If one of the remaining basal points be moved up to join the seven coincident ones, one of these fixed lines becomes indeterminate, and the system of cubics through eight coincident points consists of a fixed line through the eight coincident points and the ninth fixed point together with any two lines of the pencil through the eight points. If the nine basal points coincide, any three lines through it form a cubic of the system. UNICURSAL QUARTICS. The inverse of a conic from any point not on the curve is a nodal bicircular quartic. This is shown by inverting the general equation of the conic ax² + 2hxy + by² + 2gx + 2fy + c = 0; x y by substituting for x and y, ————————— and —————————— , (x² + y²) (x² + y²) we get the equation ax² + 2hxy + by² + 2(gx + fy)(x² + y²) + c(x² + y²)² = 0. The origin is evidently a double point on the curve, and is a crunode, acnode, or cusp according as the conic is a hyperbola, ellipse, or parabola. The factors of the terms of the fourth degree, viz: (x + iy) (x + iy) (x - iy) (x - iy), show that the two imaginary circular points at infinity are double points on the quartic, which is thus trinodal. Hence this nodal, bicircular quartic can be projected into the most general form of the trinodal quartic. Trinodal quartics are unicursal. If the conic which we invert be a parabola, the quartic has two nodes and one cusp. If the conic be inverted from a focus, the quartic has the two circular points at infinity for cusps. This is best shown analytically as follows: let the equation of the conic, origin being at the focus, be written x² y² 2aex b² ——— + ——— + ———— - ——— = 0. a² b² a² a² Inverting this we have x² y² 2aex(x² + y²) b²(x² + y²)² ——— + ——— + ——————————————— - ———————————————— = 0. a² b² a² a² Now transform this equation so that the lines joining the origin to the circular points at infinity shall be the axes of reference. To do this let x + iy = x₁ and x - iy = y₁; x₁ + y₁ x₁ - y₁ ∴ x = ———————— and y = ———————— . 2 2i. Making these substitutions and reducing we have (dropping the subscripts), (x² + 2xy + y²) (x² - 2xy + y²) 4aexy(x + y) b²x²y² ————————————————— - ————————————————— - ———————————— - ————————— = 0. a² b² a² a² Making this equation homogeneous by means of z, we have ┌ ┐ │ (x² + 2xy + y²) (x² - 2xy + y²) │ 4aexyz(x + y) b²x²y² z² │ ———————————————— - ———————————————— │ - ————————————— - ————————— └ a² b² ┘ a² a²