REDUCTION TO A NEY^IH 
23 7 
of the parallelograms about the diagonal of the complete 
parallelogram, we see that 
BC.CB = BF.ED 
= BF. FG. 
Consequently G lies on a hyperbola with BF, BA as 
asymptotes and passing through D. 
Thus, in order to effect the construction, we have only to 
draw this hyperbola as well as the circle with centre D and 
radius equal to the given length. Their intersection gives the 
point G, and E, F are then determined by drawing OF parallel 
to DC to meet BC produced in F and joining AF. 
(J3) The vevens equivalent to a cubic equation. 
It is easily seen that the solution of the vevens is equivalent 
to the solution of a cubic equation. For in the first figure on 
p. 236, if FA be the axis of x, FB the axis of y, FA = a, 
FB = h, the solution of the problem by means of conics as 
Pappus gives it is the equivalent of finding a certain point 
as the intersection of the conics 
xy — ah, 
{x — a) 2 + (y — h) 2 — 4 (a 2 + 6 2 ). 
The second equation gives 
[x + a] (x— 3a) = {y + b) (3b-y). 
From the first equation it is easily seen that 
(x + a): (y + b) — a: y, 
and that {x—3a)y = a{b—3y)\ 
therefore, eliminating x, we have 
a 2 {b— 3y) = y 2 {3b-y), 
or y 3 —3by 2 —3a 2 y + a 2 b = 0. 
Now suppose that Z ABC = 6, so that tan 6 = h/a; 
and suppose that t = tan BBC, 
so that y = at. 
We have then 
a i t 3 —3ba 2 t 2 ~3a 3 t + a 2 b = 0, 
m