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 = ah/c), we have
to find the point B on BG such that AR produced solves the
problem by making PR equal to k, say.
Let DR = x. Then, since BR. RG = PR. RA, we have
((a — x)(b + x) = k V (c 2 + x 2 ).
An obvious expedient is to put y for V(c 2 + x 2 ), when
we have
(a — x){b + x) = ky, (1)
and y 2 = c 2 + x 2 . (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 BG given in position in R, and if RQ be drawn
at right angles to BG and of length bearing a given ratio
to AR, the locus of Q is a hyperbola.
For draw AD perpendicular to BG and produce it to A'
so that
QR:RA = A'D:DA = the given ratio,
c c 2