of H alley and Fregier. 155 



Hence KG = In, 



and AJ=IK = IK+In + NG = AM + NG. 



Therefore JQ is the normal at Q. 



And JHQ is a straight line, because 



HK.NG = PN.In = AI.QM=KJ.QM. 



4. Let (on the same side of the axis with H) be the centre 

 of the circle APpQ and Om its ordinate. The point m is found 

 from AK by taking an abscissa AZ equal to 2a and bisecting KZ. 



For the diameters bisecting Pp and AQ meet the axis in 

 points U, V such that 



IU = ±AM+2a = AV, 



and Om bisects UV. 



Therefore 2Am = AM + 4a - AI 



= AK + 2a. 



To determine Om from HK we have 



Om\\AI = \QMj2a, 



the ordinate of the middle point of AQ being equal to \QM. 



Therefore Om = \HK. 



The circle described with centre and radius OA determines 

 the points P, p, Q, and thus the normals HP, Hp, HQ. 



5. A short proof, by means of another circle and a second 

 parabola, of Halley's theorem that A, P, p, Q lie on a circle is 

 given by Messrs Milne and Davis in their Geometrical Conies. 



The focal perpendiculars SP', Sp', SQ' to the normals PG, pg, 

 QJ bisect them, and S, P', p', Q' lie on the parabola whose 

 equation is 



y 2 = a{x — a). 



This, by its intersection with the circle on SH as diameter, 

 determines the points P', p', Q', and thereby the points P, p, Q. 



The algebraic sum of the ordinates of P', p', Q' being zero, 

 the same follows for P, p, Q, which therefore lie on a circle 

 through A. 



But as a practical construction it is of course simpler to draw 

 this circle and find P, p, Q in Halley's way. 



6. After finding he gives a construction for the limiting 

 point h on the ordinate of H such that, according as HK is less 

 or greater than hK, two normals or none can be drawn from H 

 across the axis. From h one normal only can be so drawn. 



