Solution of a Problem in Fluxions. 73 
at its — and p/= the parameter of its axis. Now as in (7) and 
Or p is constant, and =0 in (8), it is evident that I shall have 
* Or 1 C74 ) 
F>=— a. — rs Snes 
2 “r* sin. 2) 2 (ry 
dr 
for the form of F in 
~s 
= rp! 
each of these cases; .*. in each of these curves the law of force to 
1 
the focus is F is as = (sinee C’, p’ are invariable in each of them) ; 
Tr 
2 
if in (9) I change the sign of a I have the case of the particle 
moving in an hyperbola, and acted on by a central force in the focus 
of the opposite hyperbola, and I have F= — aa . Fisas Bit the 
sign minus shows the force to be repulsive. (See Prin. b. I. sec. iii. 
prop.’s 11, 12, 13.) ‘The same results may be 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—aw= the supplement of the 
angle contained by r and2C. Then by a known theorem (which is 
easily derived from Euc. 2. 13.) in trigonometry, I have (2a— 7)? = 
=r? +4C?+4Cr cos. (v—@) (10); or by reduction a? — C? =ar+- 
+Cr cos. (v—a) eee al iw ) 
(P= B)5 oe P= oe RES 
a+Ccos. (v—a) ee cos. (v—=) 
” 1—e? C ; 
aS ws (11) (e=—); (11) is the same equation that La 
Cc : ‘ 
Place has given at the place cited above; aime being the eccentri- 
city divided by half the greater axis; also a (1 —e?)= the semipa- 
rameter. 
Vou. XVH.—No. 1. 10 
