PreCalculus
posted by Lucy .
Find the equation of the ellipse whose semimajor axis has length 6 and whose foci are at (3,2+/sqrt11).
I have the focal constant = 2a which means that a=3 but don't know how to proceed.
Thanks.

Well you know the center is halfway between the foci, at (3,2)
I agree that a = 3
The form is:
(y+2)^2/a^2 + (x3)^2/b^2 = 1
we need b
the distance from center to focus is sqrt(a^2b^2)
so
11 = 3^2+b^2
b^2 = 2
b = sqrt 2
so
(y+2)^2/9 + (x3)^2/2 = 1