Conic Parameters

The constructions in the General series jump directly from the given objects to the conic section, and do not rely on any of the other parameters (foci, diameters, etc.). Now, although those objects were not needed for the construction, they should still appear in the completed sketch. For this purpose the Sketchpad document Conic Challenges.gsp has a custom tool named Conic Parts. The same tool is used on ellipses and hyperbolas. The given objects are three points on the curve and the tangent line through each of those points. The curve itself should already be completed, but it is not actually used in the construction. I created the custom tool because I had to effect those same constructions so many times. By all means make use of it, but do try to understand what lies beneath.

The first part of this construction is identical in the ellipse and hyperbola cases, so those sketches are shown together here. Let the given points on curve be A, B, and C, with respective tangent lines a, b, and c. Construct the line through the midpoint of chord AB and the point of intersection of tangent lines a and b. This line must be a diameter (Conica (II, 29)). Use the same procedure to construct the diameter through the midpoint of BC. The two diameters must meet at O, the center of the section.

Produce AO to point D, such that O is the midpoint of AD. Draw line AOD. Construct BP parallel to a, meeting AD at P. Line AD is a diameter, and BP is an ordinate to that diameter.

On AD construct point G such that PG is the geometric mean of PA and PD. That is, PG² = (PA)(PD). (Here use either of the two points satisfying that condition for G.) Construct point K such that KO is parallel to BP and KD is parallel to BG. Produce KO to point L, such that O is the midpoint of KL.

By Conica (I, 21), KO is an ordinate with respect to diameter AD, and since it stands on the center, KL and AD are conjugate diameters.

With that we have conjugate diameters of the section, but they generally are not the axes. From here the constructions for the ellipse and hyperbola cases diverge, as they do not align at all. They will have to be taken separately.

Ellipse Case

What follows is the Rytz construction, named for 19th-century Swiss mathematician David Rytz, who first published it. The Web Sketch below starts with a fresh ellipse. It has center Q, and the axes are given, with semi‑axes colored red and blue. Drag point H, which is the midpoint of a line segment equal in length to the sum of the semi‑axes. Note that the red end is tethered to the horizontal axis, and the blue end to the vertical axis. As H is dragged, J traces the ellipse. Confirm this by considering parametric equations, with the direction QH as the angle parameter. In fact, this construction represents the mechanism of an ellipsograph drawing instrument.

Vertex J is the end of a diameter. With the conditions given here, construction of the conjugate diameter is only a matter of effecting this same construction, but with QH rotated 90° in either direction. Press the Show Conjugate button to see what results when H is rotated 90° left about the center. Point N is a vertex of the conjugate diameter. Notice that the colored line segments being transformed here, taken with the axes, form congruent right triangles. If point N is rotated 90° back right to Nʹ, this image must fall on the first line segment, but it will not fall on J or even on the curve. In fact, H is the midpoint of J and Nʹ.

That last relationship will prove very useful. Returning now to the first ellipse, a pair of conjugate diameters, which we have already constructed, can be used to define the axes.

Begin with the same ellipse used earlier, with center O and conjugate diameters AD and KL now in place. Rotate point A by 90° about O to Aʹ. Let M be the midpoint of AʹK. Draw the line AʹK, and draw the circle centered on M and passing through O. Let this line and circle meet at R and S. Lines OR and OS are on the axes of the ellipse. Point K divides RS into segments equal to the semi-axes. The semi-axis on OR is equal to KS, and the semi-axis on OS is equal to KR.

Here the axes TU and VW are constructed through R and S, respectively.

Having all four vertices of the ellipse, there should be little trouble constructing the foci and directrices.

Hyperbola Case

The hyperbola case is quite a lot simpler. We have already constructed center O, along with diameter AD, the tangent line a through point A, and second diameter KL. Through K and L construct lines parallel to AD, and let them meet line a at H and J. By Conica (II, 3), lines OH and OJ are the asymptotes of the hyperbola. We now have the asymptotes, point A on the curve, and tangent line a. See Hyperbola 5 or Hyperbola 6 for construction of the axes, axial vertices, foci, and directrices.

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