Solution of a Problem in Fluxions. 73 



at its focus, andp'= the parameter of its axis. Now as in (7) and 

 (9) — , is constant, and =0 in (8), it is evident that I shall have 



F= — 7,-d—r-- — TT— — 7i' di — , \ =—7-7 for the form of F in 



2 ?•- sin. ^vj^" 2 \rp')~r'^p' 



dr 

 each of these cases; .".in each of these curves the law of force to 



1 . 



tlie focus is F is as -7 (since C, p' are invariable in each of them) ; 



.2 

 if in (9) I cMnge the sign of — I have the case of the particle 



moving in an hyperbola, and acted on by a central force in the focus 



C" . 1 

 of the opposite hyperbola, and I have F= r~r •' • F is as -^ the 



sign minus shows the force to be repulsive. (See Prin. b. I. sec. iii. 

 prop.'s 11, 12, 13.) The same results maybe found after the man- 

 ner of La Place, (Mec. Cel. Vol. I. p. 114,) by using the polar 

 equation of the ellipse and transforming it into the equations of the 

 parabola and hyperbola as he has done. As the polar equation of 

 the ellipse is not usually given in treatises on the conic sections; I 

 will conclude this communication by the following investigation of it. 

 Put 2C~ the distance of the foci; 2a= the sum of the lines from 

 the foci to any point in the curve =- the transverse axis; r= one of 

 these lines; then 2a -r= the other; v—-^= the supplement of the 

 angle contained by r and 2C. Then by a known theorem (which is 

 easily derived from Euc. 2. 13.) in trigonometry, I have (2a - r)^ = 

 =r2+4C--f 4Crcos. (v— 'z^) (10); or by reduction a ^ — C2=ar-{- 



+Crcos.(.-t.); .•.^-^+c^^7Z^= C ' = 



H cos. (v — -^) 



a ^ ■' 



a(l-e^') . C 



~l+ecos. (i; -^ (^^) ^^"^a^' (H) '^the same equation tliatLa 



C 



Place has given at the place cited above; — =e being the eccentri- 

 city divided by half the greater axis; also a (1 — e^)— the semipa- 

 rameter. 



Vol. XVII.— No. 1. 10 



