt {15} \lt 4 \text{ (which is equal to } \sqrt {16})$, then the vertices of the ellipse are $(0, \pm 4)$, which means that $a=4$, while the end-points of its minor axis are $(\pm \sqrt {15}, 0)$, which means that $b=\sqrt {15}$.
Substitute these values into the equation above to obtain:
$$\dfrac{x^2}{(\sqrt {15})^2}+\dfrac{y^2}{4^2}=1 \\ \dfrac{x^2}{15}+\dfrac{y^2}{16}=1$$
