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Parabola problem

  1. Sep 5, 2005 #1

    I have a math problem that does not seem to work out. It is a word problem about a suspension bridge. The bridge's main cable is in the shape of a parabola, and the entire bridge, the road, is 200 ft long. The ends of the bridge, the longest virtical cables on the bridge, are 50 ft high. the lowest point in the main cable is 10 ft off of the road with a virtical cable coming straight down. there are 4 additional cables inbetween the middle cable, and the longest cable on either side. i have to figure out the lengths of the virtical cables. i have put this diagram onto a graph, where the lowest part of the main cable is at the origin, the road is at (0,-10) as the directrix, and the focus would then be at the point (0,10). the 4 cables are evenly spaced out every 20 ft along the length of the bridge. since i know that the length from the focus to a point on the parabola is the same distance as that same point to the directrix, i used the distance formula and put the two distances equal to each other.

    (A) sqrt ((X2-X1)^2 + (Y2-Y1)^2)
    (X2,Y2)= (20,?) (X1, Y1)= (0,10) or the focus

    sqrt ((20-0)^2 + (?-10)^2)

    sqrt ((20)^2 + (?^2 -20? +100))

    sqrt (400 + ?^2 -20? +100)

    sqrt (?^2 -20? +500)

    (B) sqrt ((X2-X1)^2 + (Y2-Y1)^2)
    (X2,Y2)= (20,?) (X1, Y1)= (20,-10) or the directrix

    sqrt ((20-20)^2 + (?+10)^2)

    sqrt ((0)^2 + (?^2+20?+100))

    sqrt (?^2 +20? +100)

    (A=B) sqrt (?^2 -20? +500) = sqrt (?^2 +20? +100)

    (?^2 -20? +500) = (?^2 +20? +100)

    400 = 40?
    10 = ? so the distance from the directrix to the top of the first virtical cable is 20 ft since the top is at point (20,10)

    however, this does not work for the rest of the cables. the second cable, if im doing my math correctly, is 40 ft tall and the fifth cable that should b 50 ft tall is over 111 ft...

    How can i solve this?
  2. jcsd
  3. Sep 6, 2005 #2
    I would have done it differently, and left out the whole directrix, focus stuff. I would imagine the tallest cable which is 50 ft tall is at the origin so that its coordinate is (0,50)
    I would then find the h, and k value by the given information. I would obtain (100,10) since the cables are 20 ft apart and the K value is given. I would then just find the value for (a) by substituting in the point(0,50), and the h,k values (100,10). By the way I use f(x)=a(x-h)^2+k. (those are the letters here I don't know if they are different in other places) So after finding the a value you could then write the function rule. I got f(x)=(1/250)(x-100)^2+10. After having this useful little thingy I would then solve for the different Y values along each x interval of 20 ft by setting the y value to be 20 at first and then 40, 60, 80, 100

    Anyways I got these lengths for the cables.

    First major one (given) 50 ft
    second: 35.6 ft
    third: 24.4 ft
    fourth: 16.4
    fith: 11.6
    sixth:(given) 10

    Hey uhh I don't even know if this is right by the way. You better wait until a professional helps you before you submit your answers or drill something in your head. I may be too tired and forgot something or something.

    I had trouble understanding the part about the main cable I took it as if it were in the vertex and that it was 10 ft long.
    Last edited: Sep 6, 2005
  4. Sep 6, 2005 #3


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    Set up a coordinate system so that the vertex of the parabola is at x= 0. The "ends of the bridge", where the parabola is highest, will be at +100 and -100. Since the cables are 50 ft at those points, you have y(100)= 50 and y(-100)= 50. At the center of the bridge the cable is only 10 ft high so y(0)= 10. That is enough to get the coefficients in the general expression for a parabola, y= ax2+ bx+ c.
    Actually, because you chose x=0 at the vertex, by symmetry, you have y= ax2+ c and only need y(0)= 10, y(100)= 50 to determine a and c.

    Once you have the equation, just find y(20), y(40), y(60), and y(80).
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