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