489J AND THE (2, 2) CORRESPONDENCE OF POINTS ON A CONIC. 19 
viz. the components are 
(aa + 6/3 (ac — b 2 ), 
(ab + 3/3 (ad — be), 
(olc + /3 (ae + 2bd — 3c 2 ) + ^K, 
where as before 
I assume 
ab + 3/3 (ad — be), ac + /3 (ae 4- 260 — 3c 2 ) + ^ K\x, l) 2 , 
ac + /3 (ae + 260 — 3c 2 ) + -fa K, ad 4- 3/8 (be — cd)\x, l) 2 , 
ad + 3/3 (be— cd), ae 4- 6/3 (ce — d 2 ) \x, l) 2 , 
2 (a 2 - 3//3 2 ) 
/3 
/3 = — f, a = 4c -66», iT=3{(4c- 60) 2 -f/}. 
aa + 6/3 (ac — b 2 ) = a (4c — 60) — 4 (ac — 6 2 ) = 46 2 — 6a0, 
a6 + S/3 (ad —bc) = b (4c — 66) — 2 (ad — be) = — 2a0 4- 6bc — 6bd, 
ad 4- S/3 (be - cd) = d (4c — 60) —2 (be — cd) = — 2be 4- 6cd — 6dd, 
ae 4- 6/3 (ce — d 2 ) = e (4c — 60) — 4 (ce — d 2 )= 4d 2 — 6e0, 
ac + /3 (ae + 260 — 3c 2 ) — = c (4c — 60) — § (ae + 260 — 3c 2 ) — ^ {(4c — 60) 2 — f/} 
= — -|ae — 260! 4- f c 2 — §0 2 , 
ac 4- /3 (ae + 260 — 3c 2 ) +y%K 
= c (4c — 60) — § (ae + 260 — 3c 2 ) + \ {(4c — 60) 2 — f 7} 
= — ae 4- 9c 2 — 18c0 + 90 2 , 
agreeing with the former result. 
I return to the general form 
y 2 (a , b , c $>, l) 2 
+ 2y (a', 6', d \x, l) 2 
+ (a", b", c"J_x, 1) 2 = 0, 
giving 
dx 
V[(a, 6, c\x, l) 3 (a // , b", c"\x, l) 2 —{(a', b', c'\æ, l) 2 } 2 ] 
~~ V[(a, a', a ,r $y, l) 2 (c, 0, c"$y, l) 2 -{(6, V, b"^y, l) 2 } 2 ] ’ 
Operate a linear transformation on the x, say