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This paper is very useful in how it explains the mapping of any coordinates $(x,y)$ across an ellipse with the function $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

to $$\left(\frac{a^2b^2x}{a^2y^2+b^2x^2},\frac{a^2b^2y}{a^2y^2+b^2x^2}\right)$$

But what if that ellipse is a rotated one, as described here?

I can't even figure it out if I compare it with a circle, but if anyone could help me convert $(x,y)$ to the inverted coordinates, I'd be very happy.

Hosein Rahnama
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soupynoodles
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  • I think the deduction method should be the same as for the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ but a little more complicated for the calculations. Optionally, you can transform the ellipse rotated to its simplest form and then turned back to the original. Am I wrong? – Piquito Dec 05 '15 at 13:12
  • hi @Ataulfo I like the idea of changing the ellipse of inversion, but I'm not really allowed to do that :( I'm also not sure of the method should be the same for the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$. It looks like calculations are going to be more complicated, but how is the deduction method same or different? – soupynoodles Dec 05 '15 at 13:15
  • hi @amd would it be possible if you could help? thanks! – soupynoodles Dec 05 '15 at 16:54

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