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 + (x-3)^2/b^2 = 1
we need b
the distance from center to focus is sqrt(a^2-b^2)
so
11 = 3^2+b^2
b^2 = 2
b = sqrt 2
so
(y+2)^2/9 + (x-3)^2/2 = 1
Find the equation of the ellipse whose semi-major 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.
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