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Ellipses question

  1. Aug 1, 2008 #1
    1. The problem statement, all variables and given/known data

    A bridge is built in the shape of a semielliptical arch. It has a span of 118 feet. The height of the arch 25 feet from the center is to be 8 feet. Find the height of the arch at its center?

    2. Relevant equations

    not sure if the 25 feet from the center is the focal axis or not?

    3. The attempt at a solution

    with the given info i know that i have the variable a in the equation of an ellipse:

    (x^2/a^2) + (y^2/b^2) = 1 (a>b)

    and i know that i am looking for the minor axis or semiminor axis to be exact. The Foci:

    (+ or - c,0) where c^2 = a^2 - b^2 I have, a, which is the span 118/2 = 59 for the semimajor axis. i believe i have, c, which is 25, but when i plug them in and solve for, b, i do not get the right answer. I believe i may need to do something with the 8 feet, but i have not seen it. If anyone could help point me in the right direction, I would be greatly appreciated.

  2. jcsd
  3. Aug 1, 2008 #2
    I think the statement "The height of the arch 25 feet from the center is to be 8 feet" means that at [tex]x=\pm{25}[/tex], [tex]y=8[/tex] giving the coordinates [tex](\pm{25},8)[/tex]. So try putting those coodinates into the equation [tex]{\frac{x^2}{25^2}}+{\frac{y^2}{b^2}}=1[/tex] and solve for [tex]b^2[/tex].
  4. Aug 1, 2008 #3


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    Staff Emeritus
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    Was that a typo? We are told that the span is 118 feet and you haven't used that. Put x= 25, y= 8 into
    [tex]\frac{x^2}{118^2}+ \frac{y^2}{b^2}= 1[/itex]
    and solve for b.
  5. Aug 1, 2008 #4
    Oh yeah whoops :S But span=2a so a=59, so I believe the equation is [tex]\frac{x^2}{59^2}+ \frac{y^2}{b^2}= 1[/tex].
  6. Aug 2, 2008 #5
    yes, thank you for your quick replies, when i put the values in for x,y and a, (59) I get:

    (118 * sqr root(714)) / 357 ---> which breaks into a cool +/- 8.83. which what do you

    know, is exactly the right answer :) Thanks again for the help with the problem, I was thinking I had to use the 8 somewhere, once again thanks for all your help, and steering me in a right direction.
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