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Math Help - ParaBola Help please

  1. #1
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    ParaBola Help please

    I nee help the help in finding vertex, focus, and directrix of the follwoing parabola,

    3x+y^2+8y+4=0

    I get -52/3 vertex, -4,-52/3 and directrix -100/3/ I am not gettind right according to my teacher and the book. Am I even close, Thanks for any answers, help
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  2. #2
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    Quote Originally Posted by creatureman View Post
    I nee help the help in finding vertex, focus, and directrix of the follwoing parabola,

    3x+y^2+8y+4=0

    I get -52/3 vertex, -4,-52/3 and directrix -100/3/ I am not gettind right according to my teacher and the book. Am I even close, Thanks for any answers, help
    3x + y^2 + 8y + 4 = 0

    3x + 4 = -(y^2 + 8y)

    Now complete the square:
    3x + 4 = -(y^2 + 8y + 16 - 16)

    3x + 4 = -(y^2 + 8y + 16) + 16

    3x - 12 = -(y + 4)^2

    Now factor out a 3.

    3(x - 4) = -(y + 4)^2

    Now write this as
    4*(3/4)(x - 4) = -(y + 4)^2

    Compare this to the form of a leftward opening parabola with a vertex at (h, k):
    4p(x - h) = -(y - k)^2
    where p is the distance from the vertex to the focus, and the distance from the focus to the directrix.

    So p = 3/4, and the vertex is at (4, -4).

    The focus is a point p = 3/4 units to the left of the vertex:
    (4 - 3/4, -4) = (13/4, -4)

    And the directrix is a vertical line p = 3/4 units to the right of this point:
    x = 4 + 3/4 = 19/4

    -Dan
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