Question

Question 3 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.

Class X1 - Maths -Conic Sections Page 264

Answer 1

see the attachment,
Let distance between focus to point P is K (see in y-axis)
Let the equation of parabola is in the form of
x² = 4ay

Here, you can see in attachment , focus is at the middle of the cable and shortest and longest vertical support are 6m and roadway in 100 m long.
here it is clear that Q ( 50, 24) will satisfy equation of parabola
${x}^{2} = 4ay$
so, ( 50)² = 4a × 24
$a = \frac{2500}{96}$
hence, equation of parabola is
${x}^{2} = \frac{2500}{24}y$
now ,
P(18,k) also will satisfy equation of parabola so,
${18}^{2} = \frac{2500}{24} \times k \\ 324 = \frac{2500}{24}k \\ k = \frac{324 \times 24}{2500} \\ = \frac{1944}{625} = 3.11$
now, required length of supporting wire attached to the roadway 18m from the middle = 6 + k = 6 + 3.11 = 9.11 m

Answer 2

Answer:

The length of the supporting wire is 9.11 m .

Step-by-step explanation:

It is given here that in a uniform suspension bridge, the cable hangs in the form of a parabola .

The roadway is 100 m long horizontally.

First , let us find what type of equation this parabola will have.

If we suppose that this lies on a cartesian plane, the cable is facing upwards along the vertical axis.

Thus, the equation of the parabola becomes x^2 = 4ay

Now, as the cable is symmetrical along the axis with 50m length on each side.

The longest and shortest vertical supporting wires are 30m and 6m respectively.

If these wires are connected from opposite sides, the vertical extent of the suspension becomes 24 m.

Substituting these values , as they lie on the parabola :

50 * 50 = 4 a * 24

> a = 2500/96

Now we need to find the length of a supporting wire attached 18m from the middle.

Let this length of the wire be x meters.

So, the point , 18 and x - 6 lies on the parabola

18 * 18 = 4 * a * ( x - 6 )^2

> 18 * 18 = 4 * 2500/96 * (x - 6)^2

Solving we get x = 9.11 m .

This is the required answer.

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