Posted by Hamza on Saturday, October 6, 2012 at 9:23pm.
Suppose we draw the cables on an x-y axis with the road as the x-axis.
and the posts at (-213.5,0) and (213.5,0) , with their tops at (-213.5,50) and (213.5, 50)
That would make (0,20) the vertex of the parabola
the equation using y = a(x-p)^2 + q , where (p,q) is the vertex, would be
y = a(x-0)^2 + 20
but (213.5, 0) lies on it
0 = a(213.5)^2 + 20
a = -20/213.5^2 = appr - .000439
a model equation could be
y = -.000439x^2 + 20
BTW, having crossed that bridge several times, I would have noticed that the suspended road is actually curved, and the cables are not a parabolas but rather a "Catenary"
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