Apollonius, Conics, Book I, Proposition 11
If a cone is cut by a plane through its axis, and also cut by another plane cutting the base of the cone in a straight line perpendicular to the base of the axial triangle, and if further the diameter of the section is parallel to one side of the axial triangle, then any straight line which is drawn from the section of the cone to its diameter parallel to the common section of the ccutting plane and of the cone's base, will equal in square the rectangle contained by the straight line cut off by it on the diameter beginning from the section's vertex and by another straight line which has the ratio to the straight line between the angle of the cone and the vertex of the section that the square on the base of the axial triangle has to the rectangle contained by the remaining two sides of the triangle. And let such a section be called a parabola (παραβολή).
Consider the cone with vertex A and a plane through the axis intersecting the cone. This plane intersects the cone in the axial triangle ABC, where BC is the diameter of the base circle of the cone.
The parabola is like the section of a rightangled cone (orthotome); if the cutting plane is orthogonal to a side of the axial triangle, it must also be parallel to the other side of the axial triangle (for rightangled cones). For Apollonius, the parabola is generated by a plane cutting one side of the axial triangle such that it is parallel to the other side. This works for all cones in general, but we illustrate here with a right circular cone. This has the added advantage of matching the derivation of the section of a rightangled cone that we skipped in Section 2a. This derivation and those for the ellipse and the hyperbola essentially follow those done in [11, pp. 119121].
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Consider an arbitrary ordinate (i.e. y value) ML constructed on the axis at M. We wish to determine the relationship between ML and EM, that is, the symptom of the conic. The ordinate ML is located in a horizontal plane that cuts the cone in the circle with diameter PR. In this horizontal plane construct the segments PL and LR, which results in a right triangle inscribed in a semicircle. As we have seen before, by similar triangles this implies
or



(1) 
Next Apollonius constructs a segment EH perpendicular to EM such that

(2) 

How is this possible? All the lengths EA, BC, BA, and AC are known. He is simply finding the point H that makes the ratio true. Now why Apollonius does this is an other story. Wait and see.
Now consider some triangles in the axial plane, namely ABC, APR, and EPM. These triangles are all similar, using the usual properties of parallel lines (see previous derivations for details). From the similarity we have
Combining (2), (3), and (4) gives us
.

(5) 
Also, in triangle APR, since EM is parallel to AR we know by Elements Book VI, Prop. 2 that
.

(6) 
Substituting (6) into (5) yields
.

(7) 
It is clear that
,


so combining this with (7) gives us
.


Recall from (1) that ML^{2} = PM.MR, therefore ML^{2} = EH.EM.
In this form we can understand Apollonius' use of the word parabola for this section. He has proven that the square on ML is equal to a rectangle applied (paraboli) to a line EH with a width equal to EM. This is based on the idea of application of areas, a Greek technique used to deal with multiplication and division of lengths and areas.
Finally, if we let ML be y, EM be x, and EH be p, we have a standard equation for a parabola, y^{2} = px.
Notice once again that all that was necessary to derive the symptom of the parabola was a knowledge of similar traingles. We will see this again as we derive the symptom of the next conic section, the ellipse.