440 HITCHCOCK. 



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



1/34(527) + /35(247) } (52p)^ /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 /St a zero without destroying the zero character of ^2 

 or jSo. By use of the identity 



p(452) = i34(52p) + /36(24p) + ^2(45p) 



we find the vector takes the form 



i35(52p){(247) (52p) - (524) (72p) - (24p) (527)} 

 . (527) 



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



+ /3 



I, 



(237) (527) j 



the terms in ^a destroying each other. But we have identically 



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



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

 contrary to hypothesis. Hence its plane cannot be constant. 



Case 2. The three axes ^2, 185, and |3^ are coplanar. The form 

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



{^.(247)+/3.(457)}(23p)(45p) ^ ^ 



(237) (457) ^ \ iH5^ iyzj\ Hj, v 



where ai and &i are constants, which, like the original constants a 

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



