Posted by **Lucy** on Sunday, October 12, 2008 at 5:05pm.

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.

- Pre-Calculus -
**Damon**, Sunday, October 12, 2008 at 5:31pm
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

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