A history of Greek mathematics - Wilbourhall.org

)
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
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