1460 
cutting the parabola twice perpendicularly does not only become 
greater and greater, but it removes farther and farther from view 
and disappears at infinity, which proves that the parabola does not 
possess double normals. 
It is different with the ellipse. Here a calculation, as simple as 
the one just performed produces as equation of the nodal curve in the 
az-plane : 
272? = (x? — a’) (a? —c’), 
a curve of order + consequently, cutting the plane 8 in the vertices 
#2 == +a, and in the foci = + ¢, and being real for |z| < c, and 
iv) >a, while the points at infinity must be determined out of the 
relation 272? = 2‘, so «=O twice, and 2? = 2’. 
The two branches passing through the vertices of the ellipse are 
much like an equilateral hyperbola and form the active part, whereas 
the points at infinity are represented by the minor axis, in. fact 
therefore by a double normal; the branches passing through the 
foei on the contrary, which in the finite are in no way connected 
with the surface and are parasitic as far as they extend, approach 
the z-axis on both sides asymptotically, and have both a point of 
inflection in Z,, as follows immediately from the symmetry with 
regard to 3. In the vertical plane passing through the minor axis of 
the ellipse lies of course as well a nodal curve of order + of which, 
however, only the hyperbolical branches are real. 
§ 6. According to the two preceding sections the intersections of 
the rest-nodal curve with 2 consist of the following groups; 
a. the d nodes of k#; through each of them pass 4 branches; 
6. the x cusps of kv, through each of them pass 6 branches, 
c. the 5u — 3v + 3 — Be — 36 vertices of k”*); through each of 
them passes one branch; 
d. the 2(u—s—2) w— 2e—o) points, in which the y—2e—o tangents 
at £# out of each of the two isotropic points cut the curve; through 
each of them pass 2 branches; 
e. the (v—2«—o)* foci of fr; through each of them passes one 
branch. 
The order of the rest-nodal curve of 2 is therefore: 
d = 4d + 6x + (Sur 31-88-30) + 4 (u-s-2) (w—2e-0) + (v-2e-6)’. 
For the parabola we find from this 5 u—3 r—3 0 + (r—o)*? = 
=—10—6—3+1—=2, for the other conics 5u—3r+v?=10—6+4—8 
(which is evidently correct according to what precedes), and for the 
1) Anw. Cykl. p. 19. 
