SOLUTION: A nuclear power plant has a large cooling tower with sides curved in the shape of a hyperbola. The radius of the base of the tower is 60 meters. The radius of the top of the tower
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Question 544483: A nuclear power plant has a large cooling tower with sides curved in the shape of a hyperbola. The radius of the base of the tower is 60 meters. The radius of the top of the tower is 50 meters. The sides of the tower are 80 meters apart at the closest point located 100 meters above the ground.
Find the equation of the hyperbola that describes the sides of the cooling tower. Assume that the center is at the origin. Determine the height of the tower.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
A nuclear power plant has a large cooling tower with sides curved in the shape of a hyperbola. The radius of the base of the tower is 60 meters. The radius of the top of the tower is 50 meters. The sides of the tower are 80 meters apart at the closest point located 100 meters above the ground.
Find the equation of the hyperbola that describes the sides of the cooling tower. Assume that the center is at the origin. Determine the height of the tower.
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Standard form of equation for hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
..
For given problem:
(h,k)=(0,0)
Equation becomes: x^2/a^2-y^2/b^2=1
Length of transverse axis=80 (distance at closest point)=2a
a=40
a^2=1600
..
Finding b, using coordinates of base (60,-100)
x^2/a^2-y^2/b^2=1
(60^2/40^2)-(100^2/b^2)=1
10000/b^2=(6/4)^2-1=2.25-1=1.25
b^2=10000/1.25=8000
..
Equation of hyperbola:
x^2/1600-y^2/8000=1
..
Finding height of tower:
plug in x=50 (radius of top of tower)
x^2/1600-y^2/8000=1
50^2/40^2-y^2/8000=1
1.5625-1=y^2/8000
.5625=y^2/8000
y^2=.5625*8000=4500
y=√4500≈67.08 meters above center
ans:
Height of tower≈67.08+100≈167.08 meters
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