78 G, PLARR ON THE SOLUTION OF THE EQUATION Vpgp=0. 
and which, as we shall see, extends as far as 
ane 
e=e,° 
We will call second sheet the one which begins at 
esses 
é=—€,, 
, and which ends at { Bows F 
and third sheet the other, beginning at 
{ Et. through aah, ending at { 
eé=¢€,, 
For E = h, we have by I‘ = 0, the two values of ¢, e=¢,, e=e,; and 
the corresponding cyclical conic is transformed into two half circles, because 
the value of E gives the equation of the two planes R, R’: 
0 — (hy — hz) S?re a (A; — hy) S*vé ° 
The plane R which cuts the first octant (both planes passing through the 
axis p) is given by : 
S¢(Vh,—h, } —Vh.—hsv) = 0. 
Replacing h,—h,, by 2e,, &c., the normal x, to this plane R will be 
(64) X2 = A,Je, — vrJey. 
The normal to the plane R’ will be 
(65) x, = Ae, + ve. 
In representing by a\ + bu + cvan e,, we get 
(66) ¢ = Ve, + pbN + va/e, perpendicular to x. 
Myc’ =— dv/e, + pbN’ + vse, perpendicular to x3. 
for the ¢’ in the planes R, R’, respectively. 
The circle of radius e¢, is comprised within the circle of radius e, = e, , 
because the inequality e< e reduces itself to 0 < (3h,.)’. 
Therefore the circle « = ¢¢< forms the junction curve between the first 
sheet and the second sheet. Along this circle the normal to the surface is 
perpendicular to the plane R of the circle ; namely, we have, W being = /,: 
