480 DR G. PLARR ON THE DETERMINATION OF 



We see that A and B are of the third degree in t, and C and D of the second degree. 



The elimination of s gives us 



BC+AD=0 (V) 



consequently of the fifth degree in t. We must remember that we have put 



(-D)=-X 1 Hu^-u^{+L 1 +L ) ; 



Let us state the coefficient of t 5 in the equation (V.), 

 The coefficients of t z in A and B are respectively, 



mA, = tt 2 +(b 2 -u 2 = b 2 , 



inB ' = (^) Xl 

 The coefficients of t 2 in C, D, are respectively, 



in C , = u 2 [u 2 + (b 2 - u 2 )] xl = u 2 b\ 



inD ' = (jp) xl - 

 Therefore the coefficient of t 3 x t 2 or f in (V.) will be 



As the terms in A beside the term b 2 f are all divisible by b 2 , owing to L = b 2 (a 2 + u i ), 

 and as the same remark applies to C in respect to bH 2 , and its other terms, we shall put 



A = b 2 A! 



C = u 2 b 2 C 

 a*¥ 



Then the equations (III'), (IV) (by dividing the first by b 2 , the second by 6V) may 

 be put under the forms, 



A' + ^B'=0 . . . (HI") 



C'-~D'=0 .... (IV") 



a 4 6 6 



and the s eliminated in 



B'C'+AT>' = 0, (V") 



where A', B', &c, are of the form, 



A=A +A 1 *+A 2 < 2 +* 8 



B=B +B 1 t + B 2 t 2 + t* 

 G = C + C l t + t 2 

 T> = J) + T> 1 t + t 2 . 



