Algebra II-Please check

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Please help-What is the equation of the ellipse with foci (0,6), (0,-6) and co-vertices (2,0),(-2,0)

Please explain the steps because I have 5 to do for homework-Thank you-I'm really stuck on this
I thought the answer would be x^2/4 + y^2/40 = 1 but that can't be because the choices are:
x^2/1 + y^2/40 = 1 or x^2/1 + y^2/36
I'm really confused

  • Algebra II-Please check -

    F1(0,-6), F2(0,6).

    The focus points show that the x-coordinate is constant @ zero, and the
    y-coordinate varies from -6 to +6. Therefore, the focus points are on a vertical line. Which means we have a y-ellipse and a^2 is under y^2 in the Eq.

    X^2/b^2 + Y^2/a^2 = 1.

    b = +-2.
    F^2 = a^2 - b^2,
    F^2 = a^2 - 2^2 = 6^2,
    a^2 - 4 = 36,
    a^2 = 40,

    Eq: X^2/4 + Y^2/40 = 1.
    Check: Let Y = 0 and solve for X:
    X^2/4 + 0/40 = 1,
    X^2/4 = 1,
    Multiply both sides by 4:
    X^2 = 4,
    X = +-2 = b = co-vertices = x-intercepts. In your 1st choice, the "1"
    should be a 4.

    NOTES:

    1. "a" is always greater than b.

    2. If this was an x-ellipse, a^2
    would be located under X^2.

    3. "a" is always on the major(longst)
    axis.

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