388 
PAPPUS OF ALEXANDRIA 
Measure DA" along DA equal to DA'. 
Then, if QN be perpendicular to AD, 
(.AR 2 -AD 2 ):{QR 2 -A'D 2 ) = (const.), 
that is, QN 2 : A'N. A"N = (const.), 
and the locus of Q is a hyperbola. 
The equation of the hyperbola is clearly 
x 2 = y{y 2 -c 2 ), 
where y is a constant. In the particular case taken by 
Archimedes QR = RA, or y = 1, and the hyperbola becomes 
the rectangular hyperbola (2) above. 
The second lemma (Prop. 43, p, 300) proves that, if BO is 
given in length, and Q is such a point that, when QR is drawn 
perpendicular to BG, BR. RC = k . QR, where k is a given 
length, the locus of Q is a parabola. 
Let 0 be the middle point of BG, and let OK be drawn at 
right angles to BG and of length such that 
OC 2 = k.KO. 
Let QN be drawn perpendicular to OK. 
Then QN' 2 = OR 2 
= OG 2 — BR. RG 
= k. (KO — QR), by hypothesis, 
= k.KN'. 
Therefore the locus of Q is a parabola. 
The equation of the parabola referred to DB, DE as axes of 
x and y is obviously 
which easily reduces to 
(a — x) (b + x) = ky, as above (1). 
In Archimedes’s particular case k = ah/c. 
To solve the problem then we have only to draw the para 
bola and hyperbola in question, and their intersection then 
gives Q, whence R, and therefore ARP, is determined.