296 
ON CURVATURE AND ORTHOGONAL SURFACES. 
[519 
principal cone meeting the tangent plane in two lines, the principal tangents, such 
that at a consecutive point P' on either of these the normal meets the normal at P; 
viz. we have the principal tangents at the tangents of the two curves of curvature 
through the point P. 
The plane through the normal and a principal tangent is termed a principal 
plane; we have thus at the point of the surface two principal planes, forming with the 
tangent plane an orthogonal triad of planes. 
11. I proceed to further develop the theory, commencing with the following lemma: 
Lemma. Given the line Xu + Yv + Zw = 0, and conic 
{a, b, c, /, g, h\u, v, wf= 0, 
then, to determine the coordinates (и г , v 1} w 2 ), (u 2) v. 2 , m 2 ) of the points of intersection 
of the line and conic, we have 
(a,..^YZ-Z v , ZZ-XZ, Xtj — Ygy 
= (bh + V v l + fWi) (fw 2 + t]V 2 + £w 2 ), 
or, what is the same thing, we have 
(a,...^-Z v , ZZ-XZ, Xr) — Ff) 2 = 0 
as the equation, in line coordinates, of the two points of intersection. The proof is 
obvious. 
12. Making the equations refer to a plane and a cone, and writing throughout 
Ь t], £ as current point coordinates, the theorem is: 
Given the plane X% + Yr) + Z% = Q, and cone 
(a, b, c, /, g, v, 0* = 0 5 
then, to determine the lines of intersection of the plane and cone, we have 
(a,..bY£-Z v , ZZ-X& Xt] — Ybf = 0 
as the equation of the pair of planes at right angles to the two lines respectively. 
13. Denoting the coefficients by (a), (b), &c., that is, writing 
(a,..lYZ-Z v , ZZ-XZ, X v -YZy 
= ((a), (b), (c), (/), (g), (A)£f, v, 0 2 , 
the values of these are 
(a) = bZ 2 + cY 2 — 2/YZ, 
(b) = cX 2 + aZ 2 — 2gZX, 
(c) = aY 2 + bX 2 -2hXY, 
(f) = — aYZ — fX 2 + gXY + hXZ, 
( g ) =-bZX +/YX-gY 2 + hYZ, 
(A) = - cXY+fZX + gZY - liZ\