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 A 2 — Af) A 4 — (Ao A 3 — Ai A 2 ) A3 + (Ai A3 — A|) A 2 , 

 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. 



