ON POLYZOMAL CURVES.
529
414]
viz., the tangents from I are £ = 9az, where 6 is any root of this equation. Similarly,
if we have
4k (<f> 2 -1) [ka 2 ft 2 (p - 1) - b/3c}> - c} + (ecf> + |J = 0,
the tangents from J are 77 = <f)/3z, where c/> is any root of this equation.
147. The two equations may be written
f 24k 2 a 2 ft 2 ,
— 6kaa,
— 8k 2 a?ft 2 — 4 kc + e 2 ,
Gaa.
24k 2 o. 2 ft 2 -f- 24>kc + 6
h <K l) 4 = 0,
+ Se-,
24 k 2 o?ft 2 ,
— Gkbft,
— 8k 2 a. 2 ft 2 — 4 kc + e 2 ,
6 kbft
+Se
i<t>, 1)‘ = 0,
/3’
24 k 2 a. 2 /3 2 + 24/cc + 67^
ft 2
which equations have the same invariants; in fact for the first equation the invariants
are found to be as follows, viz., if for shortness G = — 8k 2 a 2 ft 2 — 4>kc + e 2 , then
I = oIGttfft* + 57GJc 3 ca 2 ft 2 + lUk 2 (a 2 a. 2 + b 2 /3 2 ) + 72kab + SC 2 ,
J=G {576/ i 2a 4 /9 4 + o7GMft 2 + 144/c 2 (a 2 a 2 + b 2 /3 2 ) + SGkeaft - C 2 }
— 864<k 3 eaba. 2 ft 2 — 216/r 2 e 2 (a 2 a 2 + b 2 f3 2 ) — 21 Gk 2 a 2 b 2 ,
and then by symmetry the other equation has the same invariants. The absolute
invariant / 3 vJ 2 has thus the same value in the two equations, that is, the equations
are linearly transformable the one into the other, which is the before-mentioned
theorem that the pencils are homographic.
148. The equations will be satisfied by 9= cf) if only aa = bft, that is, if a, b = mft, mot;
or by $ = — </> if only aa. = — 6/3, that is, if a, b — mft, — mat: the equation of the curve
is in these two cases respectively
k (£ 2 — a 2 z 2 ) (7f - ft 2 z 2 ) + ez 2 tji7 + rnz 3 (ft$j + arj) + cz 4 = 0,
k (£ 2 — a 2 z' 2 ) (r) 2 — ft 2 z 2 ) + ez 2 grj + mz 3 (ftij — arj) + cz* = 0.
If to fix the ideas we attend to the first case, then the equation in 9 is
24>k 2 a 2 ft 2 ,
— Gkmcnft,
- 8k 2 a 2 ft 2 — 4>kc + e 2 , $9, 1) 4 =0;
Gkmaft 4- 3me,
24& 2 ot 2 /3 2 + 24/rc + Gin 2
and we may take as corresponding tangents through the two nodes respectively £ = 9az,
77 = 9ftz\ the foci A, B, C, D, which are the intersections of the pairs of lines (£=-9 x az i
c. vi. 67
