240 
PROFESSOR M. J. M. HILL ON THE LOCUS OF SINGULAR POINTS 
Similarly 
Hence 
if 
2e (y - b) [Sar ( x - a) + ey (y - b )] , 
| ^ = 2 K.£ n ~ l (ySvr — fiSey) — 2/^S (ASx — pey) , 
= 2^’C" -1 (ae 3 3/ - P$cx) + 2#e (A&c - gey). 
DJ> / f]/ _ / HZ 
D.c / Ite Dy / I)?/ ’ 
(a — «) [&£" _1 (aey — £Sz) + # (AS® — ^ey)] 
= (y - b) \k£ n ~ l (ySx - fey) - h (hSx - gey)]. 
Making use of the values of a: — a, y — b, which satisfy the equations which have 
been taken to determine them, and which are solved above in (A), it is necessary to 
show that 
[(A/3 — gy) £ + ey (hSx — gey)] [k£ u ~ l (aey — fex) + g (hSx — gey)] 
= [(gfi — ha) £ — Sx (hSx — gey)] \Jc£ n ~ l (ySx — fey) — h (hSx — gey)], 
i.e., to show that 
K n [(A/3 - gy) (aey - fex) - (g/3 - ha) (ySx - fey)] 
+ k£ n ~ l (hbx — gey) [a eh/ 2 — 2 feexy + ySbc 3 ] 
— £ (hBx - gey) ( ah 3 — 2/3 gh + yg 2 ) — (h$x — geyf = 0, 
^.e., 
(hSx - gey) </> 
= 0. 
Hence this is satisfied. 
Therefore 
Dip / Df _ D <f> 
/ W 
Dx / I)x Dy 
/ % 
It remains to prove that each of the equal quantities 
(x — a) [&c (x — a) + ey (y — 6)1 (; y — b) [&c (x — a) + ey (y — 6)] 
k£ n ~ J (ySx — fiey) — li (hSx — gey ) k£ n ~ l {aey — fex) + g (hSx — gey) 
is equal to 
a(x — «) 2 + 2/3 {x — a) {y — b) + y (y — b) 2 + 2 g (x — a) + 2 h(y — b) + nk£ n ~ l 
nk£ ,l ~ l (ay — fe) + (n — 1) k£ n ~ 2 («e 2 y 3 — 2 fthexy + yS 2 x 2 ) — ( ah 2 — 2{3gh -f- yg 2 ) 
Multiply numerator and denominator of the first ratio by §x, of the second by ey, 
