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Thread: Word problem on ellipses

  1. #1
    Junior Member
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    Mar 2008
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    Word problem on ellipses

    Hi I was wondering how you would set up a problem like this:

    An arch for a bridge over a highway is in the form of half an ellipse. The top of the arch is 20 feet above the ground level (the major axis). The highway has four lanes, each 12 feet wide; a center safety strip 8 feet wide; and two side strips, each 4 feet wide. What should the span of the bridge be (the length of its major axis) if the height 28 feet from the center is to be 13 feet?

    Any help would be greatly appreciated!

  2. #2
    Elite Member
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    Jun 2007
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    12,635

    Re: Word problem on ellipses

    Was there a picture with this problem?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    Junior Member
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    Mar 2008
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    87

    Re: Word problem on ellipses

    No there wasn't a picture of it, which made it harder to figure out.

  4. #4
    Elite Member
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    Jan 2005
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    Lexington, MA
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    Re: Word problem on ellipses

    Hello, Violagirl!

    An arch for a bridge over a highway is in the form of half an ellipse.
    The top of the arch is 20 feet above the ground level (the major axis).
    The highway has four lanes, each 12 feet wide; a center safety strip 8 feet wide;
    and two side strips, each 4 feet wide.
    .
    all unnecessary!

    What should the span of the bridge be (the length of its major axis)
    if the height 28 feet from the center is to be 13 feet?
    Code:
                  * * * * *
              *       |       *
           *          |          *
                    20|          | *
        *             |        13|  *
                      |          |
       *--------------+----------+---*
                           28

    [tex]\text{We have an ellipse with a semi-minor axis of 20.}[/tex]

    . . [tex]\text{Its equation is: }\;\frac{x^2}{a^2} + \frac{y^2}{20^2} \:=\:1[/tex]


    [tex]\text{The point }(28,13)\text{ is on the ellipse.}[/tex]

    . . [tex]\text{Hence: }\:\frac{28^2}{a^2} + \frac{13^2}{20^2} \:=\:1[/tex]

    [tex]\text{Solve for }a\!:\;\;a^2 \:=\:\frac{560^2}{231} \quad\Rightarrow\quad a \:=\:\frac{560}{\sqrt{231}} \:=\:36.84529492[/tex]


    [tex]\text{Therefore, the major axis is: }\:2a \;\approx\;73.7\text{ ft.}[/tex]


    I'm the other of the two guys who "do" homework.

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