Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., sadlerian professor of pure mathematics in the University of Cambridge (Vol. 5)

373] 
ON A SPECIAL SEXTIC DEVELOPABLE. 
515 
Consider in like manner the intersection of the torse with the surface ae — Xbd = 0, 
where A is a given constant coefficient; we have 
(iae — 4 bd) 3 = (A — 4) 
= (A — 4) 3 b 3 d 3 = ^ . 4)3 
aeb 2 d?, 
and therefore 
that is 
which gives 
if 0 l , 0., are the roots of 
(A - 4) 3 aeb 2 d 2 - 27A (ad 2 + b 2 e) 2 = 0, 
27Aa 2 d 4 + [54A - (A - 4) 3 ] ab 2 d 2 e + 27A6 2 e 4 = 0, 
ad 2 — 0J) 2 e = 0, or ad 2 — 0J> 2 e = 0, 
27A0 2 + [54A - (A - 4) 3 ] 0 + 27A = 0. 
The surfaces ae— Xbd = 0, ad 2 —0Jre = 0 have in common the two lines (a = 0, 6 = 0) 
and (d = 0, e = 0), and they intersect besides in a quartic curve. And so for the 
surfaces ae — \bd = 0, ad? — 0.h 2 e = 0. That is, the surface ae — \bd = 0 intersects the 
torse (ae — 4>bd) 3 — 27 (— ad 2 — b 2 e) 2 = 0, in the line a = 0, 6 = 0 twice, in the line 
d = 0, e = 0 twice, and in two quartic (excubo-quartic) curves. The two quartic curves 
become identical, if 
(54A) 2 = {54A - (A - 4) 3 } 2 , 
that is 
and therefore, if either 
±54A= 54A — (A — 4) 3 , 
(A - 4) 3 = 0, 
which gives the cuspidal curve; or else if 
(A - 4) 3 - 108 A = 0, 
that is 
A 3 - 12A 2 - 60A - 64 = (A + 2) 2 (A - 16) = 0. 
(A + 2) 2 =0 or A = — 2 gives the nodal curve: A —16=0 gives ae— 16bd = 0, a surface 
which intersects the developable in the line a = 0, 6 = 0 twice, in the line d — 0, e = 0 
twice, and in the two coincident quartic (excubo-quartic) curves given by the equations 
ae — 16bd=0, ad 2 — b 2 e = 0. As a verification, I remark, that the surface ad 2 — b 2 e = 0 
combined with the developable gives 
(ae — 46d) 3 — 27 (ad 2 4- b 2 e) 2 = (ae — 46cZ) 3 — 108 ab 2 d 2 e = 0, 
that is (ae+ 2bd) 2 (ae — lGbd) = 0, or it meets the developable in its curve of intersection 
with ae+2bd = 0 twice, and in its curve of intersection with ae — l6bd = 0; that is, 
in the line a = 0, 6=0 three times, in the line d = 0, e = 0 three times, in the nodal 
quartic ae+2bd = 0, ad 2 — b 2 e = 0 twice, and in the quartic ae — lGbd = 0, ad 2 — b 2 e = 0 
once; 3 + 3 + 8+ 4 = 18, the order of the complete intersection. 
Greenwich, January 4, 1864. 
65—2
	        
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