Mr. T. S. Davies on Pascal's Mystic Hexagram. 39 



The general form of the equation of the conic section re- 

 ferred to the axes O B, O D is 



ay* + bxy + cx* — dy — ex +/ = (1.) 



The equations of the other four sides of the hexagon will be 



(A F) x x (y - a) = x{y x -a) (2.) 



(BE) y(* 2 -/3) = y 2 (*-/3) (3.) 



(CF) y(x l -y)=y l (x-y) (4.) 



(DE) * 9 (y-*) = *(y 8 -*) (5.) 



Let (Xj Y x ) be the intersection G of the lines A F, B E de- 

 noted by equations (2.) and (3.) ; and (X 2 Y 2 ) be that H of 

 C F, D E denoted by (4.) and (5.) : then we readily find 



1 *!& - (*2 - 0) 55 - «) 



2 a£JI - K - y) (y« - 8 ) 



(a *, + /3j/ x -«/3) ?/ 2 



Y x = 



Y 2 = 



*\V% - (*i - Z 3 ) (3/1 - «J 



(&^ 2 + 7.y 2 -7 g ).yi 

 •*' 2 yi - (*i - 7) (y 2 - 8). 



(6.) 



Write D t and D 2 for the denominators of X 15 Yj and X 2 , Y 2 ; 

 and find the values of X 2 Yj and X ll Y 2 , disregarding for the 

 present the common denominator D x D 2 . To effect this, ac- 

 tually multiply the values of X 2 Y x in (6.), and likewise those 

 of X t Y 2 : then these two expressions are respectively, 



{byy*? + (* s + &y)xzy*+ a ^x 2 2 



DiD 2 



■ fiy(a + 'S)y 2 -a'S((Z + y)x 2 + u[Zytyx 1 y l 



D,D, 



= X,Y, 



X X Y 3 



Now since A, B, C, D are the points of intersection of the 

 curve denoted by (1.) with the axes of coordinates, we get by 

 putting x and y successively equal to zero, the values of a, /3, 

 7, 8 in terms of the coefficients of (J.), as follows: — 



_ d + Vd?-baf_ & _ 



2a 



8 = 



d- VcP-baf 



2« 



y = 



2c 



Insert these values in the former of the equations marked (7.): 

 then there results as the value of X 2 Y x . T) x D 2 , 



ed- Vtf*-*qf)(e*^icf) 



(7.) 



Wi 2 - 



2/ 



X\V\ + cx^—dy —ex x +/}* 2 y 2 . 



