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Question 398068: Hi! Im studying for an exam, im having a problem with this problem, i tried to input the values in the formula(x^2=4py) but cant get the answer.
The problem is:
the distance between two towers is 150m, the points of support of the cable on the towers are 22m above the roadway, and the lowest point on the cable is 7m above the roadway. Find the vertical distance to the cable from a point in the roadway 15m from the foot of a tower.
Can you show me how to get the ordinate of the focal parameter being asked? I tried for almost an hour but still cant get the correct answer. the answer stated in the book is 16.6m
THAN YOU VERY MUCH!!!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! The problem is:
the distance between two towers is 150m, the points of support of the cable on the towers are 22m above the roadway, and the lowest point on the cable is 7m above the roadway. Find the vertical distance to the cable from a point in the roadway 15m from the foot of a tower.
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Assume the cables hang like a parabola. The problem, then, is to find the equation of this parabola.
Set the origin (0,0) on the surface of the roadway in the middle between the two towers. You now have (x,y) coordinates for 3 points of the parabola, (-75,22),(0,7), and (75,22). Standard form of a parabola: y=A(x-h)^2+k, (h,k) being the (x,y) coordinates of the vertex. (h,k) are the (x,y) coordinates of the vertex which is the lowest point (0,7) the cable hangs above the road. So,
now we know (h,k), all we need to find is A to get the equation of the parabola.
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y=A(x-h)^2+k
Using the (75,22)coordinates of one of the points of support on the tower,
22=A(75-0)^2+7
22-7=A(75)^2
A=15/(75)^2
equation of parabola
y=(15/(75)^2)(x-0)^2+7
at 15 feet from the tower, x=60
y=(15/(75)^2)(60)^2+7
=(60/75)^2*15)+7
=.64*15+7
=9.6+7=16.6m
ans:The vertical distance to the cable from a point in the roadway 15m from the foot of a tower=16.6 meters
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