of Halley and Fregier. 
155 
Hence KG — In, 
and AJ= IK = IN + In + NG = AM + KG. 
Therefore JQ is the normal at Q. 
And JHQ is a straight line, because 
HK.NG = PN. In = AI. QM = KJ. QM. 
4. Let 0 (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 2 a 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 TJV. 
Therefore 2 Am = AM +4 a — AI 
= AK + 2a. 
To determine Om from HK we have 
Om/^AI = iQM/2a, 
the ordinate of the middle point of AQ being equal to hQM. 
Therefore Om = \HK. 
The circle described with centre 0 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 Conics. 
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 0 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. 
