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Math Help - Need some urgent help please. Thanks

  1. #1
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    Need some urgent help please. Thanks

    Determine an equation for the parabola with focus (3,0) and directrix x=7. Sketch

    Anyone help me out? Much appreciated. Thanks
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  2. #2
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    Quote Originally Posted by Hockey_Guy14 View Post
    Determine an equation for the parabola with focus (3,0) and directrix x=7. Sketch

    Anyone help me out? Much appreciated. Thanks
    The Focus point F and the Directrix line D are located in such a way that any point P of the parabola is the same distance from the focus point F as it is from the directrix line D.

    The vertex of the parabola lies directly between the directrix and the focus point. In our case, it is at: (5, 0)

    The distance between the graph and both the focus and directix is always equal, so:

    (x - 3)^2 + y^2 = (x - 7)^2 + (y - y)^2

    x^2 - 6x + 9 + y^2 = x^2 - 14x + 49

    y^2 = -8x + 40

    y = \sqrt{40 - 8x}
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  3. #3
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    Hello, Hockey_Guy14!

    Determine an equation for the parabola with focus F(3,0)
    and directrix x=7.

    There are two types of parabola:

    . . \begin{array}{cccc}\text{Vertical:} & (x-h)^2 \:=\:4p(y-k) \\ \text{Horizontal:} & (y-k)^2 \:=\:4p(x-h) \end{array}

    where (h,k) is the vertex
    and p is the distance from the vertex to the focus.
    . . (It is also the distance from the vertex to the directrix.)

    Graph the focus and directrix.
    Code:
              * |
                |     *           :
                |         *       :
                |     F     V     :
            - - + - - o - - * - - : -
                |     3     5     :
                |         *       :
                |     *          x=7
              * |

    The vertex is midway between the focus and the directrix: . V(5,0)
    . . And we see that: . p = 2

    The parabola always "bends around" the focus
    . . and "away from" the directrix.
    This parabola opens to the left.
    . Its equation is: . (y-k)^2 \:=\:-4p(x-h)


    Therefore, the equation is: . (y-0)^2 \:=\:-4(2)(x-5)\quad\Rightarrow\quad y^2 \:=\:-8(x-5)

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