( 42 ) 
« = —(p 2 —^ 2 )(p 2 -c 2 ), 
P À 
p = —(p*-c*)(2p — a 2 ), 
y — —Ap 2 - a 2 ) (p 2 — b 2 ), 
P 
2 
Gu 4 + Hm 2 + K 
w 2 4-a (u 2 4- a) (u 2 /B) (u 2 4- y) 
Il est visible que 
G = JEA, H = 2a (j 3 -t- y), K = 2A0y = p 2 <x l 3 r y£a i (b 2 — c 2 ). 
Or : 
2A = p 4 2a 4 (6 2 — c 2 ) — p 2 2a 4 (b 4 — c 4 ) 4- a 2 b 2 c 2 Xa 2 (b 2 — c) 
= - p 4 (a 2 - b 2 ) (b 2 — c 2 ) (c 2 — a 2 ), 
XA (/3 4- y) = — (p 2 — a 2 ) (p 2 —b 2 ) (p 2 — c 2 )Xa 4 (b 2 — c 2 ) (2p 2 —b 2 — c 2 ) 
p 2 
= — 2 (p 2 —a 2 ) (p 2 —b 2 ) (p 2 — c 2 ) (a 2 — b 2 ) (b 2 — c 2 ) (c 2 — a 2 ) , 
p 2 a3 , y2 l a i (b 2 — c 2 ) =- (p 2 — a 2 ) 2 (p 2 — b 2 ) 2 (p 2 —c 2 ) 2 (a 2 —b 2 )(b 2 —c 2 ){c 2 —a 2 ); 
donc ^ 
[ 
Gu 4 4- H u 2 4- K =— (a 2 — b 2 ) (b 2 — c 2 ) (c 2 - a 2 ) 
1 
X p 4 u 4 4-2(p 2 — a 2 )(p 2 — b 2 )(p 2 -c)u 2 -\ -(p 2 — a 2 ) 2 (p 2 — b 2 ) 2 (p 
p" 
OU 
Gm 4 4- Hm 2 4-K =r-(a 2 -b 2 )(b 2 -c 2 )(c 2 -a 2 )[p% 2 4-(p 2 -a 2 )(p 2 -b 2 )(p 2 -c 2 )l 2 . 
p 4 
X. Au moyen de cette valeur, l’équation (9) devient 
p6tfco 2 = 
du- 
ii 
- [p 4 w 2 4- (p 2 — a 2 ) (p 2 — b 2 ) (p 2 — c 2 J 5 
L"' + ? 
(p 2 _b 2 )(p 2 —c 2 ) 
u 
1 
P 2 
; (p 2 — c 2 )(p 2 — et 2 ) 
1 
u 2 4— (p 2 —a 2 )(p 2 —b 2 ) 
P 2 
] 
Ainsi, l’équation différentielle de l’herpolhodie est, finalement, 
dcc = 
du 
u 
[p 4 M 2 4- (p 2 — a 2 ) (p 2 — b 2 ) (p 2 — c 2 )] [/ —\ 
1/ [p 2 u 2 4-(p 2 —b 2 )(p 2 —c 2 )][p 2 a 2 4-(p 2 —c 2 )(p 2 —a 2 ][p 2 w 2 4-(p 2 —a 2 )(p 2 —b 2 )] 
,( 10 ) 
