Answer
$\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$
Work Step by Step
The standard form of an ellipse can be written as: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1....(1)$
We are given that an ellipse with $x$-intercept at $\pm 5$ and $y$-intercept at $\pm 4$ due to the intercepts, the centre will be at origin.)
But $5 \gt 4$, then the vertices of the ellipse becomes: $(\pm 5,0)$. This yields: $a=5$, and the end-points its minor axis are $(0, \pm 4)$, that is, $b=4$.
Now, substitute these values into the equation (1) to obtain:
$\dfrac{x^2}{(5)^2}+\dfrac{y^2}{(4)^2}=1 \\ \dfrac{x^2}{25}+\dfrac{y^2}{16}=1$
