The basic hyperboloid of two sheets is given by the equation $$-\frac{x^2}{A^2}-\frac{y^2}{B^2} + \frac{z^2}{C^2} = 1$$ The hyperboloid of two sheets looks an awful lot like two (elliptic) paraboloids facing each other. It’s a complicated surface, mainly because it comes in two pieces. All of its vertical cross sections exist — and are hyperbolas — but there’s a problem with the horizontal cross sections.
This is evident in the picture below, where you can see the cross sections of $$-x^2 - y^2 + z^2 = 1$$ The horizontal cross sections are generally circles, except that there are no horizontal cross sections when \(z\) is between \(-1\) and \(1\). To see why this happens, look at the equation above. Suppose \(z=0\). Then the left hand side is definitely negative, but the right hand side is definitely positive. There’s no way to fix this, so the cross section simply doesn’t exist!
Here’s a hint about telling the two kinds of hyperboloids apart: look at the cross sections \(x=0\), \(y=0\), and \(z=0\). If they exist, then it’s a hyperboloid of one sheet. (Go back to that page and convince yourself that its cross sections all exist.) If you end up with something negative equal to something positive, then you’ve got a two-sheeter.
By now you probably expect that larger values of \(A\), \(B\), and \(C\) make for a much steeper surface, right? Wrong! It’s true that making \(C\) larger will have a dramatic effect on the surface, but use the second picture to find out what happens when you increase \(A\) and \(B\). Surprised?