42 DR. A. JOHNSON ON SYMMETRICAL 
solving, if its roots be 4, and k,, the equation of the section we are now dealing with will 
be evidently 
heth2=VE> e+ (30) 
Hence the squares of its semi-axes are give us 
ev iF ee a Pr ea 
ii CRT ae A ii = a 

Putting (30) in the shape 
¥ 
R, =F 
we obtain Euler’s formulae, &c. 
To find the directions of the principal sections at any point of the surface U = 0. 
If a, £, y, be the direction angles, we have, from equations (15), substituting ZL, M, N, 
as before for a, b,, €. 





cos? a cos’ fy = cos’ y 
b—k l M c—k M N | | a—k n L 
l b—k WN M a — k ie n b—k WM | (31) 
M N N L L M 

To find the conditions for an umbilic. 
Substituting LZ, M, N, as before in equations (19) we get 
bN*?+cM?-21MN cL+aN-2mNzLZ aM?1+b62P—2nLM 
M+ Ne = 



WEP DT 
Lines of curvature. 
The condition that the first three of equations (6) should hold simultaneously, viz: 


1 2%, ia i 1 ah, 
2 da 2 db, 2 de oe 
as, bs C3 
ay, by. € 
has already received a geometrical interpretation in the case of the section of a quadric, 
viz: that certain three lines should lie in one plane. The application of this to the cur- 
vature of surfaces is not difficult. 
For if we substitute Z, M, N, for a,, b, ¢,, and d x, dy, d z, (proportional to the direction 
cosines of the directions of maximum and minimum curvature) we get 
adzx+tndytmadz, ndz+tbdytldz, mdx+ldy+ecdz 
L M N 0 
dx dy De (33) 


the known equation of a line of curvature but which is in fact the reduced form of 
L+daL M+ dM N+anN 
L M N — 10) 
d x dy dz 


