440 



HITCHCOCK. 



The form (159) now becomes, by putting (235) = 0, 

 1/34(527) + /35(247 )}(52p)2 _^ /32(23p) (45p) ^5(524) (52p) (72p) 



(527)2 



(237) 



(527)2 



(207) 



We note that the constant a cannot be zero, since this would give ^2 

 an inflectional element of the cones (3) and /Ss could not then lie in the 

 tangent plane if Fp is irreducible. Also, (457) cannot vanish, since 

 the planes (45p) = and (25p) = taken together constitute a quad- 

 ric cone which would then contain six axes, as the limiting form of a 

 reducible vector. 



Let now the vector (207) be thrown into the form of the right 

 member of (201) by addition of the term 



- p(52p) (452) 

 (527) 



which renders jS? a zero without destroying the zero character of jSg 

 or /Ss. By use of the identity 



p(452) = /34(52p) + /3b(24p) + /32(45p) 



•we find the vector takes the form 



^b(52p){(247) (52p) - (524) (72p) - (24p) (527)} 



(527) 



+ ^2 



(23p) (45p) (45p) (52p) 



(237) (527) j 



the terms in Pi destroying each other. But we have identically 



(247) (52p) - (527) (24p) = (452) (72p) 



whence the coefficient of 185 vanishes. The vector is thus reducible, 

 contrary to hypothesis. Hence its plane cannot be constant. 



Case 2. The three axes 182, 185, and /Sy are coplanar. The form 

 (159) becomes, putting (257) = 0, 



'^'^''''^rtwl'/''"^"'"^ + ('•■^' + W.) (52pn (208) 



(237) (457) 



where ai and bi are constants, which, like the original constants a 

 and b, cannot vanish in this case, Fp being irreducible. It is not 



