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Can you help solve this difficult analytic geometry problem?
Some nuclear power plants utilize "natural draft" cooling towers, in the shape of a hyperbaloid, a solid obtained by rotating a hyperbola about its conjugate axis suppose such. Suppose such a cooling tower has a base diameter of 400 ft, and the diameter at its narrowest point, 360 ft above the ground, is 200 ft. If the diameter at the top of the tower is 300 ft, how tall is the tower?
Please give detailed solution, thank you
The description of the tower gives us three points on the hyperbola: (200,0), (100,360), and (150,y) in terms of x = distance from the central axis and y = height above the ground. Note that the point (100,360) is the vertex of one side of the hyperbola, so we know that the center is (0,360).
For now, we know that the hyperbola opens left/right and we know the center. Equation starts off as:
[(x - 0)^2 ]/a^2 - [(y - 360)^2]/b^2 = 1
A little more analysis will tell us the values of a and b.
We know the point (100,360) is on the hyperbola so
100^2/a^2 - (360-360)^2/b^2 = 100^2/a^2 = 1 --> a^2 = 100 --> a = 10.
The point (200,0) is also on the hyperbola, so
[(200 - 0)^2 ]/100^2 - [(0 - 360)^2]/b^2 = 1
200^2/100^2 - 360^2/b^2 = 1
4 - 1 = 3 = 360^2/b^2 --> b^2 = 43200 --> b = 120sqrt(3)
Now we can plug in the point (150,y) to find the height.
[(150 - 0)^2 ]/100^2 - [(y - 360)^2]/[120sqrt(3)]^2 = 1
2.25 - [(y - 360)^2]/43200 = 1
140400 = (y - 360)^2
60sqrt(39) = 374.67 = y - 360
y = 734.67 ft or 734 ft 8 inches = height of the cooling tower
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