Since, If F = An, 4) = By, 1 = I (Df A4) Of A?) Ab A"'^1Bz 1=, (F, Mn Ax I Ax 2 Axe Ax1) J The First Transvectant Differs But By A Numerical Factor From The Jacobian Or Functional Determinant, Of The Two Forms. We Can Find An Expression For The First Transvectant Of (F, ï¿½) 1 Over Another Form Cp. For (M N)(F,4)), =Nf.4Y Mfy.4), And F,4, F 5.4)= (Axby A Y B X) A X B X 1= (Xy)(F,4))1; (F,Ct)1=F5.D' 7,(Xy)(F4)1.
Referred to the asymptotes as axes the general equation becomes xy 2 obviously the axes are oblique in the general hyperbola and rectangular in the rectangular hyperbola.
If three equations, each of the second degree, in three variables be given, we have merely to eliminate the six products x, 2, z 2, yz, zx, xy from the six equations u = v = w = o = oy = = 0; if we apply the same process :to thesedz equations each of degree three, we obtain similarly a determinant of order 21, but thereafter the process fails.
(II.) Similarly in (I.), writing for c l, c 2 the cogredicnt pair -y2, +y1, we obtain axb5-a5bx=(ab)(xy)..