J9-DISCRIMINANT OF A DIFFERENTIAL EQUATION OF THE FIRST ORDER. 821 



any consecutive two) curves of this family intersect each other six times at B, six times 

 at C, and twice at A. The remaining two points of intersection for two consecutive 

 curves are infinitely near the double point. Hence the family has no proper envelope ; 

 but merely (so far as movable intersections are concerned) a node locus which counts 

 twice. 



Moreover, the quartic does not degenerate, as we shall show by working out its 

 equation in the special case where the point A is the intersection of the cuspidal tangent 

 at B and C. 



Let us use triple coordinates and take A, B, C for triangle of reference : and let us 

 write the equation to the quartic 



a z* + (ax + hjy + (ex 2 + dxij + ey*Y + (f& +gx 2 y + hxy 2 + kf)z + u 4 = (1). 



Since z = 0, which joins the cusps, must give x~ = and y 2 = 0, we see at once that 

 w 4 =x 2 i/ 2 (dropping the useless constant). Also, since x = and y = are cuspidal 

 tangents, x = must give z 3 = 0, and y = must give z 3 = Q ; hence we must have e = 0, 

 h=Q, c=0,/=0. 



The condition that the curve pass through (0, 0, 1) is obviously « = ; and, if we 

 take x + y = to be the fixed tangent at (0, 0, 1), it is easily seen that we must have 

 a = b. Our equation now reduces to 



We have 



f(x, y, z) = a(x + y)z 3 + dxyz 2 -f (gx 2 y + hxy 2 )z + x 2 y 2 = . . (2 ). 



/, = az* + dyz* + (2gxy + hf)z + 2xy 2 ; 



f v = az A + dxz 2 + ( 2hxy + gx 2 )z + 2x 2 y ; 



f z =Sa(x+y)z 2 + 2dxyz + (gx + hy)xy; 



f a = 2gyz + 2y 2 ; 



f uu = 2hxz + 2x 2 : 



f zz = 6a(x + y)z + 2dxy ; 



f yz = 3az 2 + 2dxz + 2hxy +gx 2 ; 



f zx = 3az 2 + 2dyz + 2gxy + hy 2 ; 



fry = dz 2 + 2(gx + hy)z + 4xy . 



It will be seen that the conditions for double points at (0, 1, 0) and (1, 0. 0) are 

 satisfied. In order that these points may be cusps, it is farther necessary that 

 1f xx ^ 2 + 2lfy 2 ^ shall be a complete square in £, »?, £ at the two points. Now this 

 function reduces to 2£ 2 + 2A££ and 2>? 2 + 2(/>?^ at (0, 10) and (1, 0, 0) respectively; 

 hence we must have # = 0, /i = 0. 



Our equation is now reduced to 



a(x + y)z* + dxyz 2 + x 2 y 2 = (3); 



and it remains only to find the condition that the quartic have a third node. The con- 

 ditions for a node are 



