164 



CONIC SECTIONS. 



_ . Tan. A u— Tan. A z 



Tan. A (m-—z)=: , ~ j . ^ , , 

 2 v 1 + Tan. 4 w Tan. A z 



therefore, 



Sin. w + Sin. z_ 



Sin. u — Sin. z 

 (Tan. \ u + Tan. jr z) ( 1 + Tan . A, u T an. A a) 

 (Tan. I «— Tan. A z) (1— Tan. | ?* Tan. § z)' 



Suppose now that u is a right angle, so that Sin. u—\, 

 and Tan. A w= 1 . In this case the formula becomes 



1+Sin.z _ (1+Tan.^z)' 

 HTSin. z ~ ( 1 —Tan. | z) J * 



By comparing this formula with those which express 

 the relation of t, V, I", &c. to each other, viz. 



1 w 



.i."(i_iY 



1 +a_( 1 +a) 



!_^~ fl_£V 

 a \ a J 



&c. 



it will immediately appear, that the relation of — to 



t' V t" 



— , also of — to — , &c. is the very same as that of 

 a a a 



Sin. z to Tan. A_ z. 



Hence it follows, that if we find from the trigono- 

 metrical tables the series of sines 



Sin. z, Sin. z', Sin. z", &c. 



such that 



Figs, leo, 

 101. 



Sin. z =— , Sin. z'=Tan. A z, Sin. s"=Tan. A *', &c. 



and suppose 5 to express, as before, the hyperbole sec- 

 tor, then 



Vi 



a— I (^ Sin. z' + i Sin. z" + i Sin. z"'-f, &c.) 



The series in the denominator evidently approaches 

 eontinually to a geometrical series, of which the com- 

 mon ratio is f ; and because, in an infinite geometrical 

 progression, whose common ratio is \, the sum of all 

 the terms following any assigned term is exactly \ of 

 that term, therefore any term of the series will be near- 

 ly triple of the sum of all that follow it, and this will 

 be more nearly true as the term is farther from the be- 

 ginning ; the same also holds true of the series for the 

 circle. 



Prop. V. Problem. 



To investigate a second formula that shall express 

 the area of a circle or equilateral hyperbola. 



Let the letters a, t, V, t", &c. and s\ s denote the 

 same things as in last Proposition. Then because 

 a* _ a z V_ 



~T~2t'-*- 2' 

 by taking the square of each side of the equation, we 

 obtain 



a 4 



V 



t n — ± t ti + 4, "+- 2' 



and similarly we have 



a 4 _ a 4 t"\ 



2' 



a* 



therefore, by substituting this value of ~ a m the prece 



ding equation, and taking the value of the term r?n^> and the Conic 



also multiplying its numerator and denominator by a 1 , ■_ .,_'. 



we get • ,. , , -= — — ( U — - ±:a°- —--4- - — 1. 



& lba 2 l"* f- \ 4 T 4 2 / \2 T 2.4,T 



Now, a t", being the numerical expression for the 

 double of the triangle CAG, and 2 a t" that of the po- 

 lygon s', and consequently, 16 a 1 t' n that of 4 s n , we 

 have only to substitute 4 .s' 2 in the denominator of the 

 first member of the equation, and the result will be an 

 expression from which s' may be found ; but as the 

 terms of this expression will follow the same law, what- 

 ever be the number of quantities V, I", &c. we have 

 universally 



!a* a"- t"* t'"- ' \ 



F~U + ^ + -4T + > &c ) 



the two series being supposed continued to any extent 

 whatever. Let us now suppose the series of quantities 

 t, V, t", t'", &c. to be continued indefinitely, then the 



numerical series - -f- — — -(- — — -j-, &c. being an infi- 



nite geometrical progression, whose common ratio is \, 

 its sum will be |, and the limit of s' being s, we 

 have 



a* 2 

 a 6 J F 1 ^! ° S 



Let this equation be now resolved in respect of s*, 

 then extracting the square root, and putting 



T = T + WW>^ 



43 



we have 



2s=- 



a*t 



the formula to be investigated, and in which it must 

 be observed, that the sign of the second term of the 

 radical in the denominator is -f. for the circle and — for 

 the hyperbola. 



To determine the quantities t n , t" x , &c. we have, 

 from Prop. 2. 



a:t::a % ^zt n :2at', 

 the upper part of the sign z+z applying to the case of 

 the circle, and the lower to the hyperbola. Hence, in 

 the former case, we have, by taking the squares 



a 2 : V : : a 4 — 2 a? t' 1 + t"> : 4 a 1 i' 1 , 

 and, by conversion, 



a 1 : a 1 -f- t 1 : : a 4 — 2 a 1 1" -f- /' 4 : a 4 + 2 a 1 t n +f M ; 

 therefore, taking the square roots, 



a : y(«* + f) ::a* — t' 1 : a 1 + t' 2 , 

 and, by mixing, 



x /(a t + t') + a: \/(a* + t*) — a : 2a» : <2t n : : « l : *'*, 

 hence we find for the circle, 



t n = 



« , (^(« x +o— «) 



•(«» + **) + a 

 and, by proceeding in the very same manner it will ap- 

 pear, that in the hyperbola, 



The quantities t' n , t" n , &c. are derived each from that 

 before it, exactly as t' 1 is from /V 



