899 
In each plane through s, lies however one such point 7;*; but if 
S, is the point of intersection of s, with 7;, then also the complex 
cone of S, breaks up into a pair of planes of which one compo- 
nent is of course again t,, the other being a plane through S; 7; ; so 
S, is itself a point 7,*, and the consequence of this is that 7,* 
describes a conic 4* which passes in the first place through JS, and 
in the second place, as is easy to see, through the three cone vertices 
lying in t‚; for if a plane through s, passes also through a second 
vertex, then the complex conic breaks up into the two pencils at 
T; and at that second cone vertex. 
All rays through a point 7,* of 4** cutting s, are according to 
the preceding rays of the complex; from this ensucs reversely that 
the complex cones of all points of s, in 1; have the same base curve, 
namely 4**. If now the degenerated complex cone of a point of 4° is 
to pass through s,, then that point must evidently lie also on 4** 
and of such points there exists apart from the three cone vertices 
lying in t;, only one; in the pencil | 7; | there is thus only one ray 
for which the (degenerated) complex cone of its focus passes through 
an indicated ray sj, i.e. the second components of the complex 
cones of the foci of the rays of the pencil | 7;| form a pencil 
of planes, or the rays of 1; belonging to the congruence form a 
pencil. 
The axis a of the pencil of planes must of necessity cut the curve 
k*; for, if this were not so, then an arbitrary plane through @ would 
cut 4? in two points, and then the complex curve in that plane would 
break up into three pencils (among which one at 77 is always included) 
instead of into two. This objection does not exist when « cuts the 
curve 4? in a point A; for then each plane through a cuts 4° 
besides in A in only one point 7,* more, and A itself is a point 
7;* for the plane through a which touches 4°. The ais a is simply 
that line which has the property that the complex cones of its points 
have as common base curve the conic k* itself; for, for each plane 
through a the point 7; lying on 4? must lie at the same time on 
k**, so Kk? and &*? coincide. 
For each ray of the pencil [A] lying in 7; point A is evidently 
one focus and 7; the corresponding focal plane, for each ray is cut 
in A by an adjacent one of the pencil; the other focus is the second 
point of intersection 7;* with 42 and here the second focal plane 
passes through 7;. The focal surface must therefore touch 1 along 
the conic k?; the point A itself is however a singular point, for here 
any plane through a is a tangential plane. 
For the tangent in A to £? the two foci coincide evidently with 
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