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Math Help - Parabola Searchlight

  1. #1
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    Parabola Searchlight

    A searchlight is shaped like a paraboloid of revolution. If the light source is located 2 feet from the base along the axis of symmetry and the depth of the searchlight is 4 feet, what should the width of the opening be?
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  2. #2
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    Hello, magentarita!

    A searchlight is shaped like a paraboloid of revolution.
    The light source is located 2 feet from the base along the axis of symmetry
    and the depth of the searchlight is 4 feet.
    What should the width of the opening be?
    Code:
                       4|
            * - - - - - + - - - - - *
                        |
             *          |          *
              *        2*         *
                *       |       *
                   *    |    *
          - - - - - - - * - - - - - - -
                        |

    We are expected to know this equation: . x^2 \:=\:4py
    . . where p is the distance from the focus to the vertex.

    Since p = 2, the equation is: . x^2 \:=\:8y

    When y = 4\!:\;\;x^2 \:=\:32\quad\Rightarrow\quad x \:=\:\pm4\sqrt{2}


    Therefore, the width of the opening is 8\sqrt{2} feet.

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  3. #3
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    ok....

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    Code:
                       4|
            * - - - - - + - - - - - *
                        |
             *          |          *
              *        2*         *
                *       |       *
                   *    |    *
          - - - - - - - * - - - - - - -
                        |
    We are expected to know this equation: . x^2 \:=\:4py
    . . where p is the distance from the focus to the vertex.

    Since p = 2, the equation is: . x^2 \:=\:8y

    When y = 4\!:\;\;x^2 \:=\:32\quad\Rightarrow\quad x \:=\:\pm4\sqrt{2}


    Therefore, the width of the opening is 8\sqrt{2} feet.
    Another great reply. How did you know what equation to use to solve this question?
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by magentarita View Post
    Another great reply. How did you know what equation to use to solve this question?
    An equation of a parabola with vertex at the origin and a vertical axis of symmetry is:

    x^2=4py where p is the distance from the focus to the vertex.

    See here.
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  5. #5
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    ok but.....

    Quote Originally Posted by masters View Post
    An equation of a parabola with vertex at the origin and a vertical axis of symmetry is:

    x^2=4py where p is the distance from the focus to the vertex.

    See here.
    Ok but I still would like to know how Soroban makes those pictures.
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by magentarita View Post
    Ok but I still would like to know how Soroban makes those pictures.
    I won't attempt to speak for Soroban, but I think he just uses the symbols on his keyboard to draw the objects and then puts "code" tags around it to preserve the spacing. I guess he just highlights his drawing and selects the # symbol at the top of the edit screen to put the tags on. That's how I do it when I want to make a table or something.
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