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Hello, I am sculptor of rather complex geometrical objects not a mathematician for sure.

Designing a sculpture I discovered that both foci of an ellipse can be constructed using the intersection of two circles of equal diameter. The diameter is also equal to the long axes of the ellipse. The short axes of the same ellipse is the distance between both midpoints of the circles. It's really a beautifully simple coherence, but Wikipedia doesn't mention this method. Is it really unknown?

Intelligenti pauca
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BastiaanJ.
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    Describe the method, this post is unclear. – jimjim Jun 12 '18 at 15:01
  • Did you see the picture? – BastiaanJ. Jun 12 '18 at 15:03
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    It is somewhat clear. –  Jun 12 '18 at 15:04
  • yes, they are confusing, not sure what to make of them – jimjim Jun 12 '18 at 15:05
  • Both foci are the intersection points of the circles. – BastiaanJ. Jun 12 '18 at 15:05
  • Shifting the circles with respect to each other changes the eccentricity of the ellips. – BastiaanJ. Jun 12 '18 at 15:06
  • @JefferyOpoku-Mensah : then it would be easy to add explanation for each pic as what it is showing. – jimjim Jun 12 '18 at 15:07
  • I am writing an answer. –  Jun 12 '18 at 15:09
  • It shows the following fact: Given two circles centered at points diametrically opposite on the minor axis with diameter equal to the major axis' length, the intersection of the two circles are the foci of the ellipse. –  Jun 12 '18 at 15:19
  • Look at the rectangular triangle in the figure at https://en.wikipedia.org/wiki/Ellipse#Ellipse_in_Cartesian_coordinates – Jean-Claude Arbaut Jun 12 '18 at 15:31
  • I would like to show you guys something else. Just wonder what you think of it. – BastiaanJ. Jun 12 '18 at 15:45
  • Making use of the relative motion of both circles I replaced the short axes a, with v (velocity) and the chords with c (velocity of light). This pictures the dynamic similarity of the eccentricity of an ellips with the Lorentz factor. Since the images didn't upload in this box I posted a next question to illustrate and explain this, and of course to ask for your comments. – BastiaanJ. Jun 12 '18 at 18:47

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Yes, this is simple to show.

Consider the ellipse $\mathcal E$ with points $C, D$ diametrically opposite on the minor axis. Then draw circles $\omega_1, \omega_2$ with center at $C, D$ and the same diameter as the major axis.

I claim that $\omega_1 \cap \omega_2 = \{A, B\}$, the foci of the ellipse. Let $\mathcal{E}'$ be the ellipse tangent to $\mathcal{E}$ at $E$ and $F$, with foci at $A,B$. Clearly, it suffices to show $\mathcal{E} = \mathcal{E}'$. Now

$$AE + EB = 2r = AC + CB$$

So, $C$ lies on $\mathcal{E}'$. Since $\mathcal E$ and $\mathcal{E}'$ only differ by a scaling in the $x$-direction, and they both share $C$, the scaling factor is $1$, so $\mathcal{E} = \mathcal{E}'$.

  • Nice thanks. You agree this method works. But my actual question was if this method is new? – BastiaanJ. Jun 12 '18 at 15:23
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    If it is this simple, then it is almost surely not new. It might not have a name, but anyone experienced would be able to think of it/be able to prove it. –  Jun 12 '18 at 15:24
  • Although simple I still suspect it might be new, just because it is very practical and Wikipedia doesn't mention it. It even sheds light on Einsteins calculations of e= mc² – BastiaanJ. Jun 12 '18 at 15:27
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    No, I assure you this is not new. Doing a simple internet search, I was able to find the following.

    Constructing the foci of the ellipse

     –  Jun 12 '18 at 15:28
  • OK it looks the same only reasoned from the ellipse, not from two interacting circles. – BastiaanJ. Jun 12 '18 at 15:33
  • What I showed is only half of the method. You can put to more circles of equal radius having their midpoint on a en b, to construct the opposite ellips and to describe an oscillation. I published this method on internet in 2013. – BastiaanJ. Jun 14 '18 at 08:03