Answer
$\frac{(x-5)^2}{25}+\frac{y^2}{16}=1$
Work Step by Step
Foci: $(2,0)$ and $(8,0)$
The center $(h,k)$ is the midpoint between the foci:
$(h,k)=\frac{(2,0)+(8,0)}{2}=(5,0)$
$h=5$
$k=0$
$2c$ is the distance between the foci:
$2c=8-2=6$
$c=3$
Vertices: $(0,0)$ and $(10,0)$
The distance between the vertices is equal to the length of the major axis, $2a$:
$2a=10$
$a=5$
$a^2=b^2+c^2$
$b^2=a^2-c^2=5^2-3^2=25-9=16$
$b=4$
Equation of an ellipse with center at $(h,k)$ (major axis is horizontal):
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$
$\frac{(x-5)^2}{5^2}+\frac{y^2}{4^2}=1$
$\frac{(x-5)^2}{25}+\frac{y^2}{16}=1$
