Answer
$\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$
Work Step by Step
The standard form of an ellipse is: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
Let us consider an ellipse with $x$-intercept at $\pm 5$ and $y$-intercept at $\pm 4$.
(Due to the intercepts, the center will be at origin.)
Note that the $x$-intercepts are $5$ units away from the center of tehe ellipse while the $y$-intercepts are $4$ units away. Since $5 \gt 4$, then the vertices of the ellipse are $(\pm 5,0)$. which means that $a=5$, and the end-points its minor axis are $(0, \pm 4)$, which means that $b=4$.
Substitute these values into the equation above to obtain:
$$\dfrac{x^2}{5^2}+\dfrac{y^2}{4^2}=1 \\ \dfrac{x^2}{25}+\dfrac{y^2}{16}=1$$
