56 KANSAS UNIVERSITY QUARTERLY. 



bics with seven coincident basal points. This system is made up of a 

 pencil of lines meeting in the seven coincident basal points together 

 with the two lines joining this to the other two basal points of the 

 system. These two lines are part of every cubic of the system. 



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 to- 

 gether 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-j-by2-|-2gx-(-2fy-|~c=o; 



-^ and -^— , 



by substituting for x and y, — ^— ^ — ^ and —^^ — - , we get the equation 



ax^ -(- 2hxy-j-by 2 -)- 2 (gx-|-fy) (x 2 -^y 2) _|_c(x2 -|-y 2) 2 ^o. 



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 par- 

 abola. The factors of the terms of the fourth degree, viz: (x-fiy)(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 parbola, 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 analyt- 

 ically as follows: let the equation of the conic, origin being at the fo- 

 cus, be written 



x^ y2 2aex b^ 

 7i"+ "bi + ^T" — ^ = o- 



Inverting this we have 



X2 y2 2aex(x2-^y2) b2(x2+y2)2 



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-f-iy=Xj and x — iy=yj ; < ' < x=: \ and y= -^—. — -. 



