Fkaskr — Reduction of a Qtiartic Surface to a Canonical Form. 79 



rejecting the factor C^, 



2aG + 2PF- 2AC + 2a/?C = ; 

 but 



= 0. 



- 1 



/3 



a 



a /* + A. 



/? 



h^-X b 



Eliminating /? between these two equations, we get a quartic for a, 

 and to each value of a one value, and one only, of /8. 



Hence, to each value of A, we have four conies of the system which 

 reduce to a pair of lines, that is, a conic consisting of a tangent from 

 each node of the binodal quartic. 



The node of this conic, i.e. the point of intersection of these two 

 tangents, must lie on the corresponding Jacobian. Hence, the four 

 Jacobians, as written above, are the equations of four conies, each of 

 which passes through four of the sixteen points of intersection of the 

 tangents to the binodal quartic from its nodes ; and hence, we may 

 infer that the anharmonic ratio of the two pencils of tangents is the 

 same, since the conies pass through the nodes. 



If the point xy% has the line z = for its polar with respect to 



then 

 hence 



aCz + Ci/ = 0, aCz + Cx = 0; 



'Zr y X 



y a h + X 

 X h + X b 



- 0. 



Hence, the locus of the poles of the line z = with respect to the 

 conies of the system lies on a fixed conic ; and the above is its equation 

 determined in terms of the coefB.cients of the quartic. 



Corresponding to each value of A, we have a conic. 



The Jacobian conies are 



J^ = xy ^ ^^ + A,.s- + &c. 



Consider 



K4 ^r 



xy - 



C, 



z{G^ + F,y) 



+ A, 



