Consider x : y : 1 = t2 - 2t : t : t2 - t - 1. ........................ ( i)

The curve meets the line lx +my + n = 0 in points given by

l(t2 -2t) +mt +n(t2 -t-1) =0.

For any given Z, m, n there are two values of t. Hence the curve meets any line in two points and is therefore a conic. For a given l, m, n the values of t are the roots of the equation

t2 (l +n) + t(m -2l -n) -n =0.