Next we consider the obtuseangled cone with vertex O and a plane intersecting a generating line OB at a right angle at point A. The plane intersects the cone in the amblytome with diameter AG.
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Consider an arbitrary ordinate (i.e. y value) NP constructed on the axis at N. We wish to determine the relationship between NP and AN, that is, the symptom of the conic. The ordinate NP is located in a horizontal plane that cuts the cone in the circle with diameter BC. In this horizontal plane construct the segments BP and CP, which results in a right triangle inscribed in a semicircle. (The triangle is right by Elements, Book III, Proposition 20). We also know that triangles BNP and PNC are similar (by Book VI, Prop. 8) and this implies
or



(1) 
Now consider the similar triangles NAB and NCG in the axial plane plane through O, B, and C. The triangles are similar because they each have a right angle and opposite vertical angles. This in turn implies
Combining (1) and (2) we have
.

(3) 
In the axial plane we also have similar triangles GCN and FDA, because they each have a right angle and they have another equal angle since AD is parallel to NC. (See Elements Book I, Prop. 29, a very famous proposition indeed. Follow the link to see why.) These similar triangles give us:
or


.

(4) 
Finally, in the same axial plane are the similar triangles A'NC and A'AD. The triangles are similar because AD is parallel to NC in triangle A'NC (see for example, Elements Book VI, Prop. 2). From this we have
or


and combining with (4)


.

(5) 
Rewriting (5) as and combining with (3) yields
,
using the fact that AF = 2AL.
Just as in the last section, we can make this look more like the modern equation of a hyperbola by substituting x = AN, y = NP, 2a = AA', and p = 2AL. This gives us
,
which can be rewritten as the modern form of a hyperbola. (Do you see how?)