76 ON A CERTAIN ENVELOPE DEPENDING ON A [396 
where the reciprocant in question may be calculated from the before mentioned table 
67, viz. multiplying by 3 in order to avoid fractions, the coefficients of the table are 
(a, b, c, f, g, h, i, j, k, l ) 
= (0, 0, 0, c, a, b, b, c, a, —a — b — c) 
respectively, and for the facients (£, y, £) of the table we have to write (x, y, z). 
The expression of the reciprocant is 
= b 2 c 2 x 6 + c 2 a 2 y 6 4- a 2 b 2 z 6 + &c., 
and dividing by (x + y + zf we have the equation of the envelope in the form 
b 2 c 2 x A + c 2 a 2 y* + a 2 b 2 z i + &c. = 0, 
which must of course be identical with the former result 
bcyz (cy — bz) 2 + cazx (az — cx) 2 + abxy (bx — ay) 2 4- &c. = 0. 
Instead of discussing the curve of the third class 
«£ (v - ty + h (£-£) 2 + cf - y) 2 = o, 
it will be convenient to write (x, y, z) in place of (£, y, f), and discuss the curve 
of the third order, or cubic curve 
U — ax (y — z) 2 +by (z — x) 2 + cz(x — y) 2 = 0, 
which is of course a curve having a node at the point (# = ?/ = z), or say at the 
point (1, 1, 1), and having therefore three inflexions lying in a line. The equation of 
the tangents at the node is found to be 
a (y — z) 2 + b (z — xf 4- c (x — y) 2 = 0, 
that is, at the node the second derived functions of U are proportional to 
(b + c, c + a, a+b, —a, —b, — c). 
The equation of the Hessian may be found directly, or by means of the table, 
No. 61, in my memoir above referred to. It is as follows: 
(b + c) {a (6 + c) + 26c] x 3 
+ (c + a) {6 (c + a)+2ca) y 3 
+ (a + b) {c (a + b)+ 2ab] z 3 
— (3a 2 + 2be + 2ca + 2ab) (cy 2 z + byz 2 ) 
— (3b 2 4- 2be + 2ca + 2ab) (az 2 x + czx 2 ) 
— (3c 2 + 2be 4 2ca 4 2ab) (bx 2 y 4- axy 2 ) 
4- {4 (be 2 4- b 2 c 4- ca 2 + c 2 a + ab 2 + a?b) + 6abc\ xyz = 0.