126 



Proceedings of the Royal Irish Academy. 



6. Again, let any line tkrougli tlie origin meet a normal to the 

 curye in N, and a corresponding normal to the reciprocal curve in N' . 



Prom N di'aw NL perpendicular to OP. 

 Then, if li be the radius of reciprocation, 



P'N' . PiV^sin (^ = F'W . FL = k\ 



for the first product is equal to OP'. OF sin^, 

 'w^hich is equal to k'^. 



Therefore, if the intercept F'N' of one of 

 the normals is constant, then the projection FL 

 of the intercept PiV^of the other normal on the corresponding radius 

 vector will be constant. 



7. In order, therefore, to describe a curve parallel to the reciprocal 

 of a given curve, and so that the distance between the parallel curves 

 may be equal to a given constant (c), we have the following construc- 

 tion : — 



On the radius vector to any point P of the given curve take a point 

 X, so that 



PZ = -. 



c 



Let LN di-awn perpendicular to OF meet 

 the normal in N. 



Join NO, and produce this line, so that 



NO : ON =OF:c. 



Then the locus of N is the required parallel 



curve, h being taken as radius of reciprocation. 



Applying, for example, the above to a circle 



and conic, we get a proof of the theorem that 



"In any conic the projection of the normal on a focal radius vector 

 is constant." 



8. If y be the semichord of curvature for any point on a curve, 

 y the semichord of curvature for corresponding point on reciprocal 

 curve, then yy' = rr\ r and r' being the corresponding radii vectores ; 

 or, since 



rr' sin ^ = h-, 



yy' sin ^ is constant, and equal to square of radius of reciprocation. 



