algebra 2

a parabola has a vertex v=(6,4) and a focus f=(6,1) what is the equation

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  1. from first principles, the directrix is y = 7
    let P(x,y) be any point
    then PF = PD, where D is the vertical distance from P to the directrix

    √( (x-6)^2 + (y-1)^2 ) = √(x-x)^2 + (y-7)^2 )
    square and expand
    x^2 - 12x + 4 + y^2 - 2y + 1 = y^2 - 14y + 49
    x^2 - 12x - 44 = -14y

    verification:
    http://www.wolframalpha.com/input/?i=x%5E2+-+12x+-+44+%3D+-14y

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  2. Or, knowing that the equation of a parabola with focus at y=(0,0) and focus at y = -p is

    x^2 = 4py

    we have a shifted parabola with p=-3, so the equation is

    (x-6)^2 = 4(-3)(y-4)
    x^2-12x+36 = -12y+48
    x^2-12x+84 = -12y

    Verify at

    http://www.wolframalpha.com/input/?i=parabola+%28x-6%29^2+%3D+4%28-3%29%28y-4%29

    Oops, Reiny - if you add the word parabola to your function at wolframalpha, you will see where the focus and vertex lie. Must have had a typo somewhere in the math.

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