AND ENVELOPES OF HOMOLOGY. 597 



Substituting these in (37), we have, as the envelope required, the conic section, 



3 (^- 



0- 



in" 



2\ 



n 2 ) 



($7_ 



+ 



ya. 



+ 



a/3 



) = 



= 



(40). 



Let it now be supposed that the centre of homology moves in this last curve, and 

 let it be required to find the envelope of the inverse fa isceau. 



We have now, 

 and 



3 {(Z/r 2 + (mg)-* + (nhy 2 ) - 2 { (mngh)-^ + (nZA/)- 1 + {ImfgY 1 } = 0, 



which, with the exception of the numerical coefficients, are identical with (36) 

 and (40), the variables to be eliminated being merely inverted, and /, m, n also 

 inverted. Moreover, the process reverses the change of numerical coefficients, 

 as exhibited between (35) and (40), and the result is 



I 2 a 2 + m 2 8 2 + n 2 y 2 + mnBy + nlya. + hnaft = . . (41); 



a conic, and somewhat singularly related to the original curve, namely (35). 



X. If the centre of homology move in a conic circumscribing the original 

 triangle, required the locus of the intersection of the direct and inverse lines of 

 homology. 



As the point whose locus is required is on both the lines, we have simultane- 

 ously, 



«/ + Pff + 7h = 



and 



a B 

 f 9 



+ 



= 



Also the equation of the circumscribing conic is, 



If + mg + nh = 



From (42) and (44), we have, 



/ 9 h 



and, by (43), 



m, B 





n, y 





I, a 





n, y 





I, a 





m,B 





a 



P 7 



m, 8 



+ 



n, y 



T 



I, a 



n, y 





I, a 







m, 



8 



(42), 

 (43). 



(44). 



(45), 



= 



(46), 



Or, -(a'-/S 8 - 7 2 )+— (^_ 7 a-a8)+ Z 3-( 7 a_ a 2_ /3 2) + ^ +TO 2 + n 2 == o (47); 



ycc ' ap v 



'By 



which is the locus required— a curve of the third order. If a = 0, we have 



(nB - my) . (8 2 - y 2 ) = . . . . (48); 



