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| 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)
      (Show Source): 
You can put this solution on YOUR website! 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. ---------------------
 3 points on the parabola are:
 (-10,1.5), (0,0), (10,1.5)
 ---
 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
 ----
 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
 ==================
 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
 
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