Let's create the equation of your cables using the following data
let the vertex be at (0,10)
two other points are (300,110) and (-300,110)
the equation must be
y = a(x-0)^2 + 10
= ax^2 + 10
but (300,110) lies on it
110 = a(90000) + 10
a = 100/90000 = 1/900
y = (1/900)x^2 + 10
so when x = 0 , y = 10
when x = 50 , y = (1/900)(2500)+10 = 12.77..
when x = 100 , y = (1/900)(100^2) + 10 = 21.111..
when x = 150 , y = (1/900)(150^2) + 10 = 35
when x = 200 , y = (1/900)(200^2)+10 = 54.44..
when x = 250 , y = (1/900)(250^2)+10 79.444..
when x = 300 , y = (1/900)(300^2)+10 = 110
because of the symmetry, the cables on the left would have the corresponding same lengths
BTW, the cables would not hang as a parabola, they would hang as a catenary with equation
y = (a/b)(e^x + e^-x)
If you have time for some extra reading:
https://www.google.ca/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=catenary%20vs%20parabola
The main cables of a suspension bridge uniformly distribute the weight of the bridge when in the form of a parabola. The main cables of a particular bridge are attached to towers that are 600 ft apart. The cables are attached to towers at a height of 110 ft above the road. The cables are 10 ft above ground at their lowest points. If vertical support cables are at 50 ft intervals along the level roadway, what are the lengths of these vertical cables?
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