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.
[Math] Graphically locate the axes or foci of an ellipse from 5 arbitrary points on its perimeter.
conic sectionsgeometric-construction
Related Solutions
Hint Let $O$ denote the centre of the ellipse.
Pick an arbitrary point $A$ on the ellipse. Draw a circle with the centre at $O$ and passing through $A$.
This circle will intersect the ellipse in 4 points* $A,B,C,D$. Show that $ABCD$ is a rectangle, with the edges parralel to the axes. Since the axes go through $)$ it is easy to construct them.
*If the circle intersects the ellipse in only two points, the two points are the intersection between the ellipse and one of the axes.
Of course, if the ellipse is degenerated (a circle) the construction above will fail, since there will be infinitely many points of intersection. But in that case, there is no axis to construct.
P.S. You don't even need the center to solve the problem. Given an ellipse, you can use the fact that the line passing through the midpoints of two parralel cords passes through the centre of the ellipse.
So given an ellipse without the centre, you start by doing the following:
Pick an arbitrary chord $EF$. Pick another point $G$ on the ellipse and draw the parralel through $G$ to $EF$. Let $H$ be the intersection between this parralel and the ellipse.
Pick another cord $IJ$ which is NOT parralel to $EF$. Draw another chord as above $KL \parallel EF$.
Now, if $M,N, P,Q$ are the midpoints of $EF, GH, IJ, KL$ then $MN \cap PQ= \{O\}$ is the centre of the ellipse.
@Blue's comment shows you the way to go. Here is what your diagram looks like, squeezed horizontally by a factor of $\frac{1}{\sqrt 3}$:
Your rhombus with unit side lengths has become a square of side $\frac{1}{\sqrt2}$, so your ellipse has become a circle with diameter $\frac{1}{\sqrt2}$. Now stretch it back out; the minor axis remains unchanged, and the major axis increases by a factor of $\sqrt3$.
Best Answer
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).
Also see Rytz's construction.
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.