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K1235

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X(2), X(6), X(20), X(3070), X(3071), X(42942), X(42943), X(43465), X(43466), X(43511), X(43512)

X(43465), X(43466), Q1 = X(43511) and Q2 = X(43512) lie on the KHO-hyperbola, see preamble just before X(43457) in ETC and further details in page K1234

Geometric properties :

K1235 is a crunodal KHO-cubic, see K1191 for explanations and CL075. See also K1232, a very similar cubic.

Its KHO-equation is equivalently

(1) : 8 x^2 y - 3 (2y - z) (2y + z)^2 = 0,

(2) : 8 y (x^2 - 6 y z +3 z^2) - 3 (2y - z)^3 = 0, where x^2 - 6 y z +3 z^2 = 0 is the Kiepert hyperbola,

(3) : 8 y (x^2 + 2 y z - z^2) - (2y - z) (2y + 3z) (6y + z) = 0, where x^2 + 2 y z - z^2 = 0 is the KHO-hyperbola.

X(20) is a node with two real nodal tangents passing through the KHO-points (±√6,1,0) on the line HK.

X(6) is a point of inflexion with tangent the Brocard axis and harmonic polar the Euler line.

X(2) is a sextactic point with tangent passing through X(6) and sextactic conic the Kiepert hyperbola.

Parametrization :

For any real (sometimes complex) number t or infinity (giving X2), the KHO-point P(t) = (6 t^2 - 1, 3 t^3, 2 t (3 t^2 - 1)) lies on K1235. This gives a lot of simple points on the curve. Note that P(t), P(-t) and X(6) = P(0) are collinear for every t.

More generally, three points P(t1), P(t2),P(t3) are collinear on the cubic if and only if t1 + t2 + t3 + 6 t1 t2 t3 = 0.

Taking t3 = ∞, we see that P(t1), P(t2), X(2) are collinear if and only if 1 + 6 t1 t2 = 0.

Taking t1 = t2 = t, we see that the tangential of P(t) is P(T) where T = - 2t / (1 + 6 t^2).

It easily follows that P(t1) ≠ P(t2) share the same tangential if and only if 1 - 6 t1 t2 = 0.

Taking t1 = t2 = t3 = t, we obtain 3t (1 + 2 t^2) = 0, giving the three points of inflexion, namely X(6) = P(0) and two imaginary points (±8i√2,3,10).