Show That The Perimeter Of The Evolute Of The Ellipse Is $4(\frac{ a^2}{b} - \frac{b^2}{a})$.

Question

According to the question we have to find the perimeter of evolute of an ellipse.

For an ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, the equation of its evolute is

$$(ax)^\frac{2}{3}+(by)^\frac{2}{3} = (a^2 - b^2)^\frac{2}{3}$$

To get the points of intersection of Ellipse and evolute we have to solve these two Equations. How can we solve? Plus, Can you help me to find whole sum?

Answer 1

solving the ellipse equation for $$y^2$$ we get $$y^2=b^2-\frac{b^2x^2}{a^2}$$ and the evolute as $$b^2y^2=((a^2-b^2)^{2/3}-(ax)^{2/3})^3$$ then you can eliminate $$y^2$$

Answer 2

Assuming $a>b$,

\begin{align*} \mathbf{r} &= \begin{pmatrix} \frac{a^2-b^2}{a} \cos^3 t \\ \frac{b^2-a^2}{b} \sin^3 t \end{pmatrix} \\ \dot{\mathbf{r}} &= \begin{pmatrix} -\frac{3(a^2-b^2)}{a} \cos^2 t \sin t \\ \frac{3(b^2-a^2)}{b} \sin^2 t \cos t \end{pmatrix} \\ \frac{ds}{dt} &= \frac{3(a^2-b^2)}{ab} |\sin t \cos t| \sqrt{a^2 \sin^2 t+b^2 \cos^2 t} \\ P &= \frac{3(a^2-b^2)}{ab} \int_{0}^{2\pi} |\sin t \cos t| \sqrt{a^2 \sin^2 t+b^2 \cos^2 t \,} \; dt \\ &= \frac{12(a^2-b^2)}{ab} \int_{0}^{\frac{\pi}{2}} \sin t \cos t \sqrt{a^2 \sin^2 t+b^2 \cos^2 t\,} \; dt \\ &= \frac{12(a^2-b^2)}{ab} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2t}{2} \sqrt{\frac{a^2+b^2}{2}-\frac{a^2-b^2}{2}\cos 2t \,} \; dt \\ &= \frac{12(a^2-b^2)}{ab} \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{a^2+b^2}{2}-\frac{a^2-b^2}{2}\cos 2t \,} \; d(-\cos 2t) \\ &= \frac{12(a^2-b^2)}{ab} \left[ \frac{1}{3(a^2-b^2)} \left( \frac{a^2+b^2}{2}-\frac{a^2-b^2}{2} \cos 2t \right)^{3/2} \right]_{0}^{\frac{\pi}{2}} \\ &= \frac{4(a^3-b^3)}{ab} \end{align*}