proving the major and minor axis of an ellipse bisect each other?

Question

My question came from a friend of mine the other day, and I am looking for an answer.

how do you prove the major and minor axis of an ellipse bisect each other?

Thanks

Answer 1

Because an ellipse can always be viewed as a foreshortened circle, the foreshortening does not break symmetry. That is, whatever foreshortening occurs about an axis of rotation through the center of the circle, it always occurs equally either side of that axis (the major axis). The location of the minor axis does not shift from being centered and orthogonal to major axis through that rotation.

This is an explanation rather than a proof. A more formal proof would show how rotation by angle $\alpha$ would reduce the radius of the circle to $r\cdot \cos\alpha$, the minor axis, equally both above and below the axis of rotation while maintaining its center position on the major axis.

Answer 2

If you are allowed to use affine plane ideas then all ellipses are affinely equivalent to each other and particulary to all circles. In any circle all diameters bisect each other and thus in any ellipse, all line segments passing through the center, including the major and minor axis, bisect each other. A point being in the middle of a line segment is an example of an affine combination of points and thus is affinely invariant.

Answer 3

If your ellipse is defined by a standard equation (a quadratic form) then you can translate and rotate the coordinate axes so that the equation becomes $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. $$ Then the endpoints of the axes are $(\pm a, 0)$ and $(0, \pm b)$ and the origin where they cross clearly bisects both.

Edit in response to comment.

You can read about that rotation here: http://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS/upfiles/topics/ess_at_13_ra_stu.pdf

You might also get something from

https://en.wikipedia.org/wiki/Ellipse

https://www.maa.org/external_archive/joma/Volume8/Kalman/General.html