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Math Help - Ellipse Equation

  1. #1
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    Ellipse Equation

    My instructor writes this on our review sheet, "An ellipse with vertices (-7,2) and (1,2) passes through (-1,-1). Write the equation."

    Since the vertices have y-coordinate 2, I know the major axis is horizontal. I use:
    (x-h)^2 (y-k)^2
    a^2 + b^2

    So I find the center (-3,2)
    I also find a= 4

    So, how do I find b^2? My instructor did not provide the minor axis or "c."

    So far I have:

    (x+3)^2 + (y-2)^2 Thank You, Kay
    16 b^2
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  2. #2
    o_O
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    \frac{(x+3)^{2}}{16} + \frac{(y-2)^{2}}{b^{2}} = 1

    Since you are given the point (-1, -1) just plug in and solve:
    \frac{(-1+3)^{2}}{16} + \frac{(-1-2)^{2}}{b^{2}} = 1
    \frac{(2)^{2}}{16} + \frac{(-3)^{2}}{b^{2}} = 1

    etc. etc.
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  3. #3
    Moo
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    Hello,

    Ok, so far it's correct.

    \frac{(x+3)^2}{16}+\frac{(y-2)^2}{b^2}=1

    There is just one thing to do :

    (-1,-1) is on the ellipse... So replace x and y, and find b !
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