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Given a set of N > 2 (two-dimensional and coplanar) ellipses, all centered at the origin, how do I find the ellipse with the minimum area which encloses all of them?

Background:
Thanks to Will Jagy for answering my similar question re: two ellipses. I had initially naively thought I could apply this solution to a set of more than two ellipses, all centered at the origin, simply by iterating through the set, using the previously calculated enclosing ellipse and the next ellipse in the set to arrive at the solution I wanted. However, as Will pointed out to me, this algorithm is not guaranteed to give the minimum-area enclosing ellipse for every set of ellipses. And thus, my new question.

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  • $\begingroup$ By the way, Dave, there is no need to create a new account for each question. In fact, the way MO works, as you post questions and answers on a single account, you accrue reputation points which grants you more power over MO. $\endgroup$
    – Lee Mosher
    Commented Feb 11, 2013 at 14:54
  • $\begingroup$ @Lee I actually haven't created an account. If I have the need to ask another question or want to answer another question I come across in this forum I will adhere to my rule of three and create one. $\endgroup$
    – Dave
    Commented Feb 11, 2013 at 15:45
  • $\begingroup$ This looks like a duplicate of mathoverflow.net/questions/119663 $\endgroup$
    – Misha
    Commented Feb 11, 2013 at 17:05
  • $\begingroup$ @Misha Actually yes it is. However, the solution provided in the other question only is valid if the set has only two ellipses, at least from what I can tell. I guess this question can be deleted if the other can be solved more completely. $\endgroup$
    – Dave
    Commented Feb 11, 2013 at 20:44
  • $\begingroup$ @Dave: Actually, we mean different questions, I was referring to the one asked by the OP named "Its" (I do not think it was you!). Its' original question was sloppy as he did not ask for the least volume ellipse, but in a comment he provided the clarification that this is what he meant. $\endgroup$
    – Misha
    Commented Feb 11, 2013 at 21:35

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