Answer
$\dfrac{x^2}{16}-\dfrac{y^2}{20}=1$
Work Step by Step
The standard form of the equation of the ellipse when the major axis is horizontal can be expressed as: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ in which $(h,k)$ 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: $\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$ in which $(h,k)$ 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 $2a$:
$b=\sqrt{a^2-c^2}=\sqrt {6^2-4^2}=\sqrt {20}$
$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$
or, $\dfrac{x^2}{16}-\dfrac{y^2}{20}=1$
