140 Proceedings of the Royal Society of Edinburgh. [Sess. 
P 1 and N 1 are the points corresponding to P on the first positive and 
negative pedals, then 
arc BPj = (n + l)(arc BN t + straight line N X P). 
Dem. 
By Prop. XIV (fig. 40) 
that is 
Now 
and 
Hence if 
Thus 
and 
= (n + 1 or -(^R, = (n + 1 )QR, 
p rp 
PiQi = (w+ 1)QR. 
Pj^Qj^ =| ds x if s x = arc BP lf 
QR = QNf - RNf = QN/ - PN/ = QN/ - PX^ + X/ N r 
«r 1 = arc BN 1? 
QR = d . PN X + dcr v 
ds 1 = (n + 1 )(d . PN X + do-]) 
arc BP X = (n+ l)(arc BN, + PXj). 
[The following analytical proof* may be given. 
Let ( r , 0) be the polar co-ordinates of any point P on the curve, and 
let P x and N x correspond to P on the first positive and first negative pedals 
respectively. Let the positive direction of the arc s be such that s increases 
as 0 increases, and let the positive direction of the tangent PPj be that of 
the curve at P. Also let \fs be the angle from the positive direction of OP 
to that of PP r Then 
OP = r ; ds = dr sec i fr. 
Let 
and 
so that 
Then 
p = OPj = r sin i fr 
t = P X P = — PP L - r cos if/, 
t 2 + p 2 = r 2 . 
dt = -dr-JLdp 
t t y 
= dr sec if/ — —dp sec if/ 
r 
= ds — ^-ds„ 
where ds x is the element of arc at P r 
Similarly, if p, or, t correspond on the first negative pedal to p, r, s, t, 
* Suggested by Professor G. A. Gibson. 
