
274 MRE. SANG ON THE CURVES PRODUCED BY REFLECTION. | 
If OZ be drawn to bisect the straight. line AB, the perpendiculars Aa and — 
B£ drawn to it are of equal length and 
_ the tracing point is at‘an infinite dis- — 
tance along it in either direction. Now 
it will be shown that the perpendiculars 
from C and D upon the same line are 
also alike, or that OZ bisects the straight 
line CD, while OV drawn to bisect the 
angle AOB, also bisects COD; where- 
fore the angle VOZ formed by OZ and 
OV is the same whether derived from 
AOB or from COD; it is thus the 
characterising angle of the curve. . 
If, for shortness, we put a, 6, c, d, for the lengths of the four lines OA, OB, 
OC, OD, we easily obtain 
ab.sin COD 

Higa 
ab. sin COD 



OF asin BOD+0 sin AOD > pile ~6.sin BOD+a sin AOD * 
Otsewite that sin AOD?—sin BOD*?=sin AOB. sin COD, we convert these 
into 
big ed.sin AOB ci It ed .sin AOB : 
@= sin BOD+d.sin AOD’ °~ —d.sin BOD+e.sin AOD? 
which are just what would have been obtained on supposing ¢ to be the source 
of light and D to be the position of the eye. ; 
The angle i as found from AOB, has for its tangent 2= =— = tan AOV; as i 
found from CoD the expression becomes ae Ta 
for c and d their values in terms of @ and 6, we find these expressions to be — 
identic ; wherefore OZ bisects the straight line CD. From this it is obvious 
that the character of the locus depends on the angle VOZ. 
Taking OZ for the axis Z of rectangular continates, let us put Z os cm 
tan COV ; and on substituting 
ZO B, and Aa=-—X,=BB= +X;=p, then 
~ tan AOZ= , tan BOZ= a 
whence | tan (AOZ— BOZ) =tan 2VOZ=. ewe. 
‘“ > pitty 6 for the varying angle ZOD , 
a.sin AOD=a.sin 6+,. cos 0; a.cos AOD=a. cos #—p. sin 8 
5. sin BOD=8. sin 0—p.cos 6; 6.cosBOD=8. cos A+p.sin 8, 

