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I have the following system:

$a b x_0 y - \sqrt{a^2 - x_0^2} (a^2 x - a^2 x_0 + b^2 x_0) = 0$

$ x_0^3 ( a^2-b^2) = a^4 x$

I need to solve it and get $(ax)^{2/3} + (by)^{2/3} = (a^2- b^2) ^{2/3}$

This is the only part I have left on a exercice about computing the evolute of the ellipse using the normal rects (I have to do it that way)

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1 Answer 1

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$x_0(a^2-b^2)=\dfrac{a^4x}{x_0^2}, abx_0y=\sqrt{a^2-x_0^2}(a^2x-x_0(a^2-b^2))=\sqrt{a^2-x_0^2}\left(a^2x-\dfrac{a^4x}{x_0^2}\right) \to $

$(by)^2=(ax)^2\left(\dfrac{a^2}{x_0^2}-1\right)^3 \to (by)^{\frac{2}{3}}=(ax)^{\frac{2}{3}}\left(\dfrac{a^2}{x_0^2}-1\right) \to (ax)^{2/3} + (by)^{2/3}=\dfrac{a^{8/3}x^{2/3}}{x_0^2}=\left(\dfrac{a^8x^2}{x_0^6} \right)^{1/3}=\left(\dfrac{a^4x}{x_0^3} \right)^{2/3}=(a^2-b^2)^{2/3}$

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