Parabolic cable of a 60 m portion of the roadbed of a suspension bridge are positioned as shown below. Vertical Cables are to be spaced every 6m along this portion of the roadbed. Calculate the lengths of first two of these vertical cables from the vertex.

#### Solution

From the diagram,

The equation is x^{2} = 4 ay and it passes through C(30, 13)

Equation of Parabola x^{2} = 4ay.

30^{2} = 4a × 13

4a = `30^2/13`

∴ Equation of the parabola is

x^{2} = `30^2/13 y`

**(i)** Let VG = 6 and GE = y

∴ E is (6, y) and it lies on the parabola

36 = `30^2/13 y`

⇒ y = 0.52

Gable from the road = 3 + 0.52

= 3.52 m

**(ii)** Let VH = 12 and HF = y

∴ F(12, y) lies on the parabola

12^{2} = `30^2/13 y`

⇒ y = `(144 xx 13)/900`

= `208/100 y`

y = 2.08

Cable from the road = 3 + 2.08 = 5.08

The heights of the first two vertical cables from the vertex are 3.52 m and 5.08 m