## Algebra and Trigonometry 10th Edition

$\frac{(x-3)^2}{1}+\frac{y^2}{9}=1$
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 $2a$: $2a=3-(-3)=3+3=6$ $a=3$ $2b=2$ $b=1$ $\frac{(x-3)^2}{1^2}+\frac{(y-0)^2}{3^2}=1$ $\frac{(x-3)^2}{1}+\frac{y^2}{9}=1$