ROTATIONS IN HYPERSPACE. 687 



The intersection of the conic (56) with n is given by aX + 6/i = 

 and hence the conic is tangent to /•]. The same argument shows that 

 it is also tangent to ro and r^. Hence: The locus of the ends of the 

 curmturc vectors of curves perpendicular to a given direction is a conic 

 tangent to the three sides of the triangle determined by the vectors x, y, z. 

 The center of the conic (56) can be determined from the middle of 

 the segment into which the conic projects on the x-, y- and 2-axes. 

 One end of each segment is at the origin. The projection on OX is 



c'Vx 



r = 



c^'(X- + M-) + (oX + hisf 



The value of - which makes the denominator a minimum will make r 



M 

 a maximum. Similarly we can determine the projection of the 

 ellipse on OY and OZ. Hence the center of the ellipse is the end of 

 the vector 



^ (6^ + c')x + (g^ + c')y + (a^ + 6^)2 

 ''' 2(a2 + 62 + c2) 



The point of the curvature triangle corresponding to the direction 



a/V(J/i.r)2, h/V{M.-r)\ c/V~{M^^ is 



P2 = 



From which we have 



a" + 62 _|_ ^2 

 2pi + P2 x-\- y -\- z 



The right side of this last equation is the median center of the curva- 

 ture triangle. Hence the center of the curvature triangle is a point 

 of trisection of the line joining the center of the conic to the point of 

 the triangle corresponding to the direction a/y/{M\-ry, hl^ {Mi-r)"^, 



c/y/{Mz'r)-. It can be shown that two perpendicular directions 

 among those satisfied by (54) correspond to points at opposite ends 

 of a diameter of the conic (56). Hence the points of the curvature 

 triangle which correspond to three mutually perpendicular directions 

 through the point form a triangle whose median center coincides with 

 the median center of the curvature triangle. The points of the conic 



