220 
PROFESSOR M. J. M. HILL ON THE LOCUS OF SINGULAR POINTS 
Although, to find the discriminant in the usual way, it is necessary to substitute 
each set of values of X, p in this expression, and then to multiply the results together, 
it is possible to determine more readily which value of p will lead to the factor z 2 by 
expanding p in ascending powers of 2 . 
Now 
p — -— (c + eX) ± - (c + eX) | 1 — 
afjZ 
c (c + e \) 3 2 c 2 (c + eX)* 
Taking the upper sign 
c + eX 
2c (c + eX) 3 
Substituting this value of p in 
[p 3 (c + eX) 2 + 4yp (c + eX) 2 + (3 g 2 — h 2 ) 2 3 ] Z 2 , 
the coefficient of 2 2 in the bracket is (— h 2 ), there being no lower power of 2 . 
This being true for each value of X, the factor 2 4 is accounted for. 
The other value of p will lead to a factor, in which there is a term independent of 2 . 
The elimination will now be completed. 
It is necessary to substitute the values of X and p from 
k = dr 
and 
ap 2 + 2c (c -f eX) p + 2 cgz — 0, 
in 
p 2 (c + eX ) 3 + 4yp (c + cX) 2 + (3g 2 — A 2 ) z 2 . 
Substituting first for p 3 , and multiplying by a, this becomes 
p (c + eX) [— 2c (c + cX ) 3 + 4gaz] + [— 2 cgz (c + eX ) 2 -j- az 2 (3g 2 — A 2 )]. 
Substituting both values of p in this, multiplying the results together, and multi¬ 
plying by a, the result is 
2cgz (c -j- eX ) 3 [—2c (c + eX ) 3 + 4 ^a 2] 3 
— 2c (c + eX) 2 [— 2c (c + eX) 2 -f- 4 got.z] [— 2 cgz (c + eX) 2 + az 2 (3 g 2 — A 2 )] 
+ a [— 2cgz (c +eX ) 3 + az 2 (3g 2 — A 2 )] 2 . 
This reduces to 
— 4c 2 aA 2 (c 4- eX) i z 2 
4- 4cpa 3 (3A 2 — g 2 ) (c 4- eX) 2 z 3 
p a 3(3g 2 - Wfz\ 
