Answer
$\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
Work Step by Step
Given: Vertices:$(\pm 5,0)$
Consider the standard equation for the ellipse :$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ...(1) ; $a\geq b \geq 0$
with Vertices: $V(\pm a,0); Foci: F(\pm c,0)$ and $c^2=a^2-b^2$
In this problem, we have $a=5; a^2 =25$; Foci: $F(\pm 4,0)$
Then, we have $c=4$
Now, $c^2=a^2-b^2= (4)^2=(5)^2-b^2$
or, $b^2=25-16 \implies b^2=9$
Hence, we have $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
