Given the cable of a uniformly loaded suspension bridge hangs in the form of a parabola. 

The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m.

The longest wire being 30 m and the shortest being 6 m.

This is shown in the above figure.

Now from the figure,

BC and OA are the longest and shortest wires respectively and 

DE is the supporting wire attached to the roadway 18 m from the middle.

Now OC = 50 m, BC = 30, OA = 6

Let DE = h

Coordinate of B is (50,30) and coordinate of D is (18,h).

Vertex of the parabola is A(0,6)

Now the equation of the parabola is

(x-0)² = 4a*(y-6)

=> x² = 4a(y-6) .............1

Since point B (50,30) lies on the parabola, then from equation 1 

(50)² = 4a(30-6)

=> 2500 = 4a*24

=> 4a = 2500/24

=> 4a = 625/6

Put this value in equation 1, we get

x² = (625/6)*(y-6)

Again D (18,h) also lies on the parabola, 

(18)² = (625/6)*(h-6)

=> 324 = (625/6)*(h-6)

=> h - 6 = (324*6)/625

=> h - 6 = 1944/625

=> h - 6 = 3.11

=> h = 6 + 3.11

=> h = 9.11

So the length of a supporting wire attached to the roadway 18 m from the middle is 9.11 m