Math
posted by Vanna .
Cooling towers for nuclear reactors are often constructed as hyperboloids of one sheet because of the structural stability of that surface. Suppose all horizontal cross sections are circular, with a minimum radius of 200 feet occurring at a height of 600 feet. The tower is to be 800 feet tall with a maximum crosssectional radius of 300 feet. Find the equation of the surface.

Math 
drwls
Make the axis of symmetry the y axis and the minimum cross section be the y = 0 plane. The general equation for the hyperbola that generates the surface is
x^2/a^2  y^2/b^2 = 1
where a is the minimum radial distance from the y axis, which in this case is 200 feet. To get b, require that x = 300 when y = 200 feet. (That is, y = 200 feet above the "throat" where x = 200.
(300/200)^2  (200/b)^2 = 1
(200/b)^2 = 1.25
b/200 = sqrt(4/5)
b = 178.9 ft
(x/200)^2  (y/178.9)^2 = 1
The domain of y is 600 to +200 
Math 
Vanna
However, the equation is for a hyperboloid of one sheet, not a hyperbola. I need to find a equation that follows the general equation of a hyperboloid of one sheet: ax^2 + by^2  cz^2 where a,b,c are > 0.

Math 
drwls
The hyperbola that I described, when rotated about the y axis, in an x,y,z coordinate system. becomes the hyperboloid of one sheet. I leave the rest up to you
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