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parabola with vertex at (3,-2) and a directreix of x=2. how do i wrtie this as an equation....PLEASE HELPPPPPPPP

  • math -

    Let's use basic concepts for this one
    let P(x,y) be any point on the parabola
    the focal point will be F(4,-2), (remember the vertex is midway between the focal point and the directrix)

    Then PF = distance from P to directrix

    √((x-4)^2+(y+2)^2) = √(x-2)^2
    square both sides and expand

    x^2 - 8x + 15 + y^2 + 4y + 4 = x^2 - 4x + 4

    y^2 + 4y + 20 = 4x

    from here you could complete the square to get
    x = 1/4(y+2)^2 + 3 to confirm that the vertex is (3,-2)

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