412 ON A SIMPLE PROOF OF LAMBERT'S THEOREM. [52 



The coordinates of any point in the hyperbola referred to its axes 



may be represented by 



x = a csh (u), 



y = a*/e* I snh (u). 



If u, u' denote the values of u corresponding to the two extremities 

 of the arc, we have 



r = a(e csh (u) I), r' = a(e csh (u'} 1 ), 

 k- = cr [csh (u) csh (u')J + a 2 (e 2 1 ) [snh (u) snh (u')J ; ' 

 r + r' I , u + u'\ , u u' 



/ , i* ~r w \ it* t* 



or = I e csh - I csh I , 



2a \ 2 I 2 



.7 = snh 2 e~ csh 2 1 . 



Also twice the area of the sector limited by r and r' 



= cf v/cr 1 [(e snh u u) (e snh u' w')] 



[/ /\ / I 



/ y/ "T* *?/ \ ?v ^~* ?f 



2 ( e csh - ) snh (u u'} \ , 

 \ 2/2 J 



and twice the area described in a unit of time is 



\/fta (e 2 l). 



_ , , , ,. 



Hence *-( 2 ecsh- -snh- --(w-V) ; 



XA 1 / 



and therefore if a be given, then r + i 1 ', k, and i are functions of the two 

 quantities e csh - and w u'. 



n i I fly* 



Let u u'='2a, and ecsh--- = csh(/3), which is always possible since e 

 is greater than 1. 



Then T -^- = csh (0) csh (a) - 1 , 



.Zft 



= snh () snh (a) ; 

 therefore - = csh (/8 + a) 1 , 



^Cft 



r + r'-/; 

 and = csh ( a) 1. 



2a ; 



