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b,b’ of 6*. So the locus of the points Q lies on a cubic seroll 4° 
with double line /. 
In the plane bl two points Q coincide in S,, two other ones in 
S,, whilst Q,, lies in / and Q,, in b. If P moves along /, q=Q,,Q,, 
describes a cubic scroll ®* with double line /; for through Q,, pass 
two lines g.g’ to the points Q,, of the chords 6,6’ concurring in the 
point P corresponding to Q,,. 
The scrolls 4’, ® have the trisecants tt, of of passing through 
S, and S, in common. For if P coincides with S,, 4 becomes a 
chord 6 and, as Q,, coincides then with S,, at the same time a 
line g. 
As / is nodal line for both serolls, these surfaces have still a 
twisted cubic 4° containing the points @,, in common. In the planes 
touching o* in S, and S, the point Q,, coincides with the point of 
contact; so 9,5, is a chord of 4°. This curve intersects 6* in the 
two points the tangents of which intersect S,S,; it has for chords 
the single director lines of the scrolls 4°, ®°. 
So by the transformation (PQ) the chord / passes into the system 
consisting of 7 itself, the tangents s,,s, and a twisted cubic. 
Evidently a trisecant t is transformed into that line to be counted 
thrice and the tangents in the three points it has in common with o*. 
If / touches of in S,,, the scroll ®* becomes a cone with nodal 
edge /. In the osculating plane of of in S,, q lies along /; so this 
plane is common tangential plane of A? and ®*, having still in 
common the trisecant through S,,. The residual intersection A* touches 
in S,, the tangent of 6*. 
$ 10. If 7 is unisecant of of in S the curve 4° breaks up into 
the tangent s of of in S and a curve 4°. The polar line /’ of / 
becomes chord of 4°, s being one of the three chords cutting / and 
l’. The plane /S touches H° in S and is therefore polar plane of 
P=S; it contains the tangent s and the trisecant of o* on which 
S lies. Of the three variable points (2 common to 4° and a plane 
through /’, two coincide with S and only one lies outside JS. 
Any plane through / contains besides S three points Q and has 
therefore in S two points with 4° in common. Also the plane /S 
not passing through / has in S two points in common with 4°; so 
S is a node of 45. The plane ds contains beside S only one point 
Q; so it passes through the nodal tangents of the node. So to a 
unisecant corresponds a twisted quintic with a node. 
The curve is of rank eight, through / passing four common 
tangential planes of o* and «4°, 
