Here is the way I know how to do it..

Consider the general form of the quadric: .

Denote and

You have with (so the quadric is without unique centre) and with (so the quadric in non-degenerated).

Hence, it can only be an elyptic paraboloid or a hyperbolic paraboloid.

The roots of the characteristic polynomial are . The sets of the respective eigenvectors are .

Denote . They are already orthogonalized, so we only normalize them: is a orthonormal basis in the next transformation is an isometry:

By that, the quadric becomes: , which is indeed a hyperbolic paraboloid, like you had noticed already

I guess there could be a simpler way though, but I don't see it now..