• Basic theorem for orthotomes (section of a right angled cone, a.k.a. parabola)

by Henry Mendell (Cal. State U., L.A.)

Problem A:  To construct the orthotome

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

Theorem D:  If there is an orthotome HFG with diameter FK of a cone with vertex at A, where AF is the line from the vertex of the cone to the vertex of the orthotome, then for any ordinate SN parallel to the tangent at F from the section to the diameter and abscissa FS from the vertext F to the ordinate:  the square on SN is twice the rectangle FS * AF.  This may also be expressed as FS : SN = SN : 2AF.

Problem A:  To construct the orthotome
p. 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).

p. 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.

p. 3

Let there be a tangent to the cone at right angles to the generating side at some arbitrary point F.  For future reference, AF is 1/2 the latus rectum of the curve.  The tangent at F is clearly parallel to the base of the cone.

p. 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 as a parabola.  For brevity, we shall call it an orthotome.

p. 5
Theorem B:  The ordinates with right orientation to the same point on the diameter are equal.

First draw a line FK perpendicular the tangent on the orthotome.  Take any other point P on AB and draw the diameter PR of the circle MRNP of the cone at P.  Clearly FK intersects RP at some point S. proof:  Let PSR be the diameter and join SF, with S on the section.  It is enough to show that if we draw SF, it is at right angles to AP and the tangent at F. Clearly, SF lies on the section and so is perpendicular to AB. The tangent at P perpendicular to AB is perpendicular to RSP.  The tangent at F is also perpendicular to FP and so is parallel to the tangent at P.  Hence, any line from a point on RSP is perpendicular to the tangent at F.  (A line drawn from a perpendicular to a line to parallel to the line will be perpendicular to it). This line FK is the principal diameter of the section. Let the base  HKG of the orthotome be parallel to the tangent at F, i.e. at right angles to the diameter BD of the base circle from the side of the cone AF.  Similarly, take any other line connecting points on the section MN parallel to the tangent at F.  Clearly, HK = KG and MS = SN.  Each of these is called an ordinate. proof:  MSN lies on a circle of the cone MRNP.  Assume N is on the circle.  We will show that M is too.  MN is parallel to the tangent at F, which is also parallel to the tangent at P perpendicular to AP.  This tangent lies in the plane of circle MRNP.  Hence so does MSN. MSN lies on a circle MRNP with PSR is a diameter.  Hence MSN is a chord or diameter of the circle divided by a diameter at S. The line part of the diameter from the vertex to the ordinate is called the abscissa.  FS is the abscissa for ordinates MS and SN, and FK is the abscissa for ordinates HK and KG.

p. 6
Theorem C:  If there is an orthotome HFG with diameter FK of a cone with vertex at A, where AF is the line from the vertex of the cone to the vertex of the orthotome, then for any ordinate SN parallel to the tangent at F from the section to the diameter and abscissa FS from the vertext F to the ordinate:  the square on SN is twice the rectangle FS * AF.  This may also be expressed as FS : SN = SN : 2AF.

Fundamental Theorem:  The square on the ordinate is equal to twice the rectangle formed by the abscissa and the line from the vertex of the cone to the vertex of the orthotome. That is:  KG2 = 2 AF*FK.  Note that 2 AF is the latus rectum of the curve.

p. 7

Since RNP is a semicircle and NS is perpendicular to PR, it follows that  SN2 = RS * SP

p. 8

Now, since angles RAP = SFP = a right angle and AR = AP (all lines drawn from the vertext of a cone to a circle of the cone are equal), it follows that angle FPR is 1/2 right angle. and right triangles SFP = RAP (since APR is common to both) and are isosceles,  FS = FP.

p. 9

Since angles SFP and RAP are right angles, it follows that: AF : FP = RS : SP Hence, AF : FP = RS : SP = RS*SP : SP2 (since FS = FP, it follows that SP2 = SF2 + FP2 = 2FP2) RS*SP : SP2 = RS*SP : 2FP2 But, AF : FP = AF * 2FP : 2FP2 Hence, AF * 2FP : 2FP2 = RS*SP : 2FP2 Or 2 AF * FP = RS*SP Since, RS*SP = SN2 SN2 = 2 AF * FP And since FP = FS, SN2 = 2 AF * FP Q.E.D.