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How one get from this ellipse equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ that ellipse equation $$\frac{x^2+y^2}{F(\phi)^2}=1,$$ where $$F(\phi)=\frac{ab}{\sqrt{(b\cos\phi)^2+(a\sin\phi)^2}}$$ and $$\phi=\phi(x,y)=\arctan(\frac{y}{x}).$$


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In general, we have $x=r\cos\phi$ and $y=r\sin\phi$. Substitute in the usual equation of the ellipse. We get $$\frac{r^2\cos^2\phi}{a^2}+\frac{r^2\sin^2\phi}{b^2}=1.$$ Multiply through by $a^2b^2$, and simplify a little. We get $$r^2\left(b^2\cos^2\phi+a^2\sin^2\phi\right)=a^2b^2.$$ Now solve for $r$. On the usual assumption that $a$ and $b$ are positive, and $r\ge 0$, we get $$r=\frac{ab}{\sqrt{b^2\cos^2\phi+a^2\sin^2\phi}}.$$ This is equivalent to your expression.

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