430 



Problem. To find the system of rectangular conjugates. This of 

 course is cardinal, and is treated everywhere : but is made far easier 

 by Eliminants, as follows. Let us inquire after that diameter, com- 

 mon to two given concentric surfaces, which shall have its conjugate 

 planes the same for both. 



Take the centre for the origin, and x=mz, y=nz for the common 

 diameter sought. Then the central planes conjugate to it in the two 

 surfaces are , 



To identify these two planes, let 



Aw-f D+E _ Dm+Bn+F 



Eliminate m, n, and you find that p is to be determined by the very 

 same eq. as in the preceding ; and since its eq. is of the third de- 

 gree, it has always one real value. 



Next, let the second surface be a sphere, and you find at least one 

 diameter of the first surface perpendicular to its conjugate plane. 

 Make this diameter the axis of x, and take for the axes of y and z 

 the two principal diameters of the section in the conjugate plane. 

 Then D=0, E=0, F=0 ; so that the general eq. is reduced to 

 Aa? 2 -J-By 2 -!- Cs 2 +G=0. Moreover, the system of axes is now rect- 

 angular : hence the axis of y, and that of z, equally with that of x, 

 are each perpendicular to its conjugate plane, and the eq. for p must 

 have three real roots, corresponding to these three axes. 



We might similarly investigate " the conditions of contact for two 

 concentric surfaces ;" which, when one of them is a sphere, gives the 

 cubic whose roots are a 2 , 6 2 , c 2 , principal axes of an Ellipsoid. 



Problem. To discuss the results of Tangential Co-ordinates. 

 [This expression is employed as by Dr. James Booth in an original 

 tract on the subject.] 



Put P 



Then P.r + Qy + Rer + S:=0 is the eq. to the surface, and 



= is the eq. to the tangent plane at (xyz). Hence if 



