( 654 ) 
and from this ensues the equation 
my.” —(2emn + 1l)u+ no = 0, 
giving the correspondence between g and u. Each value for g furnishes 
two values for u, and for each value of u we find one value for g. 
The curve /, can therefore be determined by means of an invo- 
lutory pencil of conics (1) and a projective pencil of rays (2). It 
is easy to see, that these two pencils have «7. =0 as a corresponding 
common element and that consequently the generated curve of order 
five breaks up into 2,=O and the curve 4, All those conics of the 
pencil (1) have two tangents 2,0, z, == 0 in common, and the 
vertex (/?) of the pencil of rays is situated on the first of these 
tangents. 
3. The points of contact D, and D, on the double tangent d, 
are projected out of the cusp A by two right lines, the equation of 
which will be obtained by eliminating x, out of (4,) and (d): so from 
[Amr + wx, (wx, — ba,)]? — 16m’x, a, (aa, — bz,) = 0 
we find 
AD, , AD, = Am?’ — a, (a's, — be) = 0. 
By eliminating x, (a’r,—br,) out of the latter equation and (#,, 
we have 
(mx,* + nv,x,)? — 4m?a,‘ = 0 
or 
maz,” + nv,a, + 2ma,? — 0, 
therefore 
NED, ME Zen gets so tN 
net, +3mr,?=—0 . BO ee os. ee 
On these conics lie the points of intersection of 4, with the pair 
of lines AD,,AD,. The first conie (4) gives by combination with 
the equation (#,) again the double tangent (d), and the second conie 
(5) furnishes by eliminating 7,’ out of (4,) and (5) the equation 
om (2nz,2,) J Onze, (a's, — Bep 
or 
4mnz, -|- 3 (ea, bej SE - . -: ee 
On this line lie the two points D',, D, of intersection of £, with 
the projecting rays AD,, AD,. 
The line D',D', cuts the curve £, again in two other points L,, Z, 
and bears four projecting rays out of the cusp A. This quadruple 
of rays will be obtained by eliminating iv, out of (6) and (4:,), namely 
