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K1232

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X(4), X(6), X(15), X(16), X(20), X(3068), X(3069), X(6560), X(6561)

X(43769) → X(43802)

other points below

Geometric properties :

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

Its KHO-equation is (1) : x^2 (2y + 3z) - 3z (2y + z)^2 = 0 or (2) : 12 z^3 - (2y + 3z) [x^2 - 3z (2y - z)] = 0.

X(20) is a node with two real nodal tangents and polar conic 2 x^2 - 3 (2y + z)^2 = 0.

X(6) is a point of inflexion with tangent passing through X(3146) and harmonic polar the Euler line. Note that the polar conic of X(3146) splits into the lines {6,20} and {6,3091}. This latter line meets the cubic again at Q3, Q4 with KHO-coordinates (±4√2,3,2).

X(4) is a sextactic point with tangent passing through X(6) and sextactic conic the Kiepert hyperbola.

K1232 meets the line {4,15} again at Q1 = (6,-5,6) and the line {4,16} again at Q2 = (6,5,-6). These points also lie on the lines passing through X(20) and X(398), X(397) respectively.

Parametrization :

For any real (sometimes complex) number t or infinity (giving X4), the KHO-point P(t) = (2 (3 t^2 - 2), 3 t (t^2 - 1), 2 t) lies on K1232. 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.

Generalization :

For every real number t, the KHO-cubic with equation

(1) : x^2 [2 y + (1 - 4 t) z] - 12 z (y - t z)^2 = 0 or (2) : [2 y + (1 - 4 t) z] [x^2 - 3z (2 y - z)] - 3 (2 t - 1)^2 z^3 = 0

is a crunodal cubic with

• node N = (0,t,1) on the Euler line,

• point of inflexion X(6) with inflexional tangent 2 y + (1 - 4 t) z = 0,

• sextactic point X(4) with tangent passing through X(6) and sextactic conic the Kiepert hyperbola.

When t = 1/2, it splits into the line GK and the Kiepert hyperbola.

When t = -1/2, it is K1232 and when t = 0, it is K1233.

When t = -1, the node is X(30) and the cubic has two real asymptotes parallel at X(11485), X(11486) to the Euler line.

When t = -1/4, the inflexional tangent is parallel to the Euler line, the node is X(376) and the cubic contains X(395), X(396).

When t =1/4, the inflexional tangent is the Brocard axis, the node is X(631) and the cubic contains X(590), X(615), these two points on the line GK and on the Evans conic.