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Given a general ellipse (with axes not necessarily parallel to the x- or y-axes), is there a compass-and-straightedge method for constructing the major and minor axes?

Just to clarify the question: if you know the equation of an ellipse -- say, something like $4x^2 + 9y^2 - 10xy = 40$ -- it is of course possible using algebra and trigonometry to calculate the angle of rotation of the ellipse's axes relative to the x- and y-axes. But if you instead have only a graph of the ellipse (with, let's say, the center identified), rather than its equation, is there a purely geometric way to construct the axes?

It occurs to me as I write this that if you knew where the foci were, the problem would be trivial. So I suppose an equivalent formulation of the problem would be: Given an ellipse, is there a compass-and-straightedge method for constructing its foci?

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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.

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  • $\begingroup$ Ah, lovely and simple. Thanks! $\endgroup$ – mweiss May 30 at 2:17
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    $\begingroup$ @Blue TY, I was just finishing adding this when you wrote the comment, great minds think alike, eh? :) $\endgroup$ – N. S. May 30 at 2:24
  • $\begingroup$ @mweiss You don't even need the centre, check the P.S. $\endgroup$ – N. S. May 30 at 2:24
  • $\begingroup$ Indeed! :) .... $\endgroup$ – Blue May 30 at 2:25

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