# Thread: Suspension Bridge Parabola Problem

1. ## Suspension Bridge Parabola Problem

The cable of a suspension bridge hangs in the form of a parabola when the load is evenly distributed horizontally. The distance between the two towers is 150m, the points of support of the cable on the towers are 22m above the roadway, and the lowest point on the cable is 7m above the roadway. find the vertical distance to the cable from a point in te roadway 15m from the foot of the tower.

2. Set up a coordinate system so the x-axis is along the bridge roadway, the y-axis is vertical, and the origin, (0, 0), is half way between the two towers. Since the two towers are 150 m apart, they are at x= 75 and x= -75. The cable attaches to the towers 22 m above the roadway (y= 0) so the cable passes through the points (75, 22) and (-75, 22). Since the parabola is symmetric, we must have y= ax^2+ b. Also, the minimum is at (0, 7). Put x= 74, y= 22, and x= 0, y= 7 into equation to solve for a and b. Then, of course, evaluate y at x= 75- 15= 60.

3. The bridge 150m long, think of this as your X axis. The height of the tower is 22m, that's your Y axis. The lowest point drops down the middle of the bridge which is at (0,7) which is your vertex. That makes your domain (-75, +75) Which also means that at X=-75, Y=22. And at X=75, Y=22. I like to set this up as a vertex Y=a(x-h)^2+k Where h and k is the vertex. We have X,Y so we can do 22=a(75-0)^2+7, solve for a by subtracting 7, squaring 75, then dividing by 75 which will give you a. Then you want to know the what x will be when y is 15, so do 15=a(x-0)^2+7. a will be what you got from the last part, and just solve the equation Algebraically.

I'm pretty sure that's how you go about doing it. Please correct me if I am wrong!

-Nichole

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# the cable of a suspension bridge hangs in the form of a parabola

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