math

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

asked by Miley
  1. 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)

    posted by Reiny

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