129] 
AND ON AN IRRATIONAL TRANSFORMATION &C. 
147 
We have, in fact, identically 
ly x z x (¡3n 2 + ym 2 ) + mz x x x (yl 2 + an 2 ) + nx 1 y 1 (am 2 + /31 2 ) 
= (amnx 1 + /3nly x + ylrnzd) (lx x + my x + nz x ) — Imn (axd + /3yd + yzd), 
aydz-d (J3n 2 + уm 2 ) 2 + /3z x 2 xd (71 2 + an 2 ) 2 + yxdyd (am 2 + /31 2 ) 2 
— a(3y { — Pxd — mdyd — n 3 zd 
+ ( m yi + nz d) + { nz i + 2#i) ndyd + {lxj + myd) n 2 yd — 2lmnx 1 y 1 z 1 ] (1х г + my x + nzd) 
— (l 4 /3yxd + ndyayd + n 4 a/3zd) (axd + fiyd + 4 z i) 5 
which show that if 1х г + my x + nz x = 0 and axd + ¡3yd + y x z 2 = 0, then also lx» + ту» + nz» = 0 
and axd + /3yd + 7z 2 2 — 0 : this is, of course, as it should be. 
I shall now consider l, m, n as given functions of x ly y x , z x satisfying identically the 
equations 
lx x + my x + nz x = 0, 
l 2 bc + m 2 ca + n 2 ab = 0, 
equations which express that lx + my + nz = 0 is the tangent from the point (х 1г y x , z x ) 
to the conic ax 2 + by 2 + cz 2 = 0. And I shall take for a, /3, 7 the following values, viz. 
a = axd + byd 4- czd — a (xd + yd + zd), 
/3 = axd + byd + czd - b (xd + yd + zd), 
7 = axd + byd + czd - c (xd + yd + zd) ; 
so that x 1} y x , z x continuing absolutely indeterminate, we have identically axd + ¡3yd + yzd = 0- 
Also taking © as a function of x 1} y x , z 1} the value of which will be subsequently 
given, I write 
X» = %y x z x (/3n 2 + 7m 2 ), 
y 2 = ®z x x x (71 2 + a n 2 ), 
z» — ®x x y x (am 2 + ¡3 l 2 ); 
so that x ly y x , z x are arbitrary, and x 2 , у», z 2 are taken to be determinate functions 
of x ly y x , z x . The point (x 2 , у», zd) is geometrically connected with the point (x lt y x , zd) 
as follows, viz. (x 2> y 2 , zd) is the point in which the tangent through (x ly y x , zd) to 
the conic ax 2 + by 2 + cz 2 = 0 meets the conic passing through the point (x Xi y x , zd) and 
the points of intersection of the conics ax 2 + by 2 + cz 2 = 0 and x 2 + y 2 + z 2 — 0. Con 
sequently, in the particular case in which (x ly y u zd) is a point on the conic 
x 2 + y 2 + z 2 = 0, the point (x 2 , y 2 , z 2 ) is the point in which this conic is met by the 
tangent through (x 1} y x , zd) to the conic ax 2 +by 2 + cz 2 = 0. 
It has already been seen that lx x + my x + nz x = 0 and axd + /3yd + yzd = 0 identically ; 
consequently we have identically lx 2 + my 2 + nz 2 = 0 and axd + ¡3yd + yzd = 0. The latter 
equation, written under the form 
(axd + byd + czd) (xd + yd + zd) - (xd + yd + zd) (axd + byd + czd) = 0,