## Calculus 8th Edition

$\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$
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, $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$