Concyclic Points of Two Ellipses with Orthogonal Axes

What is this about?

In the applet there are two ellipses each defined by its two foci and a point on its boundary. All six are draggable.

7 July 2013, Created with GeoGebra

Problem

Given two ellipses with orthogonal axes.

four concyclic points on two ellipses

If there are four points of intersection, all four are concyclic.

Hint

Analytic geometry makes a simple but powerful application for this problem. Just choose a suitable system of coordinates. In addition, it makes sense to check various equations that an ellipse can be defined with.

Solution 1

Ellipse can be defined ad the locus of points whose distance to a given line (called directrix) times a given number from $(0,1)$ equals the distance of the point to a given one (called focus.) Ellipse also has a long axis and a short axis, and the directrix is perpendicular to the long which also house its foci.

So let's choose a Cartesian coordinate system with the axes along the long axes of the two ellipses. Let their foci be respectively at $(\bar{a},0)$ and $(0,\tilde{a}),$ directrices at $y=-\bar{b}$ and $x=-\tilde{b}$ such that one is given by the equation $|(y+\bar{b})\bar{e}|=\sqrt{x^{2}+(y-\bar{a})^{2}}$ while the other is defined by $|(x+\tilde{b})\tilde{e}|=\sqrt{(x-\tilde{a})^{2}+y^{2}},$ where $\bar{e}$ and $\tilde{e}$ are their eccentricities. The four points of intersetion $(x_{i},y_{i}),$ $i=1,2,3,4,$ satisfy both equations. The reason I chose these equations is that a single squaring eliminates all radicals. We thus have

$(y_{i}+\bar{b})^{2}=[x_{i}^{2}+(y_{i}-\bar{a})^{2}]/\bar{e}^{2}$ and $(x_{i}+\tilde{b})^{2}=[(x_{i}-\tilde{a})^{2}+y_{i}^{2}]/\tilde{e}^{2},$

$i=1,2,3,4.$ Each of the four points satisfies both equations and, therefore, the sum of the two. The sum of the two equations is of the second order. I am mostly concerned with the coefficients by $x^2$ and $y^2.$ Taking all the terms to the right, the sum of the two equations is

$x^{2}[(1/\bar{e})^{2} + (1/\tilde{e})^{2} - 1]+ y^{2}[(1/\bar{e})^{2} + (1/\tilde{e})^{2} - 1] +\ldots = 0.$

The coefficients by $x^2$ and $y^2$ are equal and differ from 0, implying that the equation is that of a circle.

Solution 2

In the first solution I assumed that the directrices of the two ellipses are perpendicular. On closer inspection, this assumption is not necessary. Let there be two ellipses:

$\displaystyle\bar{f}(x,y)=\frac{(x-\bar{u})^2}{\bar{a}^2}+\frac{(y-\bar{v})^2}{\bar{b}^2}-1=0,\space$ and
$\displaystyle\tilde{f}(x,y)=\frac{(x-\tilde{u})^2}{\tilde{a}^2}+\frac{(y-\tilde{v})^2}{\tilde{b}^2}-1=0.$

The plan is to show that there is a linear combination $\alpha\bar{f}+\beta\tilde{f}$ of the two equations that describes a circle. Naturally, our only concern is the equality of the coefficients by $x^2$ and $y^2$. Thus, we seek $\alpha$ and $\beta$ such that

$\displaystyle\frac{\alpha}{\bar{a}^2}+\frac{\beta}{\tilde{a}^2}=\frac{\alpha}{\bar{b}^2}+\frac{\beta}{\tilde{b}^2},$

or,

$\displaystyle\alpha\big(\frac{1}{\bar{a}^2}-\frac{1}{\bar{b}^2}\big)=\beta\big(\frac{1}{\tilde{b}^2}-\frac{1}{\tilde{a}^2}\big).$

Now, if either $\bar{a}^2=\bar{b}^2$ or $\tilde{a}^2=\tilde{b}^2,$ we already have a circle and there is nothing else to prove. Otherwise, we can certainly choose $\alpha$ and $\beta$ to make the equality hold.

Acknowledgment

This problem was posted by a post by Emmanuel Antonio José García at the CutTheKnotMath facebook page.

Conic Sections > Ellipse

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

 62822157

Search by google: