These proofs do not follow Apollonius, in that they are designed to go with Archimedes' treatment. This may be an incorrect assumption, but if they are in the style of Archimedes, they are not necessarily objectionable. Calling this early conic section, therefore, may be somewhat deceptive, in that they are theorems to prove the theorems listed in Archimedes as part of the conic elements.

We construct here the a parabola or orthotome, which we know as y2 = 2 a*x, where the ordinate (the line with distance y) and the abscissa (the line with distance y) are at right angles (that is, the case for right orientation).  In this presentation, however, the parameter a will not appear, Instead, we will take two ordinates y1 and y2 with corresponding abscissae, x1 and x2, such that x1 : x2 = y12 : y22. However, this way of thinking about the curve is still anachronistic. We must conceive the properties geometrically.

We use the notation of Dijksterhuis for the square on a straight-line and rectangle formed from two straight lines. T(AB) is the square on straight-line AB, while O(AB,CD) is the rectangle formed from straight-lines AB and CD (or lines equal to AB and CD).

Problem A:  To construct the orthotome and give a first definition of the diameter, ordinate, and abcissa lines.

Theorem B:  The ordinates with right orientation to the same point on the diameter are equal.

Theorem C:  If A and B are two ordinates of an orthotome and X and Y, the coresponding abcisses, then T(A) : T(B) = X : Y.

Theorem D: The tangent to the orthotome at the vertex is the line parallel to the base; otherwise, it intersects the diameter extended at a point whose line to the vertex does not overlap the abcissa and is equal to it. It also follows that there is only one tangent to a given point.

Problem A:  To construct the orthotome
 (diagram 1) We start with a right angled cone, i.e. a cone generated by an isosceles triangle, so that the angle at the vertex is right (since the angles at the base will be 1/2 right angle). (diagram 2) Take a side of the cone AB, i.e. a line of the generating right triangle.  For convenience, we'll call this line the side of the cone. (diagram 3) Let there be a tangent to the cone at right angles to the generating side at some arbitrary point F. The tangent at F is clearly parallel to the base of the cone. F is the vertex of the section. (diagram 4) Through the tangent, i.e., at right angles to the side of the cone, slice through the cone with a plane.  This creates the section of a right angled cone, also known later as a parabola.  For brevity, we shall call it an orthotome. The line FK at right angles to the lines AB and the tangent of the cone is called the (principal) diameter of the orthotome. The point F is the vertex of the orthotome. The tangent at F parallel to the base of the cone will be tangent to the orthotome and so will the be the tangent at the vertex of the orthotome.

Theorem B:  An ordinate (arranged line) is a line drawn from the diameter of the section to the perimeter of the section and parallel to the tangent at the vertex of the section. Since the diameter so-far described is perpendicular to the tangent at the vertex, the ordinate will be here perpendicular to the diameter. Ordinates from the same point on the diameter are equal.
 First draw a line FK perpendicular to the tangent and on the orthotome. Take a parallel MSN to the tangent at F on intersection of the cutting plane and the cone, with S on FK as well. I say that MS = SN. Since, MN is parallel to the tangent at F, and FS is perpendicular to the tangent at F, MSF must be a right angle. MN will clearly be a chord of circle MPNK parallel to the base HBGD of the cone. Let P also be on MPNK and AB, join PS and extend it to the circumference of circle MPNR. PR is a diameter of circle MPNR. For AB is perpendicular to both tangents at F and P, FK is perpendicular to the tangent at F. Therefore, SP must be perpendicular to the tangent at P (cf. Euclid, Elements XI, prop. 8 and take the tangent as parallel lines and FSP as the plane). Hence, PR is a diameter of circle MPNR (cf. Euclid, Elements III prop. 19). Hence, MS = SN (cf. Euclid, Elements III prop. 19). This line FK is the diameter of the section The line part of the diameter from the vertex of the section to the ordinate is called the abscissa (line cut off by the ordinate).  FS is the abscissa for ordinates MS and SN.

Theorem C (Fundamental Theorem):  the squares on any two ordinates are proportional to their respective abscissae.
 (diagram 1: initial diagram) That is:  T(HK) : T(MS) = KF : SF note: the standard relation in Apollonius will be T(KG) = O(2*AF,FK), where 2*AF is called the latus rectum of the curve. Let RMPN and DHBG be circles parallel to the base of the cone, with M, H on the intersection of the respective circles and the orthotome. Since, a perpendiculars to the diameter of a semicircle is in mean proportion between the segments of the diameter it cuts off, T(MS) = O(RS,SP) and T(HK) = O(DK,KB), when it follows trivially that T(MS) : T(HK) = O(RS,SP) : O(DK,KB). (diagram 2) Also, since RS is parallel to DK and FS is parallel to AD, RSKD is a parallelogram and hence RS = KD. Hence, T(MS) : T(HK) = O(RS,SP) : O(DK,KB) = SP : KB.But SP : KB = KF : SF (because of similar triangles KFB and SFP). Hence, T(MS) : T(HK) = KF : SF.

Theorem D: The tangent to the orthotome at the vertex is parallel to the base (obvious from the definition of vertex). A line is tangent to an orthotome at any other given point iff being extended it intersects the diameter extended outside the orthotome and the line from the intersection to the vertex of the orthotome equals the abscissa for the given point.

 (diagram 1: initial diagram) Proof: Let there be an orthotome ABG and take point G, and construct the ordinate and abcissa. Extend the abcissa to E so that EB = BD. Join EG. (diagram 2) I say that EG is tangent to the orthotome at G. For suppose that it intersects another point of the orthotome at H and construct the ordinate HZ for the point at H. Then ZB : BE = BE : BD. But BD = BE. Hence, ZB : BE = BE : BE and ZB = BE. Therefore, H is either G or A. By hypothesis it must be A. However, AG intersects the diameter at D inside the parabola and hence a different point from E, which is very absurd. Hence, EG does not touch the perimeter of the orthotome except at G and so is tangent at G. (diagram 3) Suppose that a line EG is tangent to the orthotome at G but EB ≠ BD (this would imply that there are two tangents at G). (diagram 4) Then, find a line BD : EB = EB : X. EG will intersect the orthotome at a point whose abcissa is equal to X.
top