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too complicated to be written here. Click on the link to download a text file. 

X(6), X(53), X(216), X(393), X(1609), X(2165), X(14576) X(14576) = X(4) x X(52) 

Every pivotal cubic pK with pivot H and pole Ω is an isogonal pK with respect to the orthic triangle, its pivot being the tangential of H i.e. the Ωisoconjugate of H. When Ω lies on K627, this cubic is an isogonal pK with respect to a second triangle and the cubic is said to be a biisogonal pivotal cubic. Furthermore, this pK and its hessian generate a pencil of cubics which always contains a stelloid and a focal cubic which is the hessian of the stelloid. The radial center of the stelloid is the singular focus of the focal cubic. The pencil also contains two other K+ i.e. cubics with concurring asymptotes. With Ω = X(6), X(53), X(216), X(14576) we obtain K006, K049, K044, K415. The focal cubic associated to K006 is K1092. K627 is the barycentric product of H and K044 : for any point M on K044, the point H x M lies on K627. 
