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April 18, 2014

April 18, 2014

Posted by **Hamza** on Saturday, October 6, 2012 at 9:23pm.

- Math -
**Reiny**, Sunday, October 7, 2012 at 1:05amSuppose 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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