Implementing the method I sketched earlier for a general one-sheet hyperboloid in canonical position x^2/a^2 + y^2/b^2 - z^2/c^2 = w^2 makes it easy to locate lines missing from reguli using symmetry. Four lines from each regulus are face diagonals of the cuboid with corners [+/- a, +/- b, +/- c], and base planes for the correspondence between pencils through two of them can be selected from corner-slices +/- x/a +/- y/b +/- z/c = 1 . The result is gratifyingly elegant: a general point P(s, t) on the surface has homogeneous coordinate vector [w,x,y,z] = [s*t+1, -a*(s+t), b*(s*t-1), c*(-s+t)] . It is trivial to check that P satisfies the (homogeneous) quadric equation; and being linear in s and t separately, P describes a line when either parameter is fixed. A missing subset of measure zero is reached by letting s -> oo or t -> oo . Showing their line intersection sweeping out the surface as the two moving planes rotate about their pencil axes in synchrony might make a worthwhile demonstration? Fred Lunnon On 11/23/11, James Propp <jamespropp@gmail.com> wrote:
Thanks!
It occurs to me to ask: Is there a parametrization of the one-sheeted hyperboloid that makes it manifest that it's a ruled surface?
And: Is there a parametrization that makes it manifest that it's a doubly-ruled surface?
Jim