A history of Greek mathematics - Wilbourhall.org

114 CONIC SECTIONS
But, by similar triangles,
NG:AF=NC:AD
= A'N:AA'.
Hence PlY 2 = AN.
A'N
AA f
.
^-,
AA
AN.A'N.
which is the property of the hyperbola, A A' being what we
call the transverse axis, and 2 AL the parameter of the principal
ordinates.
Now, in order that the hyperbola may be rectangular, we
must have 2 AL :
AA
f equal to 1. The problem therefore now
is: given a straight line AA\ and AL along A''A produced
equal to \ A A\ to find a cone such that L is on its axis and
the section through AL perpendicular to the generator through
A is a rectangular hyperbola with A'A as transverse axis.
other words, we have to find a point on the straight line
through A perpendicular to AA f such that OX bisects the
angle which is the supplement of the angle A'OA.
This is the case if A'O :
OA
= A'L :
LA
= 3:1;
therefore is on the circle which is the locus of all points
such that their distances from the two fixed points A' , A
are in the ratio 3:1. This circle is the circle on KL as
diameter, where A'K-.KA = A'L: LA = 3:1. Draw this
circle, and is then determined as the point in which AO
drawn perpendicular to AA' intersects the circle.
It is to be observed, however, that this deduction of a
particular from a more general case is not usual in early
Greek mathematics ; on the contrary, the particular usually
led to the more general. Notwithstanding, therefore, that the
orthodox method of producing conic sections is said to have
been by cutting the generator of each cone perpendicularly,
I am inclined to think that Menaechmus would get his rectangular
hyperbola directly, and in an easier way, by means of
a different cone differently cut. Taking the right-angled cone,
already used for obtaining a parabola, we have only to make
a section parallel to the axis (instead of perpendicular to a
generator) to get a rectangular hyperbola.
In