Answer
$\dfrac{(x-2)^2}{3}+\dfrac{(y-2)^2}{4}=1$
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 : $(2, \dfrac{0+4}{2}, 0)=(2, 2)$
The ellipse is in the horizontal axis, so the distance between the vertices is equal to $2a$:
$b^2=a^2-c^2=2^2-1^2=3$
$\dfrac{(x-2)^2}{3}+\dfrac{(y- 2)^2}{2^2}=1$
or, $\dfrac{(x-2)^2}{3}+\dfrac{(y-2)^2}{4}=1$
