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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.


  • Pre-Calculus -

    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)
    11 = 3^2+b^2
    b^2 = 2
    b = sqrt 2
    (y+2)^2/9 + (x-3)^2/2 = 1

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