The cable of a suspension bridge hangs in the form of a parabola. The supporting towels are 500ft. Apart and 50ft high and the cable at its lowest point is 10ft above the road way using the axes of the parabola as x axes. Find the equation of the parabola . Hence the length of the vertical cable 60ft from the centre of the bridge collect the nearest 10th of a foot
y = ax^2 + 10
a*250^2 = 50
Now, knowing a, find y at x=60
Thank you Sir, for your response
@Pee
To find the equation of the parabola, we will use the vertex form of a parabolic equation: y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola, and a is a constant that determines the shape of the parabola.
Given that the supporting towers are 500ft apart and the cable at its lowest point is 10ft above the road, we can determine the vertex of the parabola. The x-coordinate of the vertex will be the midpoint between the two towers, which is (500/2) = 250ft. The y-coordinate of the vertex will be the lowest point of the cable, which is 10ft above the road. Therefore, the vertex is (250, 10).
Substituting the vertex into the vertex form equation gives us:
y = a(x - 250)^2 + 10.
Now let's find the value of a. We know that at one of the supporting towers, which is 500ft from the vertex, the height of the cable is 50ft. Substituting (500, 50) into the equation gives us:
50 = a(500 - 250)^2 + 10.
Simplifying the equation:
50 = a(250)^2 + 10.
Expanding and solving for a:
50 = 62500a + 10,
62500a = 40,
a ≈ 0.00064.
Now we can write the final equation of the parabola:
y = 0.00064(x - 250)^2 + 10.
To find the length of the vertical cable 60ft from the center of the bridge, we need to substitute x = ±60 into the equation to get the corresponding y-values. By symmetry, the heights on either side will be the same, so let's calculate for one side only.
Substituting x = 60 into the equation:
y = 0.00064(60 - 250)^2 + 10,
y ≈ 10.72ft.
Therefore, the length of the vertical cable 60ft from the center of the bridge is approximately 10.72ft (to the nearest 10th of a foot).