The Parametrisation Theorem
If the point P lies on the conic C0 : xy+yz+zx = 0, and P ≠ [0,0,1],
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] = [0,0,1].

Hence, if P ≠ [0,0,1], 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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