The Parametrisation Theorem If the p-point P lies on the conic C0 : xy+yz+zx = 0, and P ≠ Z, then P has the form (t,1-t,t(t-1)) for some real t. Proof Suppose that P[x,y,z] is on C0, so that xy+yz+zx = 0. This can be rearranged as (x+y)z = -xy. If x+y = 0, then the rearranged equation shows that xy = 0, and so x or y = 0. But x+y = 0, so we must have x = y = 0, i.e. P = [0,0,z] = Z. Hence, if P ≠ Z, x+y ≠ 0. Since we can scale the coordinates in P by any non-zero factor, we may assume that x+y = 1. Then, if x = t, y = 1-t, and then, from the rearranged form of the equation, since x+y = 1, z = -t(1-t) = t(t-1), as required.

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