Let $b >0$ and $\alpha : (-a, a) \rightarrow \mathbb{R^2}$ be a plane curve parametrized by arc length. Suppose that: $$k(s) = k(-s) \quad \forall s \in (-a,a)$$ where $k(s)$ denotes the curvature of $\alpha$ at the point $s$. Show that the trace of $\alpha$ is symmetric with respect to the normal line of $\alpha$ at $0$.
My try
I denote by $\phi:\mathbb{R}^2\rightarrow\mathbb{R}^2$ the function that takes a point $b$ to his symmetric relative to $\alpha(0) + L(N_{\alpha}(0))$ (normal line at zero). Then I define the curve $\beta(s)=\phi(\alpha(-s))$ and my attempt is to prove that $\beta=\alpha$.
I know that $\beta(0)=\alpha(0)$, so I tried to show that the curves have the same curvatures $K_{\alpha}=K_{\beta}$ and the same tangents at zero $T_{\alpha}(0)=T_{\beta}(0)$, but I don't know how to show this.
Your approach seems great. So, if you differentiate carefully and correctly with the chain rule, you should find exactly that. (If we take $\alpha(0)=0$, then $\phi$ will be linear, and so $(\phi\circ\gamma)'(s) = \phi(\gamma'(s))$.) You should get $$\beta'(s) = -\phi(T_\alpha(-s)) \quad\text{and}\quad \beta''(s) = \kappa(s)\phi(N_\alpha(-s)).$$ So we have $T_\beta(0)=-\phi(T_\alpha(0)) = T_\alpha(0)$ and $\kappa_\beta=\kappa_\alpha$ (why?).