SOLUTION: A cable suspended between two posts that are the same height and 20 meters apart has a sag of 1 1/2 meter. If the cable hangs in the form of a parabola, find its equation, taking t

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A cable suspended between two posts that are the same height and 20 meters apart has a sag of 1 1/2 meter. If the cable hangs in the form of a parabola, find its equation, taking t      Log On


   



Question 1184415: A cable suspended between two posts that are the same height and 20 meters apart has a sag of 1 1/2 meter. If the cable hangs in the form of a parabola, find its equation, taking the lowest point as the origin.
Answer by Alan3354(69443) About Me  (Show Source):
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A cable suspended between two posts that are the same height and 20 meters apart has a sag of 1 1/2 meter. If the cable hangs in the form of a parabola, find its equation, taking the lowest point as the origin.
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3 points on the parabola are:
(-10,1.5), (0,0), (10,1.5)
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A parabola has the form y = ax^2 + bx + c
Sub the values and solve for a, b & c.
Do the (0,0) first:
0 = a*0 + b*0 + c ---> c = 0
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For (-10,1.5):
1.5 = a*100 - 10b
For (10,1.5):
1.5 = a*100 + 10b
1.5 = a*100 - 10b
--------------------- Subtract
0 = 20b
b = 0
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1.5 = a*100 + 10b
1.5 = a*100 - 10b
-----------------------Add
3 = 200a
a = 3/200
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y = (3/200)x^2 is the parabola