



 
Hi Elizabeth, This is the diagram I drew from your description. The parabola passes through $(0, 0)$ and there are 6 cables, 4 meters apart so the bridge is $4 \times 7 = 28$ metres long and hence the parabola passes through $(28, 0).$ The first cable is $3.072$ metres long and hence the parabola passes through $(4, 3.072).$ Substitute $(0, 0)$ and $(28, 0)$ into the equation of the parabolas, $(xh)^2=4a(yk)$ and you can then solve the resulting equations for $h.$ Do you see how to proceed from here? Penny  


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