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When you invert an ellipse across another ellipse, you get an egg-shape. What is the equation of that, in general terms?

I tried to Google it, but I only came across this (fairly) useful page.

http://www.mathematische-basteleien.de/eggcurves.htm

Nowhere does it provide the equation of an egg where the axis of symmetry is vertical, not horizontal.

Also, how would this equation differ if the less pointed end of the egg was facing upwards or downwards?

soupynoodles
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  • To change from a horizontal axis of symmetry to a vertical axis of symmetry, just interchange $x$ and $y$ in the equation. To then change which end is pointing up (that is, reflect the egg in the $x$-axis), replace $y$ with $-y$. (Note it's important to do this after interchanging $x$ and $y$ or else it will have no effect.) – kccu Dec 04 '15 at 15:06
  • Thanks @kccu I just noticed, though, that what I'm really trying to find (I'm editing the question atm) is specifically the general equation of the egg formed after you invert an ellipse across another ellipse with different eccentricities. Is it a cubic egg? Or a Wassenaar egg? – soupynoodles Dec 04 '15 at 15:09
  • how do you invert something across an ellipse ? – mercio Dec 04 '15 at 15:23
  • @mercio Um I think $\left(x,y\right)$ map to $$\left(\frac{a^2b^2x}{a^2y^2+b^2x^2},\frac{a^2b^2y}{a^2y^2+b^2x^2}\right)$$ but I am not sure. – soupynoodles Dec 04 '15 at 15:28
  • @mercio when I tried to plug that stuff into my original ellipse with equation $$\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$$ Idk the algebra was too hard. Plus I ended up with something that looked like an ellipse standard form and not an egg, which is what I get when I test values into CAS.

    Since the ellipse I'm trying to invert is not homothetic (same eccentricity) of the ellipse of inversion, the resulting ellipse must be an egg but the algebra just shows me it's an ellipse :(

     – soupynoodles Dec 04 '15 at 15:30
  • hi @mercio by any chance could you find the equation? – soupynoodles Dec 04 '15 at 15:54
  • Try starting with inversion relative a circle to simplify the manipulations, then extend that result to inversion in an ellipse, as I did when answering your old question (http://math.stackexchange.com/a/1450168/265466). – amd Dec 04 '15 at 19:46

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