154 
PROFESSOR M. J. M. HILL OH THE LOCUS OF SINGULAR POINTS 
In this case, A may be written 
R/ (x, y, z, o. l )f(x, y, z, a 2 ) = K/,/ 2 
• • • • ( 51 ) 
where a„ a. 2 are the roots of (2) which become equal when x = £ y = y, z = £. 
Therefore, 
, a/-® , I? 9 ! 
dx ~ + Uj2 \Dx + Da, dx 
fR/l 
1 /„N 
Dx+Da.,dxj ' ' ( 5 '* 
13 f Cd 
Now if it be assumed (see immediately below, under C) that the terms f 2 
|~w» 0(/^ 
f, 2 0 2 vanish, then when x — y = y, z = £, it follows that a, = a 2 = a, and, 
therefore,/, = 0 ,f 2 = 0, and = 0. 
Cl- -1 1 3 a 0A 
Smnlarlv, x- = 0, x- = 0. 
dv ■ dz 
Therefore, A contains E 3 as a factor. 
(C.) Examination of the term f 
v ' J Da, dx ’ 
Taking 3 a,/dx from (47) this term becomes 
Now f is of the form 
DA Df_ / py t 
Da, Da,Dx/ Da, 2 ' 
A (a — a,) (a — a 2 ) (a — a 3 ), 
where a„ a 2 , a 3 all become equal to the same thing as a ]f a 2 when x = y = y, z — £. 
Hence, taking as infinitesimal of the first order the difference in the values of the 
parameter a at the points £ y, £ and £ -f- 8^, y -j- Sy, £ + 8£, it follows that f 2 is of 
the third order of small quantities, Df/Da, of the second, and D'fjDaf of the first. 
Hence, assuming that Dy/DcqDa: is not infinite, it follows that the term under 
investigation is of the fourth order, and therefore vanishes ultimately. 
Example 4.-— Envelope Locus when two consecutive Characteristics coincide. 
Let the surfaces be 
</> (a, y, z) + bp (x, IJ, z) — aj = 0. 
(A). The Discriminant. 
The discriminant is found by eliminating a between the above and 
■— 3 [x// (cc, y, z) — «] 3 = 0, 
Hence it is [f ( x , y, z)] 3 . 
