   too complicated to be written here. Click on the link to download a text file.  X(6), X(53), X(141), X(14533), X(42007), X(46288), X(51539), X(51540), X(51542), X(51543), X(51544), X(51545), X(51546), X(51547) Y1, Y2, see Table 79 Y1, Y2 are now X(51539), X(51540) in ETC Geometric properties :   K1284 is a crunodal isocubic with pole X(32), node X(6) with nodal tangents passing through X(6) (Brocard axis) and X(25). If P, Q are two points on the Euler line, harmonically conjugated with respect to X(3) and X(25), then the lines KP and KQ meet the cubic again at two points P' and Q' which are X(32)-isoconjugated. It follows that the line P'Q' envelopes the pivotal-conic (C) which is inscribed in the tangential triangle (anticevian triangle of K), tangent at X(3053), X(154) to the nodal tangents X(6)X(3), X(6)X(25) respectively. A fairly simple parametrization of K1284 is given by S(t) = a^2(-a^2+t b^2+t c^2) / (-t a^2+b^2+c^2) : : , where t is a real number or ∞ or an expression of global degree 0 in a, b, c. Note that S(t) and S(1/t) are X(32)-isoconjugated. K1284 meets the sidelines of ABC at three collinear points A', B', C' which lie on the trilinear polar of X(1576), passing through X(i) for i in {32, 184, 211, 217, 237, 1186, 1501, 2206, 2387, 3051, 3117, 3118, 3203, etc}. This line is the X(32)-isoconjugate transform of the Kiepert hyperbola hence the tangents at A, B, C to both curves are the same. The contact-conic has perspector X(2451) and passes through X(i) for i in {64, 98, 1105, 1968, 1975, 1988, 3224, etc}. Its X(32)-isoconjugate transform is the line that passes through the points of inflexion of the cubic. This line is the polar line of K in (C). It contains X(i) for i in {154, 237, 682, 1613, 3053}. The isogonal transform of K1284 is K1285 = cK(#X2, X110). 