015 
this curve is also in (1,1)-correspondence with m2", whilst it never 
happens that Gj and J/;, coincide, the reasoning given above leads 
here anew to the order 32 of (e). 
§ 11. A twisted biquadratic curve e* of the second species lies 
on one quadratic surface 4? only. It can be considered as partial 
intersection of 4” with the cubie scroll +* generated by the bise- 
cants 4 of o* cutting a given bisecant 4,. Each point of 4, bears 
two bisecants 6,6" and the plane (4,6’) passes through the single 
director line g of &*. 
The pairs 6,6’ determine an involution on g, the double points of 
which lie in two double tangential planes of o*. 
Reversely the line vommon to any two double tangential planes 
of g' is single director line of a *; for the bisecants lying in these 
planes are cut by one bisecant 6, only and this line is the double 
director dine *)of =*. 
We now determine the number of orthogonal pairs 6, 6’. 
Any edge of a director cone A* of * is at right angles to three 
other edges; so the planes of the orthogonal pairs envelop a cone 
of class three. On A®* the pairs 6,4’ determine an involution and 
the planes of the pairs of edges pass through an edge parallel to g. 
From this it follows that g bears three *) orthogonal pairs 4, 0’. 
As the lines g form a congruence, there are a planes 2 con- 
taining orthogonal bisecants; so these planes envelop a surface 2° 
of class three. The planes intersecting ¥Y? in orthogonal hyperbolas 
are parallel to the tangential planes of a cone of the second class 
and envelop therefore a conic *2 at infinity. Evidently a common 
tangential plane of “2 and “2 cuts of in an orthoeentrie group. So: 
the planes of the orthogonal quadrangles inscribed in @* envelop a 
N 
developable of class six. 
$ 12. We consider once more the locus of the quadruples of 
orthocentres in the planes -/ through a line /. If 4 contains a point 
1) The lines g form a congruence (6,3). 
2) If £3 is represented by the equation 
(aa + by + cz + dja? = (ve + Uy Hee 4+ dy" 
we find for any pair 0, 0’ the equations 
y=, an + by + ez + d=i (aa 4+ by + cz + a). 
So the orthogonal position of the lines (2) and (—-) requires evidently 
(c'a?—c)? (L—a’) + (aad (bb) = 0. 
So there are three orthogonal pairs. 
