( 632 ) 
twice through each of those base-points; so it has with the neteurve 
” . . a . . r . . . 
cy indicated by a definite point Y yet (nn'—6) single points in 
common; here 2’=8(n—1) represents the order of H. The curve 
er having a node in D, determines with cy a pencil represented by 
a tangent of the curve (Y). From this ensues that the class of (Y) 
is indicated by hk” = 3n (n—1) — 20. 
The genus g" of this curve is also easy to find. As the points of 
(Y) are conjugated one to one to the points of H these curves have 
the same genus. So we have 
= } (n'—1) (n'— 2) — b = £ (8n—A4) (8n—5) — b. 
We shall now seek the number of nodes and the number of cusps 
of (Y). These numbers dé” and x" satisfy the relations 
2d'+3x n' (n"— 1) — k", 
J" 4x! = 4 (n'—1) (n"— 2) —g". 
From this ensues after some reduction 
d' = 2(n—1) (n—2) (3n?—3n—11) +3, 
x! = 12 (n—1) (n—2). 
The curve (Y) has nodes in the points VY, which are images of 
the curves cz possessing a node in a base-point of the net. For, to 
each right line through a point Yp a pencil corresponds, in which 
cp must be counted for two curves with node. 
Each of the remaining nodes of (Y) is the image of a curve «r, 
possessing two nodes. 
So a net N” contains * (n—1) (n—2) (8n*—3n—11) curves with 
two nodes. 
To a cusp of ()’) will correspond a curve replacing in each pencil 
to which it belongs two curves with node. According to a well- 
known property that curve itself must have a cusp. For a definite 
pencil its cusp is one of the base-points; this pencil has for image 
the tangent in the corresponding cusp of (1’). 
So a net Nr contains 12 (n—1) (n—2) curves with a cusp. 
The two properties proved here are generally indicated only fora 
net consisting of polar curves of a et We have now found that 
they hold good for every net, independent of the appearance of fixed 
points 5. 
We can now easily determine the class z of the envelope Z of 
the nodal tangents of the net. 
Through an arbitrary point P of a right line / pass z of these 
