$\begin {array} {1 1} (A)\;9.11 \: m & \quad (B)\;9\: m \\ (C)\;-9.11 \: m & \quad (D)\;-9\: m \end {array}$

- If the focus is on the positive y axis, then the parabola is open upwards, Hence the equation is $ x^2=4ay$

Step 1 :

The vertex is the lowest point of the cable.

Let us take the origin of the coordintes plane as the vertex of the parabola, so the vertical axis is along the positive y - axis.

This is as shown in the figure.

According to the fig, AB and 0C are the longest and shortest wires attached to the cable respectively. LM is the supporting wire attached to the roadway, 18 m away from the middle.

Given AB = 30 m, 0C = 6m and BC = $ \large\frac{100}{2}$$=50 m$

Step 2 :

The equation of the parabola is of the form $x^2=4ay$

The coordinates of the point A are (50, (30-6) ) (i.e) (50, 24)

Substituting this for x and y in the equation of the parabola,

$(50)^2=4a(24)$

$ \Rightarrow a = \large\frac{50 \times 50 }{4 \times 24}$$ = \large\frac{625}{24}$

Step 3 :

Hence equation of the parabola is

$ x^2 = 4 \bigg( \large\frac{625}{24} \bigg)$$y$

(.e) $x^2=\large\frac{625}{6}$$y$

$ \Rightarrow 6x^2=625 y$

Step 4 :

The x coordinates at point D is 18.

at x = 18

$6(18)^2 = 625 y$

$ \therefore y = \large\frac{6 \times 18 \times 18}{625}$

$ = 3.11 $ ( approx )

$ \therefore LP = 3.11 m$

But LM = LP + PM

= 3.11 + 6

$ = 9.11 m

Hence the length of the supporting wire attached to the roadway 18m from the middle is 9.11 m approx.

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