Answer
$\dfrac{(x-2)^2}{16}+\dfrac{y^2}{12}=1$
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 : $(\dfrac{0+4}{2}, \dfrac{0+0}{2})=(2, 0)$
The ellipse is in the horizontal axis, so the distance between the vertices is equal to $2a$:
$b=\sqrt{a^2-c^2}=\sqrt {4^2-2^2}=\sqrt {12}$
$\dfrac{(x-2)^2}{4^2}+\dfrac{(y-0)^2}{(\sqrt {12}^2}=1$
or, $\dfrac{(x-2)^2}{16}+\dfrac{y^2}{12}=1$
