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point of contact ¢, and a second through the point of contact t; if 
the variable conic describes the pencil, then the jocus will pass two 
times through the point 1; so 1 is a double point and so are 2, 
3 and 4. 
b. The points of intersection of K°® and C® are double points. 
C2 is the locus of the poles Aj, ... Ag4; im these poles two 
tangents concur; so the locus also passes two times through these 
poles. It is evident that we are dealing with points of intersection 
of C3 and K*, not lying at the same time on K?, so these are the 
six points (4,9, A31), (Big, Bsa), (Cos Czy) formerly found. 
12. It is now evident, that the curve K® has ten double points; 
so it is unicursal. Of these points six lie on C°, the remaining 4 
are 1, 2, 3, 4. The six double points representing 12 points of 
intersection of A® with C? there are still 6 points; these are 
evidently the points where C? also intersects K?, so that now all 
the points of intersection of C3 and K® are found. 
Moreover it is clear that the curve K® touches the curve A? in 
the six common points, so that it has no more points in common 
with it. 
Au additional remark is, that the 10 double points have a partic- 
ular position in reference to each other. They are situated so, that 
the points A lie in pairs with the 4 points 1, 2, 3, 4 on a conic. 
This corresponds with the geometrical truth that the ten double 
points of a curve of the 6 order cannot have an arbitrary position 
in reference to each other. 
13. The algebraic reckoning comes to a similar result. Let the 
hyperboloid be: ; 
2 2 „2 
Egt Ea 
a? b? 2 
This is rectangular, if 
1 1 1 1 1 1 
5 en or — = 
a? Ri 2 be b2 git te 
If we start from a general equation of a quadratic surface 
an €? + agg y® + agg 2? + apie, Cue =e 
