## The quadratic function in real world's problem

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
honest_denverco09
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### The quadratic function in real world's problem

I have a problem as :

The cable is hang in the shape of parallel parabola by two pillars which are erected on the ground.
The distance between two pillar is 160 ft. And these two pillars have the same heights are 75 ft.
The vertical support cable, labeled a - k, are equally spaced, and the center of the parabolic cable touches the ground at f.

What is the length of each support cable, a - k ?

Annotation : a = k = 52, b = j = 33, c = i = 19, d = h = 8, e = g = 2, f = 0 (The answer are rounded)

stapel_eliz
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The cable is hang in the shape of parallel parabola by two pillars which are erected on the ground. The distance between two pillar is 160 ft. And these two pillars have the same heights are 75 ft. The vertical support cable, labeled a - k, are equally spaced, and the center of the parabolic cable touches the ground at f.
I'm afraid I don't follow the lettering and description entirely...? (For instance, how do cables support another cable? Shouldn't the cable be hung from posts or other rigid supports? Etc.) In what follows, I will be working under the assumption that the cable in question is actually attached to pillars, rather than to other cables, and that the attachment points are seventy-five feet above the ground and eighty feet to either side of the center of the curve.

For a general description of how to find a parabola's equation from information you've been provided, try here. Once you've studied that, the following should make some sense:

Center the parabola on the y-axis, for simplicity's sake, so the vertex is of the form (h, k) = (0, k). Also, since the parabola is right-side-up, and since it touches the ground in the center, then the vertex is at the origin, so (h, k) = (0, 0).

You are given two points on the parabola, being the attachment points. Their heights are at y = 75, and their x-values, being equidistant from the center, are x = -80 and x = +80. Since the vertex form of the equation gives you y - 0 = a(x - 0)2, or y = ax2, you can then plug either of these points into the equation, and solve for "a".

Once you have the value of "a", you can state the equation of the curve.