614 KEV. HUGH MARTIN ON CENTRES, FAISCEAUX 



LXXII. Whether the centre moves in a circumscribing conic or in a tri-tangent 

 conic, the faisceau of bi-tangent conies 



envelopes the same bi-tangent conic 



Pa? = (4mn) f /3 7 (135). 



LXXIII. Whether the centre of homology moves in a self-conjugate or a tri- 

 tangent conic, the faisceau of bi-tangent conies 



envelopes the same bi-tangent conic 



(P)*a 2 = (imnffy (136). 



I have only to add, that while the general problem of this paper becomes very 

 difficult in the case of the general equation of the second degree, by reason of 

 the complicated elimination that is necessary, I have no doubt that this difficulty 

 might be evaded by reducing the general equation to one or other of the forms of 

 the conic which have engaged our attention, by an appropriate transformation of 

 the triangle of reference. If leisure permit, I should be glad to verify this suppo- 

 sition, as well as give certain extensions of the theory, in another paper. 



Postscript. — While this is passing through the press, I find that Silvester's 

 " Dialytic method of Elimination as applied to a Ternary System of Equations" 

 (see " Cambridge Mathematical Journal," vol. ii. p. 234), may be used more 

 efficiently than the Jacobian in the last problem of Section I. Writing the 

 equations (50), (52), (51), thus : — 



= I 2 ./ 2 + m 2 .cf + n 2 .h 2 + * * * [1], 



0= * * * + a.gh + &.hf+ y.fg . . [2], 



= «./ + &.g + y.h . [3]s 



we have to find four equations in/ 2 , g 2 , h 2 , gh, hf,fg. Re- write the three just 

 given (taking twice the second) thus : — 



0= (l 2 f).f + {m 2 g).g + (nVi).h . . [4], 



= (/& + yg) ./ + (yf+ah).g + ( a g + f3f).h . [5], 



0= («)./ + (/3).gr 4- (y).h . . [6]. 



Eliminating/, g, h, we have, dropping the parentheses, 



l 2 f m 2 g n 2 h 



(37i + yg yf+ ah ag + &f 



a 13 y 



= . . . [7] 



