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are the envelopes of the following set of ellipses "classical" curves of mathematics, that appear naturally as the solution of a mathematical/physical problem;

$$\mathcal{E}:=\Biggr\{\frac{(x-e)^2}{a^2} +\frac{y^2}{a^2-e^2}-1\ =\ 0\Biggr\},\quad e\in[0,a)$$

the foci of these ellipses have coordinates $(0,0)$ and $(2e,0)$

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2 Answers 2

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For the current version of the question, we can just derive the equation of the envelope in the usual manner. Using Mathematica to do the tedious algebra on my behalf, we obtain:

First[Factor[GroebnerBasis[{(x - e)^2/a^2 + y^2/(a^2 - e^2) == 1,
                            D[(x - e)^2/a^2 + y^2/(a^2 - e^2) - 1, e] == 0}, {x, y, a}, e]]]
   (4 a^2 - x^2 - 4 a y) (4 a^2 - x^2 + 4 a y) (x^2 + y^2)

where we see the envelope has three components: an isolated point at $(0,0)$ and two parabolas (which are certainly "classical curves of mathematics" as the OP asks), $y=\pm\dfrac{x^2}{4a}\mp a$:

With[{a = 1}, 
     Show[ParametricPlot[Table[{e + a Cos[t], Sqrt[a^2 - e^2] Sin[t]}, {e, 0, a, a/24}],
                         {t, 0, 2 π}], 
          Plot[{x^2/(4 a) - a, a - x^2/(4 a)}, {x, -2, 2}, 
               PlotStyle -> Directive[AbsoluteThickness[4], ColorData[97, 3]]], 
          AspectRatio -> Automatic, PlotRange -> {{-2, 2}, Automatic}]]

ellipses and their envelope

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  • $\begingroup$ Beautiful. This looks like an eye. $\endgroup$ Commented Aug 4, 2019 at 10:49
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What do you mean by the "envelope" of the $\mathcal{E}_1$ set of ellipses?


          Ellipses
          $\mathcal{E}_1$: $a=1$; $e=0,0.1,\ldots,0.9,1$.


Added after OP change, now $\mathcal{E}$:


          Ellipses2
          $a=2$; $e=0,0.1,\ldots,a$.


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    $\begingroup$ I will delete the $\mathcal{E}_1$ part of the question; I had added that as the "ellipses in standard position" counterpart to the $\mathcal{E}_2$ set of ellipses that have their left focus in the origin and the right one on the x-axis. $\endgroup$ Commented Aug 3, 2019 at 18:17

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