Answer
$\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
Work Step by Step
The general equation for the ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ...(1) ; $a\geq b \geq 0$
Here, vertices:$(\pm a,0);
Foci: F(\pm c,0)$ and $c^2=a^2-b^2$
We are given that the vertices are: $(\pm 5,0)$
This gives: $a=5$
and foci: $F(\pm 4,0)$
This gives: $c=4$
Now, $c^2=a^2-b^2 \implies (4)^2=(5)^2-b^2$
or, $b^2=9$
Thus, equation (1) becomes:
$\dfrac{x^2}{5^2}+\dfrac{y^2}{9}=1$
or,
$\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
