Answer
$\dfrac{x^2}{8556.25}+\dfrac{(y-77.5)^2}{6006.25}=1$
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 : $(0, 77.5)$
Since the ellipse is in the horizontal axis, the distance between the vertices is equal to $2a$:
$a=\dfrac{185}{2}=92.5$ and $b=\dfrac{155}{2}=77.5$
$\dfrac{(x-0)^2}{(92.5)^2}+\dfrac{(y- 77.5)^2}{(77.5)^2}=1$
or, $\dfrac{x^2}{8556.25}+\dfrac{(y-77.5)^2}{6006.25}=1$
