114 
CONIC SECTIONS 
But, by similar triangles, 
NG:AF=NG:AD 
= A'N-.AA'. 
Hence PN 2 = AN .A'N. > 
AA 
2 AL . , r t / -\ T 
= ,Z37 -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: A A' equal to 1. The problem therefore now 
is: given a straight line A A', and AL along A'A produced 
equal to \AA r , to find a cone such that Z 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. In 
other words, we have to find a point 0 on the straight line 
through A perpendicular to A A' such that OL bisects the 
angle which is the supplement of the angle A'OA. 
This is the case if A'O : OA = A'Tv: LA =3:1; 
therefore 0 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 0 is then determined as the point in which AO 
drawn perpendicular to A A' 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 rect 
angular 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.