822 PROFESSOR CHRYSTAL ON THE 



az i + dijz : + 2xf- = (4), 



az* + dxz 2 + 2x 2 y = (5), 



3a{x + y)z 2 + 2dxyz = (6). 



Setting aside alternatives which lead to 2 = 0, x = and 2 = 0, y = (double points 

 already existing), \vc find that (4) and (5) are equivalent to 



az* + dxz 2 + 2(h?=:0 (7), 



x-y = (8). 



The admissible solution or solutions of (7) and (8) must also satisfy (6). Hence, setting 

 aside x = 0, as before, we must have 



Saz+dx — (9). 



Substituting the value of z/x given by (9) in (7) we get for the necessary and sufficient 

 condition required 



# + 27« 2 = (10), 



which will be satisfied if we take a = c 3 , d= —3c" 2 . The coordinates of the third node 

 are then x = y = c, z=l. 

 For these values 



reduces to 



2c 2 £ 2 + 2<?y + 6e*f» - 6c* v £ - Q<?& + 2c^ v = 2c 2 { £ - m + (o> - 1 )c£} { £ - o, 2 > 7 + (w 2 - 1 )c£} , 



where gj is an imaginary cube root of unity. The node is, therefore, an acnode. 

 The queer tic family, 



cHx + yy-3chnjz 2 + xY = (11), 



therefore, has fixed cusps at (I, 0, 0) and (0, 1, 0), the cuspidal tangents being y=0 and 

 x = 0, a movable acnode x = y = e, z=l ; each curve passes through the intersection of 

 its cuspidal tangents and has a fixed tangent at that point. 



As a verification of the result already predicted, we may calculate the coordinates of 

 the intersections of two consecutive curves of the family. These are given by 



c%x + y)z :, — 3c 2 xyz 2 + x 2 y 2 = 0, ) . _ 



c(x + y)z :i — 2xyz 2 = \ 



Now (12) is equivalent to the system 



which gives (1, 0, 0) and (0, 1, 0) each four times, together with 



