Answer
$\frac{(x-2)^2}{4}+\frac{(y+3)^2}{9}=1$
Work Step by Step
Center: $(h,k)=(2,-3)$
$h=2$
$k=-3$
Vertices: $(2,0)$ and $(2,6)$
The distance between the vertices is equal to the length of the major axis, $2a$:
$2a=6$
$a=3$
Endpoints of the minor axis: $(0,-3)$ and $(4,-3)$. The distance between the endpoints of the minor axis is equal to its length, $2b$:
$2b=4$
$b=2$
Equation of an ellipse with center at $(h,k)$ (major axis is vertical):
$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$
$\frac{(x-2)^2}{2^2}+\frac{(y+3)^2}{3^2}=1$
$\frac{(x-2)^2}{4}+\frac{(y+3)^2}{9}=1$
