671] 
ON A SIBI-RECIPROCAL SURFACE. 
253 
where (agh) is used to denote the determinant 
ch, gi, 
к 
a2, g 2 , 
h 2 
a s , g 3 , 
h 3 
, and so for the other 
symbols. Considering the reciprocal of the line L in regard to the quadric surface 
X 2 + Y 2 + Z- + W 2 = 0, the six coordinates of the reciprocal line are 
/, g, h, a, b, c, 
and it is hence at once seen that the locus of the reciprocal line is the quadric surface 
obtained from the equation T = 0 by interchanging therein the symbolical quantities a, b, c 
and f g, h: viz. writing also (£, 77, £, w) in place of (x, y, z, w), the new equation is 
T> = (flc) f 2 + (ffca) V 2 + (hab) £ 2 + (fgh) w 2 
+ [(fgb) - (¥ c ) ] + [(№) + (hca)] vZ 
+ [(ghc) - (fga)] V o> + [(gbc) + (fab)] & 
+ i(¥ a ) ~ (gM)] + [(hca) + (gbc) ] %y = 0; 
or, what is the same thing, this equation T' = 0 is the equation of the original quadric 
surface (the locus of L) expressed in terms of the plane-coordinates £, y, £, a. 
Now considering each of the quantities a lt b 1} c 1} f, g x , h 1} a 2 , b 2 , etc., a 3 , b 3 , etc., as 
a given linear function of a variable parameter X, say a x = a x + afX, b x = b x + W'X, etc., 
the equation T— 0 takes the form 
AX 3 + 3 BX 2 + SCX + D = 0, 
where А, В, C, D are given quadric functions of the coordinates x, y, z, w\ and the 
envelope of the quadric surface T= 0 is Herr Rummer’s surface of the eighth order 
(AD - BG) 2 -4 (AC-B 2 ) (BD - C 2 ) = 0. 
In like manner the equation T' = 0 takes the form 
A'X 3 + 3 B'X 2 + 3 G'X + D' = 0, 
where А', В', G', П are given functions of the coordinates f, у, £ со; and we have 
(A'D' - B'C') 2 - 4 (A'G' - B' 2 ) (B'D' - G' 2 ) = 0, 
as the equation of the reciprocal surface; or (what is the same thing) as that of the 
original surface, regarding £, y, £, tu as plane-coordinates. 
In regard to the foregoing equation T — 0, it is to be noticed that, if a 1} b x , c 1} 
/1, 9i, K; a 2 , b 2 , etc., a s , b 3 , etc., instead of being arbitrary coefficients, were the 
coordinates of three given lines L x , L 2 , L 3 respectively; that is, if we had 
«1/1 + b x g x + cJh = 0, 
«2/2 + b 2 g 2 + cX = 0, 
a 3 f 3 + b 3 g 3 + cju, = 0,