Analytic Goemetry

Find the equation of the parabola whose axis is horizontal, vertex on the y-axis and which passes through (2,4) and (8,-2).

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  1. the general formula for a parabola with known vertex and horizontal axis is
    x = a(y-k)^2 + h, where (h,k) is the vertex

    in your case the vertex , being on the y-axis , can be called (0,k) , and the equation would be
    x = a(y - k)^2 + 0
    for the point (2,4)
    2 = a(4-k)^2 (#1)

    for the point (8,-2)
    8 = a(-2-k)^2 (#2)

    divide #2 by #1
    4 = (-2-k)^2 / (4-k)^2
    4(4-k)^2 = (-2-k)^2
    64 - 32k + 4k^2 = 4 + 4k + k^2
    3k^2 -36 + 60 = 0
    k^2 - 12 + 20 = 0
    (k-10)(k-2) = 0
    k = 10 or k = 2

    if k=2, x = a(y-2)^2
    and using (2,4)
    2 = a(4-2)^2
    a = 1/2 ---------> x = (1/2)(y-2)^2

    if k=10 , x = a(y-10)^2
    again using (2,4)
    2 = a(4-10)^2
    a = 1/18 ----------> x= (1/18)(y-10)^2

    Notice that there are two possible equations,
    btw, I checked both equations using the point(8,-2), it also satisfies both equations.

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