A history of Greek mathematics - Wilbourhall.org

388 PAPPUS OF ALEXANDRIA
Measure DA" along DA equal to DA'.
Then, if QN be perpendicular to AD,
that is, QN 2 : A'N
(AR 2 -AD 2 ):(QR 2 -A'D 2 )
.
and the locus of Q is a hyperbola.
A"N
= (const),
The equation of the hyperbola is clearly
x 2 = fi(y
2 — c
2
),
= (const.),
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.