a given Curve of the Third Order in Space. 21 



of any curve of the third order, and developable surface of the 

 fourth order. The question arises, to find the equation of the 

 cone of the third order having an arbitrary point for its vertex, 

 and passing through the curve of the third order. This may be 

 done by JoachimsthaFs method : let a, /3, 7, h be the values of 

 a, by Cj d at the vertex of the cone, and in the equations of the 

 curve, say in 



ac-62=0, bd-c^=0, 



for a, b, c, d write ua + vu, ub + v/5, uc + «;7, ud-\-vh', if from the 

 equations so obtained u, v are eliminated, the resulting equation 

 will be that of the cone of the third order. 

 The substitutions in question give 



lJu^-{-'m!uv + Wv^=0, 

 where 



L =2(flc-62), M =:«7 + ca-26^, N =2(a7~^2)^ 



y = 2{bd-c^), M' = 6S + (//3-.2cy, W = 2{^8-y^), 

 The result of the elimination is 



4(LN -^ M^) (L'N' - M'^) - (LN' + UN - 2MM02 = 0, 

 and we have 



LN -M2 =4^{by-c^){a^-boc)-(coc-ay)% 

 yW-W^ = 4^{c8-dy){by-cfi)-{bS-dy)^y 

 LN' + L'N-2MM'=4(67-C)8)2 + 4(«y8-6a)(cS-</7) 

 + 2{coc-ay){bB-d^). 

 Write for shortness, 



by — 0^=1, coL—ay = mj a^^ba,-=.n, 

 ah—dcx.=ff b8 — dff=g, ch—dy=h, 

 values which give lf-\- mg + nh=. 0. Then forming the expression 



4(4//i - rrfi) {Ul-g'^) - (4/^ + 4/iA + 2mg)\ 

 this is equal to* 



— lQl{l^-nfh + lm.g-2lnh'\-ng'^ + m^h). 



* With respect to the occurrence of the factor Z( =67—0/3), it is worth 

 noticing, that, putting b=.k^, c=ky, we have identically 



i. e. the two functions will contain a common factor if b=k^, c=ky, or 

 what is the same thing, if by — c/3=0. But if the functions contain a com- 

 mon factor, their resultant vanishes, i. e. the resultant will vanish in virtue 

 of the relation 6y— c3=0, or what is the same thing, by—c^ is a factor of 

 the resultant. 



