Answer
$\frac{(x-3)^2}{1}+\frac{y^2}{9}=1$
Work Step by Step
The standard form of the equation of the elipse when the major axis is:
- horizontal: $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ in which $(h,k)$ is the center and $2a$ is the major axis length and $2b$ is the minor axis length.
- vertical: $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ in which $(h,k)$ is the center and $2a$ is the major axis length and $2b$ is the minor axis length.
Vertices: $(3,-3)~~and~~(3,3)$
The center is the midpoint: $\frac{(3,-3)+(3,3)}{2}=(3,0)$
The elipse is in the vertical position. The distance between the vertices is equal to $2a$:
$2a=3-(-3)=3+3=6$
$a=3$
$2b=2$
$b=1$
$\frac{(x-3)^2}{1^2}+\frac{(y-0)^2}{3^2}=1$
$\frac{(x-3)^2}{1}+\frac{y^2}{9}=1$
