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Question 80217: a cross section of a nuclear cooling tower is a hyperbola with equation x^2/90^2-y^2/130^2=1. The tower is 450 feet tall and the distance from the top of the tower to the center of the hyperbola is half the distance from the base of the tower to the center of the hyperbola. Fin the diameter of the top and base of the tower.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! a cross section of a nuclear cooling tower is a hyperbola with equation x^2/90^2-y^2/130^2=1. The tower is 450 feet tall and the distance from the top of the tower to the center of the hyperbola is half the distance from the base of the tower to the center of the hyperbola. Find the diameter of the top and base of the tower.
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Draw the picture on a coordinate system. Put the center of the hyperbola
at the origin.
The top of the tower is 150 ft above the center: the bottom of the tower
is 300 ft below the center.
To get the radius of the top of the tower, let y=150 and solve for x:
x^2/90^2 -150^2/130^2 = 1
x^2/90^2= 1+(150/130)^2
x^2/90^2 = 1+ 1.33
x^2=18884.02367
x-137.419
Diameter of the top = 2x=274.838 ft.
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To get the radius of the bottom, let y=-300 and solve for x:
I'll leave that to you.
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Cheers,
Stan H.
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