MISS C. A, SCOTT ON PLANE CUBICS. 
253 
( 2 .) Oa, DGj, HgK meet in /3; 1.^/3 meets h in K'; r are the foci of OD, KK'. 
(3.) IgT meets DG^ in 8 ; IgO meets 0K in y ; goes through S (figs. 2 , 3 ), 
(4.) IgT meets HgS in jx; is the intersection of HgD with h.^; L^/x goes 
through B. 
(5.) z, ^ are the foci of TP, WH (fig. 3), 
13. Now 2 ;, C being the foci of the involution TP, WH, will be imaginary if P lie in 
the segment WDH, otherwise real. When P is at W, B is at T; and as P travels 
over WDH, B travels in the opposite direction over TH. Thus the Cayleyan is 
unipartite when B is in the segment TH, otherwise it is bipartite. Now when B is 
in TH, X (fig. 3) is in TDgl ; S is therefore in THKg ; and when B is in TDH, X is in 
TD3I, and S is in TDKg. Thus the Cayleyan changes from unipartite to bipartite 
and vice versd when the cusp passes through T and Kq ; but of these two, in the 
series here considered, Kg corresponds to the case K = H, which gives a degenerate 
cubic. . 
II. Application to Special Cubics. Figs. 4, 5. 
14. The Harmonic Cubics. —If the cubic be harmonic, let K be the one of the 
three points on h that is conjugate to T, i.e., let K, T be harmonic wuth regard to Ick. 
Then since K, K! are harmonic with regard to hK, K' now^ comes at T. In the 
general case T, S are points in which h meets a series of conic polars ; hence, T being 
K', S must be K ; i.e., for a harmonic cubic, the cusps of the Cayleyan are on the 
cubic. Conversely, if S come at K, K' must come at T, and the cubic is harmonic. 
Now in the case we are considering, the conic polar of K goes through T, hence 
the line polar of T goes through K ; i.e., the inflexional tangent to the Hessian goes 
thi’ough K ; thus P is at K. Conversely, if P be at K, i.e., if the line polar of T 
pass through K, then the conic polar of K passes through T, thus K' is at T, and as 
before, the cubic is harmonic. 
In the general case, t, r are harmonic with regard to KK', and therefore in this 
case with regard to TK, i.e., with regard to TP ; hence the Hessian is harmonic; and 
as z, ^ are harmonic with regard to TP, i.e., with regard to TS, the Cayleyan, qud 
class-cubic, is also harmonic. 
The question now is, where must K be in order that the cubic may be harmonic. 
When S comes at K, HgS coincides with HgK, therefore /x is on K/3; y is also on 
KjS, since Ay has to go through S, likewise 8, since 8y goes through K. But goes 
through D, hence 8 must be at A ’> since 0 /x goes through Ig, /x and 8 must 
coincide at A- 
The pencils [Tg.GoiSDW], {K . Go^DHg} (fig. 4) estimated on the line (I) are 
equal to 
{H 3 I 3 I 3 I 1 } and {IiHgHJTg] respectively; 
