Answer
$\dfrac{x^2}{15}+\dfrac{y^2}{16}=1$
Work Step by Step
The standard form of an ellipse can be written as: $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1....(1); $ when $b \gt a$
We are given that an ellipse with $x$-intercept at $\pm {15}$ and $y$-intercept at $\pm 4$ due to the intercepts, the centre will be at origin.)
But $\sqrt {15} \lt 4(\sqrt {16})$, then the vertices of the ellipse becomes: $(0, \pm 4)$. This yields: $a=4$, and the end-points its minor axis are $( \pm {15}, 0)$, that is, $b=\sqrt {15}$.
Now, substitute these values into the equation (1) to obtain:
$\dfrac{x^2}{(\sqrt {15})^2}+\dfrac{y^2}{(4)^2}=1 \\ \dfrac{x^2}{15}+\dfrac{y^2}{16}=1$
