A history of Greek mathematics - Wilbourhall.org

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THE COFICS, BOOK V 161
We have P'N' 2 = p
. AN' = 2 NG . AN'
;
and NG 2 = NN' 2 + NG 2 ±2NG. NN',
according to the position of N'.
Therefore P'G 2 = 2NG.AN+NG 2 + NN' 2
and the proposition is proved.
= PN 2 + NG 2 + NN' 2
= PG 2 + NN' 2 ;
(2) In the case of the central conic, take G on the axis such
that AG > \p, and measure GN towards A such that
NG:GN = p:AA / .
Draw the ordinate PN through N, and also the ordinate P'N'
from any other point P'.
We have first to prove the lemma (V. 1, 2, 3) that, if AM be
drawn perpendicular to A A' and equal to \p, and if CM,
produced if necessary, meet PN in H, then
PN = 2
2 (quadrilateral MANE).
This is easy, for, if AL(= 2AM) be the parameter, and A'L
meet PN in R, then, by the property of the curve,
PN = AN.NR
2
= AN(NH + AM)
= 2 (quadrilateral MANH).
Let GH, produced if necessary, meet P'N' in H'. From H
draw HI perpendicular to P'H'
Now, since, by hypothesis, NG :
GN
— p:AA'
= AM:AC
= HN:NC\
NH = NG, whence also
H'N' = N'G.
Therefore NG 2 = 2AHNG, N'G 2 = 2 A H'N' G.
And PN 2 = 2(MANH);
therefore PG = NG 2 2 + PN 2 = 2 (AMHG).
1523.2 M