too complicated to be written here. Click on the link to download a text file. on both curves : X(3), X(4), X(15), X(16) on K1206a : X(395), X(42147), X(42682), X(42685), X(42687), X(42688), X(42690), X(42692), X(42694) on K1206b : X(396), X(42148), X(42683), X(42684), X(42686), X(42689), X(42691), X(42693), X(42695) Geometric properties :
 K1206a and K1206b are two very similar cuspidal cubics with cusp H and point of inflexion O. See K1191 for explanations and also CL075. Their respective KHO-equations are : z^2 (4x - 15y) - 4x^3 = 0 or 15yz^2 + 4x (x + z) (x - z) = 0 z^2 (4x + 15y) - 4x^3 = 0 or 15yz^2 - 4x (x + z) (x - z) = 0 Parametrization : for any real number t or infinity, the pairs of KHO-points Pa(t) and Pb(t) lie on each respective cubic : Pa(t) = (15 t^2, 4 (t^2 - 25), ± 3 t^3) and Pb(t) = (15 t^2, -4 (t^2 - 25), ± 3 t^3). Two points in a pair are collinear with O. In both cases, three points Px(t), Px(t1), Px(t2) are collinear on K1206x if and only if (t1 + t2) t + t1 t2 = 0. Note that Px(∞) = O and Px(0) = H. Pa(t) and Pb(t) coincide when t = ± 5 corresponding to X(15) and X(16). The tangential of Px(t) is Qx(t) = Px(- t/2).