Answer
$\dfrac{x^2}{48}+\dfrac{(y-4)^2}{64}=1$
Work Step by Step
The standard form of the equation of an ellipse when the major axis is horizontal can be expressed as: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Where $(h,k)$ is the center, $2a$ is the major axis length, and $2b$ is the minor axis length.
The standard form of the equation of the ellipse when the major axis is vertical can be expressed as: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$
Where $(h,k)$ is the center, $2a$ is the major axis length, and $2b$ is the minor axis length.
The center is the midpoint of the foci: $(\dfrac{0+0}{2}, \dfrac{0+8}{2})=( 0, 4)$
The ellipse is in the horizontal axis, so the distance between the vertices is equal to $2a$:
$b=\sqrt{a^2-c^2}=\sqrt {8^2-4^2}=\sqrt {48}$
$\dfrac{(x-0)^2}{(\sqrt {48})^2}+\dfrac{(y- 4)^2}{(8)^2}=1$
or, $\dfrac{x^2}{48}+\dfrac{(y-4)^2}{64}=1$
