454 Mr Dixon, Note on Plane Vnicursal Curves. 



Note on Plane Unicursal Curves. By A. C. Dixon. 

 [deceived 28 March 1904.] 



When a plane curve U of degree n has its full number 

 ^(n — l)(n — 2) (=£>), of double points, any curve / of degree m, 

 passing through all the double points, meets U in mn — 2p other 

 points, all of which may be arbitrarily chosen. But if / does not 

 pass through all the double points the rest of the common points 

 are restricted by certain conditions. 



Since U is unicursal, we have at every point of it 



x = X.0/Z0, y=Y0/Z0, 



where X6, Y0, Z0 denote quantics of the nth. degree in the 

 parameter which varies from point to point. A node arises 

 when two values of 0, say a and ft, give the same values of x 

 and y, that is, when 



Xa _ Fa _ Zu 



Xft~~ Yft = Zft W ' 



To find the intersections of TJ, f we substitute the above values 

 in terms of for x, y in the equation /=0, and after multiplying 

 by (Z0) m get an equation 



F(0)=f(X0, Y0,Z0) = O. 



This is of the degree mn, and its roots are the values of which 

 belong to the points on U that also lie on /. Let these be 

 6i, 02 ... mn - Then F(0) = CTI (0 - r ), where C is a constant 

 coefficient. 



Now / (X0, Y0, Z0) is a homogeneous m-ic in X0, Y0, Z0, 

 and therefore 



f(Xa, Ya, Z a )/f(X/3, Yfi, Z/3), 



in virtue of (1), is independent of the coefficients in /and equal to 

 (Za/Z(3) m . 



We have then 



U(a-0 r ) F(a) (Za> 



U(/3-0 r ) F(/3) \Zj3j 



This condition must be satisfied by 1 , 2 ... mn if a, ft are 

 the parameters belonging to any node. Since there are 

 ^(n — l)(n — 2) nodes we have the same number of conditions 

 to be satisfied by 1} 2 ... 0^ mi . 



If/ passes through r of the nodes each of these counts as two 

 of the intersections, and the number of conditions to be satisfied 

 by the rest is ^(n — l)(n — 2) — r, since of those above written 

 r are satisfied already. 



