83 



But this bitangential of the sextic contains the sixth power of the 

 hessian of the sextic as an irrelevant factor. In order to free it from this 

 factor, we put 



J = (Ao A2 — AJ ) A4 — ( Ao A3 — Ai A2) A8 + ( Ai A3 — Ai ) A2, 

 and then express y in terms of x for the function J. The work involved 

 in this last step is very long and tedious. These results can be used in 

 developing a bitangential of the septic, but two additional functions 

 will have to be developed, the work in which is almost beyond the range 

 of possibility. 



