1364 
§ 2. We now propose the question how to determine the three 
functions Y,,9%;P,; in such a way that the twisted curve 
f= ¢, US eS Oe 
admits in the point P, («,, y,,2,) corresponding to the value wu, of 
u the plane (2) as osculating plane. 
The twisted curve g under discussion has to cut the plane (2) 
thrice in the point w= u, Le. the equation 
} ee ) LQ (2e ) eae (Ae sn 1)= 
5 u x, 
must admit three roots w= u. 
This gives the conditions : 
pee (4) Nl #' (4) NE (w,) 
as ae See kok oe 
yp, (w,) gp. (w‚) os (w,) 
E gy", (u,) P'2 (w) S g's (u,) . (6) 
p, (‚) ps (w) p, (u‚) 
As the equation (5) must hold for any value of wu, the first diffe- 
rential coefficient of the first member must disappear. This gives by 
taking (6) into account : 
P (4, (4) ee ol gi + sja =0 =0>. 4.5498 
NONE Gs Ce) | 
As the equations (5) and fe must hold for any value of uw, they 
lead to the two seis of solutions: 
ROE RON AOR RON AOR AC 
p q s ‘ 
(8) 
and 
Gi: G, ee Ge: pee Peg; (9) 
Pl ptgts) g(p—9 +s) 8(p+9q—s) 
By representing the three equal ratios (8) by wp (w) we find 
pv [|v (uj du 
f, ile 
which passes, by replacing f (u) du by t‚ into 
e g(a) = ot Pr Se 
i. e. into a curve of the system C(p, q, 5). 
Likewise the ratios (9) furnish 
Earl 4 YP, 
i. e. a second curve belonging to a system C(p,,q,,5,) determined 
by the relations 
