260 
determined in quite another way than was done in $ 6, viz. by 
making use of the first polar surface of Z,. This surface, of which 
the order is one unit lower than the one of @, and therefore 
(ef. § 2) amounts to 2u + 2r — 4e — 20 — 1, produces in the first 
place the “contour apparent” of 2, seen from point Zes but. it) ag 
easy to see that there can hardly be question of a real “contour 
apparent”. If namely a straight line passing through Z, touches 
2 (in a point outside 3 we will suppose), consequently intersects it 
in two points lying intinitely near to each other, it must be possible 
to describe round the foot of that perpendicular two circles cutting 
ke perpendicularly whose rays only differ infinitely little, and this 
is not impossible, for it holds good for all the points of the inflec- 
tional tangents of @, but then exclusively for them. We found in 
fact before that the generatrices passing through the points of in- 
flection of 4” were torsal lines of @ with vertical tangent planes; 
these torsal lines, to the number of 2, belong therefore to the 
intersection of 2 with the first polar surface of Z,, and in reality 
form the only “contour” of 2 for Z,, barring of course 4” itself, 
which, as a matter of course and as is clear from the simple example 
of the hyperboloid of revolution, also belongs to the ‘contour 
apparent.” 
Let us imagine a line passing through Z, drawn to a point P of hk. 
We know that along £” 2 sheets of 2 osculate each other, and now ask 
how many points the line Z,P in P has in common with 2. This 
number will amount to 4, just as when the sheets simply touched each 
other along 4”, for otherwise the two branches would possess points 
of inflection in P in an intersection with a piane passing through 
Z,, P, which is not the case. Besides, round the foot of an arbitrary 
line passing through Z, rv circles are to be described cutting Av 
perpendicularly, and round / r—2; 4 points of 2 have consequently 
coincided on Z,P in P. Of the first polar surface have therefore 
on Z,P 3 points coincided in P, and the question is now what 
is the shape of that polar surface, as it can only touch the two 
sheets of @ passing through &’. The fact is that the first polar 
surface breaks up into a surface and the plane 3, and consequently 
has ke as a nodal curve. A line Z, P now contains the point P 
of the nodal curve, and moreover a neighbouring point, and therefore 
3 indeed. 
That 8 is a part of the first polar surface of 7, ensues already 
from the symmetry of @ with regard to 3. An arbitrary straight 
line passing through Z, cuts 2 only in 2r points not coinciding 
with Zop, and consequently in 2u — 4e — 25 points that do 
