Answer
$\dfrac{(x-3)^2}{36}+\dfrac{(y-2)^2}{12}=1$
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 ellipse is in the horizontal axis, so the distance between the vertices is equal to $2a$:
$b=\sqrt{a^2-c^2}=\sqrt {a^2-(\dfrac{a}{2})^2}=\sqrt {3a^2/4}=\sqrt {\dfrac{3\cdot (4)^2}{4}}=\sqrt {12}$
$\dfrac{(x-3)^2}{6^2}+\dfrac{(y- 2)^2}{(\sqrt {12})^2}=1$
or, $\dfrac{(x-3)^2}{36}+\dfrac{(y-2)^2}{12}=1$
