of Halley and Fregier. 157 



Draw the chord OIR across and at right angles to PQ, and let 

 /, g be the focal chords parallel to OR, PQ. 



Then 01*101. IB = PI.IQjOI.IR = g/f 



Therefore 01. f= IR.g = OR/(ljf± 1/g), 



or 01 ./varies as OR, the chords /and g being at right angles. 



But OR varies as the projection of 'f upon the normal at O 1 . 



Therefore, if PQ meets the normal in n, then 01 varies as 

 01 J On, and On is constant and n a fixed point. 



2. Another proof is given as a problem in The Ancient and 

 Modern Geometry of Conies, page 122 (1881), thus, 



" 279. If PQ be a chord of a conic which subtends a right 

 angle at a given point on the curve, and MN be the projection 

 of the chord upon the tangent at 0, shew that 



^±g = a constant, 



and that PQ passes through a fixed point on the normal at 0." 



The result follows from the lemma that, if OPO'Q be a rect- 

 angle (or parallelogram), and if a line OABG be drawn across 

 PQ, OP, O'Q, then 



1 1^ J^ 

 0A~ OB^W 



In the problem, this sum or difference being constant when 

 OABG is the normal at 0, it follows that PQ passes through 

 a fixed point A on the normal. 



3. Proofs from the Circle. 



a. If be a fixed point on a conic and PQ any chord 

 through a fixed point /, then OP, OQ are conjugate lines in an 

 involution, and conversely. 



This includes Fregier's theorem as a special case. 



It may be proved by projecting the conic into a circle with 

 / as centre. 



1 Take V on the normal chord ON, draw RVO' to the curve, and let VR, VN cut 

 any parallel to 00' in r, n. Lastly let 0' coincide with 0. Then 



VR.Vr\f=VN.Vn\h, 



if f, h be the focal chords parallel to VR, VN, and in the limit 



ORjf cos NOR = ONIh. 



VOL. XI. PT. II. 12 



