212 



Mr. W. H. L. Russell. 



fDec. 22; 



Let A be the origin, Adf the axis of (x), Aec the axis of y. Let 

 y = lx be the equation to Aab, y=ax + b the equation to cd, y—nxArh 

 the equation to ca, y=cx + e the equation to ef, y=mx + e the equa- 

 tion to be, then we may find the equation to ad, 



y(l — n + a) — lax — lb = 0; 



and similarly the equation to bf, 



y(l—m+c) —lcx — le—0. 



In this way it will easily be seen that the six points, a, b, c, d, e, f, are 

 completely determined, and consequently the equation to a curve of 

 the third degree passing through them (see Salmon, " Higher Plane 

 Curves," Art. 162) is 



ab .cd . ef+ .ac .be . df+ $ . ad .bf . ce + yjr . ae .bd . ef=0. 



But since ab, df, ee pass through the origin ^=0, and the equation 

 becomes 



ab .cd . ef+0 .ac .be . df+<p . ad . bf. ce=0, 



and consequently writing down ab, cd, ef, ac, be, df, ad, bf, ce, as given 

 above, we have as the equation of the required cubic — 



(y — Ix) (y—ax— b) (y — cx — c) + Oy (y —nx — b)(y— mx — e) 



+ (/)x(y(l— n + a) — lax— lb)((l— m + c)y — lex— le)=0 ; 

 differentiating this equation, and putting x=y = to determine the 



value of — at the origin, we have 



dx 



^(1 + 0)=Z_0Z3. 



dx 



dhi 



Differentiating again, and putting —d-=0, x=y = 0, since the origin 



