A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

THE COLLECTION. BOOK IV 387
length (consistent with a real solution). The problem is best
exhibited by means of analytical geometry.
If BD = a, DC = b, AD = c (so that DE = ab/c), we have
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 )
that is, QN 2 : A'N . A"N
and the locus of Q is a hyperbola.
= (const),
= (const.),
The equation of the hyperbola is clearly
to find the point R on BC such that AR produced solves the
problem by making PR equal to k,
Let DR = x.
say.
Then, since BR.RC = PR.RA, we have
(a-x)(b + x) = k^{c 2 + x 2 ).
An obvious expedient is to put y for V(c 2 + x 2 ), when
we have
(a — x)(b + x) = ley,
{ 1
'2
and y = c2 c* + x\ l
(2)
These equations represent a parabola and a hyperbola
respectively, and Pappus does in fact solve the problem by
means of the intersection of a parabola and a hyperbola ; one
of his preliminary lemmas is, however, again a little more
general. In the above figure y is represented by RQ.
The first lemma of Pappus (Prop. 42, p. 298) states that, if
from a given point A any straight line be drawn meeting
a straight line BC given in position in R, and ii'RQ be drawn
at right angles to BC and of length bearing a given ratio
to AR, the locus of Q is a hyperbola.
For c(raw AD perpendicular to BC and produce it to A'
so that
QR :
RA — A'D\ DA = the given ratio,
cc 2
x 2 = fi(y
2 — c
2
),
where // is a constant. In the particular case taken by
Archimedes QR = RA, or fi
= 1, and the hyperbola becomes
the rectangular hyperbola (2) above.
The second lemma (Prop. 43, p. 300) proves that, if BC is
given in length, and Q is such a point that, when QR is drawn
perpendicular to BC, BR . RC = k . QR, where k is a given
length, the locus of Q is a parabola.
Let be the middle point of BC, and let OK be drawn at
right angles to BC and of length such that
0C = 2 k.K0.
Let QN' be drawn perpendicular to OK.
Then QN' 2 = OR 2
= 0C -BR.RC
2
= 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 — ab/c.
To solve the problem then we have only to draw the parabola
and hyperbola in question, and their intersection then
gives Q, whence R, and therefore ARP, is determined.
*