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Find envelope of ellipses $ \frac{x^2}{a^2} + \frac{y}{b^2}=1 $ which have $a,b$ and its associated eccentricity $e$ variable while holding its perimeter

$$ p= 4 a E(e) $$

as constant.

Expected to be asteroid-like, passing through

$$ \begin{pmatrix} 0 \\ \frac{\pi a}{2} \end{pmatrix}, \begin{pmatrix} \frac{a}{\sqrt{2}} \\ \frac{a}{\sqrt{2}} \end{pmatrix}, \begin{pmatrix} \frac{\pi a}{2} \\0 \end{pmatrix}$$

Ng Chung Tak
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Narasimham
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1 Answers1

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\begin{align*} p &= 4a E(k) \\ a(k) &= \frac{p}{4E(k)} \\ F(x,y,k) &= x^2+\frac{y^2}{1-k^2}-\frac{p^2}{16E^2(k)} \\ \frac{\partial F}{\partial k} &= \frac{2ky^2}{(1-k^2)^2}- \frac{p^2}{8E^3(k)} \left[ \frac{K(k)-E(k)}{k} \right] \end{align*}

The envelope is given by $$F=\dfrac{\partial F}{\partial k}=0$$

On solving,

$$ \begin{pmatrix} x \\ y \end{pmatrix}= \frac{p}{4kE(k)} \begin{pmatrix} \pm \sqrt{1-\dfrac{(1-k^2)K(k)}{E(k)}} \; \\[5pt] \pm (1-k^2)\sqrt{\dfrac{K(k)}{E(k)}-1} \; \end{pmatrix} \, , \quad 0<k<1 \tag{1} $$

If we adimit $a<b$, then $$ \begin{pmatrix} x \\ y \end{pmatrix}= \frac{p}{4kE(k)} \begin{pmatrix} \pm (1-k^2)\sqrt{\dfrac{K(k)}{E(k)}-1} \; \\[5pt] \pm \sqrt{1-\dfrac{(1-k^2)K(k)}{E(k)}} \; \end{pmatrix} \, , \quad 0<k<1 \tag{2} $$

A plot of the envelope for $p=4$ with the ellipses is shown below:

enter image description here

Ng Chung Tak
  • 19,693
  • Why does the envelope still depend on parameter $k$? – Narasimham Mar 08 '17 at 22:43
  • Usually, we eliminate the parameter to get an implicit equation. However in this case, it's unlikely to get an implicit equation in close form. Why not we parametrize with it? A bit jump outside the box this time. – Ng Chung Tak Mar 09 '17 at 10:39
  • Can we see a sketch of the Asteriod? can we get its natural/intrinsic equation? – Narasimham Mar 09 '17 at 12:03
  • The plot of the envelope is posted. It looks like a cushion. – Ng Chung Tak Mar 09 '17 at 14:51
  • In the first option $y$ calculates to zero, but not to the expected value $\pi a/2 $ ; Also you plotted $1.$ – Narasimham Mar 10 '17 at 12:51
  • You set $p=4\times \dfrac{\pi a}{2}=2\pi a$, if take $p=4$, then $a=\dfrac{2}{\pi}$ .i.e. $\dfrac{\pi a}{2}=1$. – Ng Chung Tak Mar 10 '17 at 13:07
  • In the meantime could the cushion be parameterized only in terms of $p,$ thus eliminating/avoiding $ k$, which should be absent in the singular solution/envelope? – Narasimham Sep 06 '20 at 07:24
  • Fitting special points, we may get an approximate relation $$\Large |x|^{\frac{\ln 2}{\ln \pi-\ln \sqrt{2}}}+|y|^{\frac{\ln 2}{\ln \pi-\ln \sqrt{2}}} \approx \frac{p}{4}$$ which holds good. – Ng Chung Tak Sep 06 '20 at 09:36