7
$\begingroup$

Five distinct non-collinear points are required to define an ellipse similar to the way that three non-collinear points define a circle and can be used to determine the center point of that circle. I have found many mechanical explanations showing how to draw an ellipse given the major axis, minor axis and/or foci as well as algebraic solutions for determining these key features given any 5 points on the ellipse. What I am trying to find is a geometric ruler and compass type method for locating the major and minor axes and foci of an ellipse when none of them are known, but given 5 points which are known to be on an arbitrarily positioned and oriented ellipse. Specifically, the ellipse's axes are not necessarily oriented in any particular relation to the X and Y coordinate axes. I am not looking for an algebraic solution to be plotted. I already have that capability. Rather, I am looking for a mechanical construction technique for locating the foci and/or the two axes from the 5 known points. Once I can locate the foci or axes of the ellipse, I can easily draw the remaining key points. Ruler and compass type construction is preferable, but pins-and-a-string or similar sort of solution would be a reasonable alternative. Can this be done? If not, can someone provide a proof that it is impossible? Thanks.

$\endgroup$
  • 3
    $\begingroup$ In general, given five points, there is a conic through them. It need not be an ellipse. Dynamic geometry tool, Geogebra can draw the conic given the five points and it will also give the equation of the curve. You can do the problem algebraically - introduce a coordinate system and identify the coordinates of the points. Assume a second degree equation and solve for the coefficients. $\endgroup$ – user348749 Dec 24 '16 at 6:45
  • $\begingroup$ Try starting here. N.B.: the very first thing the article says is, “A conic can not be constructed as a continuous curve (or two) with straightedge and compass.” (Not true in the case of a circle or a degenerate conic, of course.) $\endgroup$ – amd Dec 24 '16 at 7:01
  • $\begingroup$ Thanks for the Geogebra hint @Muralidharan. I am familiar with Geogebra, but hadn't seen the Conic command until you mentioned the capability. It still doesn't answer my question about ruler and compass construction, but it may be helpful in other related problems. $\endgroup$ – David Dec 24 '16 at 7:11
  • $\begingroup$ Thanks for the wikipedia pointer @amd, but I don't think it disproves my goal. I'm not trying to construct the actual ellipse using a ruler and compass. I'm trying to reverse engineer just the axes and/or the foci. I'm not sure that the fact that an ellipse can't be drawn with a ruler and compass necessarily means that these key points can't be located with them. I'm still hopeful. In fact, I am going to look into the idea from the second paragraph in the section you sited based on the converse of Pascal's theorem regarding opposite sides of a hexagon. $\endgroup$ – David Dec 24 '16 at 7:28
  • $\begingroup$ As mentionned by @David, have a look at Pascal's theorem which provides a mean to obtain points, one by one, on a conic (cut-the-knot.org/Curriculum/Geometry/PascalConics.shtml). Besides, it is worth knowing this theorem which is rather fundamental in the study of conics. $\endgroup$ – Jean Marie Dec 24 '16 at 9:30
2
$\begingroup$

Given just 5 points that are known to lie on the perimeter of an ellipse, and generating that ellipse using the method described in the partial answer offered here by @Ng, and then continuing with the instructions found at http://whistleralley.com/conics/conic_construction/ellipse_parts/, it is possible to construct the entire ellipse, its major and minor axes, center, vertices, foci, and directrices. Therefore, this is a valid solution to my question as posed, however the initial construction of the ellipse still requires using more than just a compass and straight edge, so I am still hoping for a ruler-and-compass only solution. Thanks again @Ng for supplying the first steps that I was missing.

$\endgroup$
2
$\begingroup$

Given five points of the conic, and using Pascal's theorem you are able to find the intersection of the conic with an arbitrary line that contains one of these. Moreover, you are able to find the tangent to the conic at any of these points.

Now, you are able to produce pairs of conjugate diameters that intersect at the center of the conic (ellipse). To do that, notice that the middle of the chords parallel to a given direction are all on the line passing through the center of the ellipse (the conjugate diameter). This produces conjugate directions, to get the sizes of the diameters one uses Prop XXII from the source below. Up to now, we haven't mentioned angles.

The rest, and most interesting part is explained in an old book that you can find here: An elementary treatise on the Geometry of Conics, page 96, ex 8. There îs one beautiful basic fact being used , that the product of the intercepts on the tangent to an ellipse by a pair of conjugate directions is constant (equals the square of the half the diameter parallel to the tangent). With this and some classical construction one is able (ex 8 mentioned above) to construct a pair of conjugate directions that are perpendicular, so the axes. One now uses a consequence of prop XXII in the same book to get the sizes of the diameters.

I noticed now that one can also use results from page 106 of the same book (ex 14 or 15).

All these results seem classical, to be known by Newton and maybe even before him (Apollonius?), but forgotten these days, since the geometry of conics is nowadays seen as a particular case of "analytic geometry", excepts perhaps the "fun things" like Pascal, Brianchon, Poncelet and very basic things.

$\endgroup$
  • $\begingroup$ Thanks for the answer @Orest. I especially appreciate the reference to "An elementary treatise on the Geometry of Conics". It appears to be a great reference and I look forward to looking through it. At first glance, I think you have probably answered my question about how to accomplish the task with just a straight edge and compass, but I want to translate your answer into a working step-by-step process before I give it the final approval. I'll see if I can work through it all tomorrow and let you know what I am able to work out. $\endgroup$ – David Dec 29 '16 at 4:50
  • $\begingroup$ @David: Glad you enjoy the book, it's indeed very nice. You will have fun going through all of the steps. For instance, how to find the points of intersection of a line through the center with the ellipse you use Prop XXII page 82. The other result used is Ex. 7 on page 96. Now,, both are easy to prove if we notice that it's all about segments of same direction and they are true if the ellipse is a circle. Now, an ellipse is the projection of a circle and by projection all length of same direction contract by the same amount. $\endgroup$ – Orest Bucicovschi Dec 29 '16 at 5:50
  • $\begingroup$ @David: I found a source on wikipedia that solves the problem, here : en.m.wikipedia.org/wiki/Rytz%27s_construction $\endgroup$ – Orest Bucicovschi Dec 29 '16 at 12:06
1
$\begingroup$

Point $G$ is a dynamic point on circle containing $C$.

enter image description here

Point $J$ forms a locus of the conic containing $A$, $B$, $C$, $D$ and $E$.

enter image description here

$\endgroup$
  • $\begingroup$ Thanks @Ng. This is a method I haven't seen for generating an ellipse without knowing the foci or axes, but it still doesn't quite answer my original question. I'm trying to graphically locate either the foci or the axes because if I can locate any of these key elements, I can easily derive the others. If I can't get an answer to that question, your method may be useful to me anyway as an alternative approach to the problem that has prompted my question, so thanks for providing the detailed drawings too. $\endgroup$ – David Dec 27 '16 at 6:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.