Proof that the evolute of an ellipse is an astroid

Question

I need to prove that the evolute of the ellipse $\gamma (t) = (a\cos t, b\sin t)$ with $ a, b > 0 $ is the astroid:

$\rho (t) = (\frac{(a^2-b^2)\cos^3 t}{a},\frac{(b^2-a^2)\sin^3 t}{b} )$

I am little bit insecure if this is right. Did I make any mistake?

\begin{align*} \text{Curvature of $\rho$:} \\ \kappa&=\frac{ab}{(a^2\sin^2t+b^2\cos^2t)^{\frac{3}{2}}}\neq0 \\ \text{ Normal:} \\ n(t)&=\frac{(-b\cos t,-a\sin t)}{(a^2\sin^2t+b^2\cos^2t)^{\frac{1}{2}}} \\ \text{Hence, } \beta(t) \\ \beta(t)&=(a\cos t,b\sin t)+\frac{a^2\sin^2t+b^2\cos^2t}{ab}(-b\cos t,-a\sin t) \\ \beta(t)&= (\frac{(a^2-b^2)\cos^3 t}{a},\frac{(b^2-a^2)\sin^3 t}{b} ) \\ \text{Evolute's trace is described by the astroid:} \\ (ax)^\frac{2}{3}+(by)^\frac{2}{3}&=(a^2-b^2)^\frac{2}{3} \\ \beta(t) \text{ is not regular for the following values of t} \\ t&=0=\frac{\pi}{2}=\frac{3\pi}{2} \end{align*}

@Bernard, ok, I was not accusing you. Thanks for the edition. It is better to read now =) — Pedro Delfino, Dec 18 '17 at 16:25
https://math.stackexchange.com/questions/1951775/show-that-the-perimeter-of-the-evolute-of-the-ellipse-is-4-frac-a2b-fr — lab bhattacharjee, Dec 18 '17 at 16:26
@labbhattacharjee: I have no idea why that reference is relevant. ... Pedro, it would help if you wrote more and included the definition of the evolute that you're using. But this looks fine. — Ted Shifrin, Dec 18 '17 at 18:30

Upax · Accepted Answer · 2017-12-28T19:55:43.557

The evolute of a curve C is the locus of the centers of curvature. Let $x:U \rightarrow \mathbb{R}^2$ be a regular parametric plane curve that is of class $C^2$ , i.e., has a continuous second derivative. Let $U'$ be a subinterval of U over which the geodesic curvature $k(t) \neq 0$. Then over the interval U, the evolute of x has the following parametrization: \begin{equation} E(t)=x(t)+\frac{1}{k} n(t) \end{equation} where $n$ is the unit normal vector. The geodesic curvature and the unit normal vector you computed are correct. Then when replacing in the evolute expression you get \begin{equation} E(t)=\big(\frac{a^2-b^2}{a} \cos^3(t) ,-\frac{a^2-b^2}{b} \sin^3(t) \big) \end{equation} this correspond to the curve $\beta(t)$. Now:

The evolute is a regular curve if along the arc of x(t) , $k(t) \neq 0$ and $k'(t)\neq 0$ hold.
At the point $x(t_0)$ where $k(t_0)=0$, a normal line to x is the asymptote to both branches of the evolute.
If at a point $x(t_0)$ where $k(t)\neq0$, but $k'(t)=0$ and $k''(t)\neq0$, then $(x(t_0),y(t_0))$ is a singular point of the evolute. At this point both regular arcs of the evolute meet: they have a common tangent line and are located in opposite half-planes from it.

Then $\beta$ is not regular the the product $\cos{(t)} \sin{(t)}$ is zero. For $t=0$ we have $k(0)=a/b^2$, while $k'(0)=0$. For $t=\pi/2$ and $t=3 \pi/2$ the geodesic curvature is $k(0)=b/a^2$, while $k'(0)=0$. For all this point $k''(0)\neq 0$.

Proof that the evolute of an ellipse is an astroid

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