## Calculus 8th Edition

$\dfrac{(x-3)^2}{12}+\dfrac{y^2}{16}=1$
Here, we have Vvertices:$(\pm 5,0)$ and the major axis length is $8$ Consider the standard form for the ellipse for the major axis length $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ (When $b\geq a$) with Center:$(h,k)$ and $c^2=a^2-b^2$ Now, we have $h=3$ and $k=\dfrac{-2+2}{2}=0$ and $2a=8 \implies a=4$ Also, Now, $c=2-k=2-0=2$ Now, $c^2=a^2-b^2 \implies (2)^2=(4)^2-b^2$ $\implies b^2=16-4=12$ Consider the standard form for the ellipse for the major axis length $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ (When $b\geq a$) $\dfrac{(x-3)^2}{12}+\dfrac{(y-0)^2}{4^2}=1$ Hence, $\dfrac{(x-3)^2}{12}+\dfrac{y^2}{16}=1$