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. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Asked by Pragya Singh | 1 year ago | 122

We know that the vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y –axis.

Diagrammatic representation is as follows:

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadways, 18m from the middle.

So, AB = 30m, OC = 6m, and BC = 50m.

The equation of the parabola is of the from x^{2} = 4ay (as it is opening upwards).

The coordinates of point A are (50, 30 -6) = (50, 24)

Since, A(50, 24) is a point on the parabola.

y^{2} = 4ax

(50)^{2} = 4a(24)

a = \( \dfrac{ (50×50)}{(4×24)}\)

= \( \dfrac{625}{24}\)

Equation of the parabola, x^{2} = 4ay

= 4×\(
(\dfrac{625}{24})y\) or 6x^{2} = 625y

The x – coordinate of point D is 18.

Hence, at x = 18,

6(18)^{2} = 625y

y = \( \dfrac{(6×18×18)}{625}\)

= 3.11(approx.)

Thus, DE = 3.11 m

DF = DE +EF = 3.11m +6m = 9.11m

Hence, the length of the supporting wire attached to the roadway 18m from the middle is approximately 9.11m

Answered by Abhisek | 1 year agoAn equilateral triangle is inscribed in the parabola y^{2} = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

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