Proof that the evolute of an ellipse is an astroid 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*}
 A: 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 $k''(0)\neq 0$.
