A Generalization for Conic 1.1

On Any Five Points There Is Exactly One Conic

To prove this we need to show that it wouldn't have mattered what points on the conic we picked. This reduces to showing that given four points A, B, C, D on a conic, the cross ratios (PA, PB; PC, PD) and (QA, QB; QC, QD) are equal for any points P and Q on the conic. In other words, we need to prove that

sin(APC)   sin(BPC)   sin (AQC)   sin(BQC)
-------- : -------- = --------- : --------
sin(APD)   sin(BPD)   sin (AQD)   sin(BQD)

In the case of a circle, this is clear: LAPC= LAQC, and so sin(APC)=sin(AQC); the two sides of the equation are identical. Moreover, once we establish that any conic is projectively equivalent to a circle, this proof will suffice. (There is a proof of this which does not make recourse to affine or Euclidean concepts; however, it is much less intuitive.)

We also need to exclude some collinearities and coincidences from our initial set of five points or five lines. It is still the case that any set of five points determines a unique point conic, and that any set of five lines determines a unique line conic. However, we can only say that any five points no three of which are collinear will determine a line conic, and similarly that any five lines no three of which are coincident will determine a point conic.

It is worth considering the point conic that arises when we take five points (say ABCDE) three of which (say ABC) are collinear. Then our conic will be two intersecting lines, ABC and DE. In affine terms, this is just a degenerate hyperbola.

Rather stranger is the line conic that arises when we take five lines three of which are coincident. This is just the dual of the previous case, so our conic is one line lying on two points.