Answer
$\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
Work Step by Step
The the standard form for the ellipse is given as:
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ...(a) ( when $a\geq b \geq 0$)
with Vertices:$(\pm a,0); Foci: (\pm c,0)$ and $c^2=a^2-b^2$
From the given problem, we have vertices:$(\pm 5,0)$
and Vertices:$(\pm a,0)=(\pm 5,0) \implies a=5$
and $Foci: (\pm c,0)=F(\pm 4,0) \implies c=4$
$ (4)^2=(5)^2-b^2$
and $16=25-b^2 \implies b^2=9$
The the standard form for the ellipse is given as:
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ...(a) ( when $a\geq b \geq 0$)
$\dfrac{x^2}{(5)^2}+\dfrac{y^2}{9}=1$
Hence, $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
