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Math Help - Parabola, vertex-focus distance

  1. #1
    Junior Member Greymalkin's Avatar
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    Parabola, vertex-focus distance

    A spotlight has a parabolic cross section that is 4 ft wide at the opening and 1.5 ft deep at the vertex, how far is the vertex from the focus?

    I'm quite confused as my textbook gives no illustration, im assuming that the light is 4ft wide and 1.5ft deep from the opening/plexiglass.
    Saying the light is on a cartesian coordinate system with vertex (0,0) gives me no clue as to how this question is solved.

    Answer is 2/3, how did you get such an answer??
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  2. #2
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    Re: Parabola, vertex-focus distance

    Hello, Greymalkin!

    You are expected to know this form of a parabola: . x^2 \:=\:4py
    . . where p is the distance from the vertex to the focus.


    A spotlight has a parabolic cross section that is 4 ft wide at the opening and 1.5 ft deep at the vertex.
    How far is the vertex from the focus?

    Code:
                    |
     (-2,1.5)       |         (2,1.5)
        * - - - - - + - - - - - *
                    |
         *          |          *
          *         |         *
           *        |        *
             *      |      *
                *   |   *
      --------------*--------------
                    |
    We have the equation: . x^2 \:=\:4py

    We are told that when x = 2,\:y =\tfrac{3}{2}

    Hence: . 2^2 \:=\:4p\left(\tfrac{3}{2}\right) \quad\Rightarrow\quad 4 \:=\:6p

    Therefore: . p \,=\,\tfrac{2}{3}
    Thanks from Greymalkin
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