Answer
$\frac{(x-2)^2}{16}+\frac{(y-3)^2}{9}=1$
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: $(-2,3)~~and~~(6,3)$
The center is the midpoint: $\frac{(-2,3)+(6,3)}{2}=(2,3)$
The elipse is in the horizontal position. The distance between the vertices is equal to $2a$:
$2a=6-(-2)=6+2=8$
$a=4$
$2b=6$
$b=3$
$\frac{(x-2)^2}{4^2}+\frac{(y-3)^2}{3^2}=1$
$\frac{(x-2)^2}{16}+\frac{(y-3)^2}{9}=1$
