Finding the equation of a one sheeted hyperboloid Was working on calc 3 homework assignment and couldn't find out how to solve this question. the answer doesn't matter any more,i just really want to find out how to do it since my book seems to skip over it. the problem goes like this.
"A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. The diameter at the base is 280 m and the minimum diameter, 500 m above the base, is 220 m. Find an equation for the tower. (Assume the center is at the origin with axis the z-axis and the minimum diameter is at the center.)"
I know the equation of a hyperboloid is $\frac{x^2}{a^2} +\frac {y^2}{b^2} - \frac{z^2}{c^2} = 1$ but i don't know how to find this with the parameters given.
thanks.
 A: You are  meant to assume that the horizontal cross-sections are circles, so in your equation you can take $a=b$. Also you are given that the minimum diameter is 220, so provided we take the origin at that level (which they recommend), we will have $a=b=110$ (because then the four points $(\pm110,\pm110,0)$ will lie on the surface). Now your only problem is finding $c$. You are given that the cross-section at $z=-500$ is a circle diameter 140, so for example, the point (140,0,-500) lies on the surface. That is enough to find $c$.
A: Official Web Assign answer 
If we position the hyperboloid on coordinate axes so that it is centered at
the origin with axis the $z$-axis then its equation is given by 
$$\left(\frac xa\right)^2+\left(\frac yb\right)^2-\left(\frac zc\right)^2=1.$$ 
Horizontal traces in $z=k$ are 
$$\left(\frac xa\right)^2+\left(\frac yb\right)^2=1+\left(\frac kc\right)^2,$$ a family of ellipses, but we know that the traces are circles so we must have $a = b$. 
The trace in $z = 0$ is 
$$\left(\frac xa\right)^2+\left(\frac ya\right)^2=1$$ or $x^2+y^2=a^2$ 
and since the minimum radius of 110 m occurs there, we must have a = 110. The base of the tower is the trace in $z = −500$ given by $(x/a)^2+(y/b)^2=1+((-500)^2/c)^2$ but $a = 110$ so the trace is 
$$x^2+y^2=110^2+ \frac{55\,000^2}{c^2}.$$
We know the base is a circle of radius $140$, so we must have $110^2 +55,000^2/(c^2)= 140^2$ therefore $c^2=55,000^2/(140^2-110^2)=1210000/3$ and an equation for the tower is
$$\left(\frac x{110}\right)^2+\left(\frac y{110}\right)^2
-\frac{z^2}{1210000/3}=1\quad
\text{or}\;\; x^2+y^2-\frac{3z^2}{100}=1\quad   (-500\leq z\leq 500).$$
