460 
A SUPPLEMENTARY MEMOIR ON CAUSTICS. 
[359 
12. It is to be observed that for m < V2 the nodes are both imaginary; for 
m = V2 they coincide together at the point ; for m > VI they are both real: 
it is to be further noticed that 
node, x = \ (m + Vm 2 — 2), corresponds to cos 0 = ^ (m — Vm 2 — 2), 
where (m being > VI) the point (cos 0, sin 0) is a real point on the circle a? + y 2 = 1; 
in fact for m< § (that is, m = VI to m = f) we have |(m —W—2)<|m, that is, 
cos 0 < §; but m = or > f, then cos 0 = | (m — Vm 2 — 2) = —^ — is = or < |, and 
m + v m 2 — 2 
node, x = ^ (m — Vm 2 — 2), corresponds to cos 0 = ^ (m + Vm 2 — 2), 
where (m being > VI) the point (cos 0, sin 0) is a real point on the circle x- + y 2 = 1 so 
long as m is not > f, that is, from m = VI to m = f; but if m > f, then the point in 
question is an imaginary point on the circle—whence also the node x = ^ (m — Vvn- — 2) 
is an acnode or isolated point. 
In the case m = f we have 
node, # = 1, corresponding to cos 0 = | or 0 = 60°, 
„ a; = „ cos 0=1 or 0 = 0°, 
the last-mentioned point x — ^ being in fact the point of union of two cusps in the 
case m = f now in question. Hence in this case we have at (x = y = 0) a triple 
point equivalent to two cusps and a node; visibly, there is only a single branch 
cutting the axis of x at right angles. 
In the case m = V2, the nodes coincide as above mentioned at the point x = j- 
on the axis; for this value of m the coordinates of the cusps are 
x = VI(=||.-t- VI, which is < 1 -T- VI) ; y=± Vt- 
13. Starting from the equation 1024 (x — m) 2 U = S 3 — T 2 = 0, it is clear that the 
cusps are included among the intersections of the curves S ■— 0, T = 0; these two 
curves intersect in 24 points which lie 9 + 9 at the circular points at infinity, 2 + 2 at 
the points x — m,y 2 — l=0, and 1 + 1 are the cusps, or points x = m — fj m 3 , y 2 = (1 — f m 2 ) 3 . 
To verify this, writing for a moment 
S' = (x 2 + y 2 — to 2 — l) 2 + 6m (x — m), 
T = 2 (x 2 + y 2 — m 2 — l) 3 + 18m (x — m) {x 2 + y 2 — m 2 — 1) — 27 (x — m) 2 , 
then we have 
T' — 2 (x 2 + y 2 — m 2 — 1) S' = 6m (x — m) (x 2 + y 2 — m 2 — 1) — 27 (x — m) 2 , 
= 3 (x — m) {2m (x 2 + y 2 — m 2 — 1) — 9 (x — m)};