Prof. Newman on Determinants. 393 
In a non-centric surface, where V=0, we readily find that the 
former of these eliminants has the same sign as (D°— AB) ; and con- 
sequently, that non-centrie surfaces cannot have sections of opposite 
species. It also appears, that to determine in a non-centrie surface 
the parabolic sections, we must take Z7mn such as to verify one’ of 
the three eqq. 
ADE ADE lL mn 
DB Er) Os te ms ele =0) 12D. BaF. | —0: 
l_mn EFC EFC 
Problem. To determine the circular sections, when they exist. 
Result. Take the larger question, of ascertaining when two sur- 
faces of the second degree intersect in a plane curve. Denote the 
coetlicients of the second surface by accents. Put a=Ap—A'; 
B=Bp—B'; y=Cp—C'; Ke. and determine p by the eq. 
aoe 
op @ | =0; 
FE od Y 
which involves ¢ in the third degree. 
Then Z mn will be determined (when the surds are real) by the 
proportion 
Li:m:n=WV(—ay)te: V(y’—By)+¢: y. 
To apply this to the problem of circular sections, it is only necessary 
to suppose the second surface to be a sphere. 
The surface becomes one of Revolution, if (with oblique axes) 
either system of three eqq. is fulfilled : 
G1) aB=8, ay=e, By=¢*, 
(2) ag=de, Pe=9s, yd=ed. 
If out of each triplet we eliminate p? and p, (for it seems easiest to 
treat these as independent variables,) the result is two eqq. (expres- 
sible by eliminants), which are the two general conditions for a sur- 
face of revolution. 
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 a=mz, y=nz for the common 
diameter sought. Then the central planes conjugate to it in the two 
surfaces are 
(Am+ Da+E)x+(Dm+Bn+ F)y + (Em+Fna+C)z=0 
(A’m+ D'n+ E!)2+(D'm+ Blat+ Py +(Bin+ Pn 0)e=0. b 
To identify these two planes, let 
Am+Dan+k _— Dm+Bnt+F _— Em+Fn+C ved 
A'm+ D'n+ 2! D'm+Bla+F"” E'm+Fat+ Cp’ 
or am+on+e=bm+ Bn+o=em+on+y=0. 
Eliminate m, 7, and you find that p is to be determined by the very 
