> (', l) 4 = ; 



62 PROFESSOR CAYLEY ON POLYZOMAL CURVES. 



absolute invariant I 3 -=- J 2 has thus the same value in the two equations, that is, 

 the equations are linearly transformable the one into the other, which is the 

 before-mentioned theorem that the pencils are nomographic. 



148. The equations will be satisfied by 6 = (p if only aa = Ip, that is, if 

 a, b = mft, ma ; or by = — <p if only aa = — &/3, that is, if a, b = ?w/?, — ma : 

 the equation of the curve is these two cases respectively — 



&(ga _ B V)(, 2 - /3V) + «?*& + ms»(|8g + «,) + as* = , 

 l{% 2 - a 2 2 2 )(„ 2 - <3V) + ^ 2 £l + mz\i3% - «,) + cz i = . 



If to fix the ideas we attend to the first case, then the equation in 6 is 



C 24r-a 2 ,3 2 , 



— 6Jc7na(3, 



- 8/.V/3 2 - 4/cc + c 2 , 

 Qkmafi +3me, 



24/. 2 a-,3 2 + 24/,c + 6m a 



and we may take as corresponding tangents through the two nodes respectively 

 £ = 6az, t] — 6/3z ; the foci A,B, C, D, which are the intersections of the pairs of 

 lines (£ = O^az, n = Qflz), &c, lie, it is clear, in the line f3£ — a>j = 0, which is one 

 of the nodal axes of the curve. Similarly, in the second case, if 6 be determined 

 by the foregoing equation, we may take as corresponding tangents through the 

 two nodes respectively £ = Oaz, >/ = — Qfiz ; the foci (A,B, C, Z>), which are the 

 intersections of the pairs of lines (£ = B 1 az i n = — 0,/3z), &c, lie in the line 

 /3£ -(- ar) = 0, which is the other of the nodal axes of the curve. In either case 

 the foci A, B, C, D lie in a line, that is, we have the curve symmetrical ; and, 

 as we have just seen, the focal axis, or axis of symmetry, is one or other of the 

 nodal axes. 



149. In the case of the Cartesian, or when a = 0, (3 = 0, viz., the equation aa = bft 

 is satisfied identically, and this seems to show that the Cartesian is symmetrical ; 

 it is to be observed, however, that for a = 0, /3 = the foregoing formula? fail, and 

 it is proper to repeat the investigation for the special case in question. Writing 

 u = 0, (3 = the equation of the curve is 



ifcgV + cz 2 & + -J (a£ + hi) + cz* = 0, 



and then, taking £=6bz for the equation of the tangent from 7, we have 



- 2 Wf- 



T 



+ m ■ b{e6 + 1) 

 + z 2 . aid + c = 0. 



and the condition of tangency is 



Uf'(ab6 + c)-(c6 + l) 2 = ; 

 viz , we have here a cubic equation. Similarly, if we have n = Qaz for the equa- 

 tion of a tangent from J, then 



4Aj» 2 (ab<p + c) - (cf> + l) 2 = 0. 



