66 G. PLARR ON THE SOLUTION OF THE EQUATION Vpop=0 5 
The equation Fe’=0 assigns one or three real values to e=Te for every 
value of E, that is, for every one of the cones just spoken of. It appears there- 
fore that the surface [=0 is composed of a series of spherical conics, of which 
the radius of the sphere corresponding varies in function of E. 
Between E=/,, and E=E, the surface contains three curves on every cone, 
and therefore presents three sheets in the directions comprised between the 
cones corresponding to the adduced limits of E. 
The curves corresponding to E=h, are circles, as they are situated in the 
planes R and R’ respectively. 
The curve corresponding to E=E, and to the two equal roots e,, corre- 
sponding to it, presents a singular character, and we shall treat of it separately. 
The calcul of the values of e for the five values of E following will give us 
the means of conceiving the outer shape of the surface. 
Let us adopt the following designations, easily understood : 

B=h 
6 = @y 

The calcul of @ e@, e, can be effected in one case, and the others be 
deduced by permutation. We have for E=/, 
7 2 4\9 
0=(R+P) + Z(ah,—QY. 
Replacing P’, Q’ in function of the roots, 
P= shh,; —O=h24,,; 
we get 0=(& + holy —h2) + LIE + hah)? 
we divide by (¢,+/,h;)*, and introduce 
h? 
1 
> 
CR +hohs 
This gives : 
27 
0=(1—w)?+ Zw. 
This equation has the root w=4 single, and w=-5 double, because the 
derivate —3(1—w)?+ a , 1s annulled by w= = 
