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?

- Go back to the equation and figure out why larger values of \(A\) and \(B\) make the hyperboloid flatter, not steeper.
- Does there always need to be a gap between the two sheets, or could they touch?
- Fix that \(A = B = C\) and suppose we make \(A\) very small. What other quadric surface is very similar to the resulting picture? Is it reasonable to say that the this family of hyperboloids “limit” to the other surface? Is there a way to justify this algebraically?