Answer
$\dfrac{(x-3)^2}{12}+\dfrac{y^2}{16}=1$
Work Step by Step
We are given that the vertices are $(\pm 5,0)$ and the major axis length is $8$
Thus, $h=3$ and $k=\dfrac{-2+2}{2}=0$
We have the general equation for the ellipse for the major axis length: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ ...(1) ; when $b\geq a $
Here, the center is: $(h,k)$ and $c^2=a^2-b^2$
$2a=8 \implies a=4$
Now, $c=2-k=2-0=2$
and $c^2=a^2-b^2 \implies (2)^2=(4)^2-b^2$
or, $b^2=12$
Thus, equation (1) becomes:
$\dfrac{(x-3)^2}{12}+\dfrac{(y-0)^2}{4^2}=1$
or,
$\dfrac{(x-3)^2}{12}+\dfrac{y^2}{16}=1$
