£>-DISCRIMINANT OF A DIFFERENTIAL EQUATION OF THE FIRST ORDER. 809 



If we use our original notation for the coefficients, and remember that, since (x, y) is 

 supposed to be a point on the ^-discriminant, we have <p = 0, <p 2 r—0, we get 



a = = , d = }(V0* + 2Xfji<p xv + (ffa,) , 



h = ^<P* + M0« , e o = ^ 2 0*» + V(0M/ — <P*x) — M 2 0x;/ > 



c o = - M0* + X0„ . /o = K^-Vw — 2 V0*» + M 2 0«) , 



*i = <*& = (), « 2 = K^0 2) + a 2 ^) = |a 2 0M< • (23). 



&i = a(X^ par -f-^ OT ), 

 c i = a( — /*&>* + X0 W ) , 



The general form of the condition 6 = is, therefore, X^ + ju, <£,, = (), i.e., <p x +p<p y — 0- 

 Hence the necessary and sufficient condition in the most general case that the -^-discri- 

 minant furnish an envelope singular solution is that the three equations in x, y, p 



= 0, 0, = O, <!> x +p<j>* = (24) 



have a onefold infinity of common solutions. These common solutions furnish the 

 envelope. 



If, as will in general happen, the three equations have only a finite number of 

 solutions in common, then the corresponding points will be trie-points at which the tivo 

 touching branches touch the -^-discriminant locus. 



Since b = 0, c = are equivalent to (X 2 + jti 2 )<£ z — 0, (X 2 + ju, 2 )^ = 0, that is, to (p x = 0, 

 <p y = 0, it follows that, in the most general case, the ^-discriminant furnishes a tac-locus 

 when the four equations in x, y, p 



= 0, 2 =O, I = O, 0„ = O (25) 



have a onefold infinity of common solutions, the corresponding points being the 

 tac-locus. 



If, as may happen in special cases, the four equations have a finite number of 

 common solutions, the corresponding points are tac-points, at which the touching 

 integral branches do not in general touch the ^-discriminant, but at ivhich the ^-dis- 

 criminant itself has double points. 



* It is easy to see in another way that ^ x +p<f> !) = is simply the condition that the ^-discriminant is an integral of 

 the equation : for at any point of the ^-discriminant dx : chj is given by 



<f> x dx + <f> v dy + <p p dp = 



4> px dx + <t> pv dy + <t> pp dp = . 

 Since 4> P =0, the first of these equations gives 



<px + <p y dyldx=0. 



The condition that the p-discriminant be a solution is p=dyjdx ; hence, 



( t>z+p<P y =0. 



