First, apologize for using $ \alpha $ and $ a $ together, then
Let $a \in \mathbb{R}^+$, and let $\alpha : (-a, a) \rightarrow \mathbb{R}^2$ be a curve parameterized by arc length (p.b.a.l.) with $k_{\alpha}(s) = k_{\alpha}(-s)$ for each $s \in (-a, a)$. Prove that the trace of $\alpha$ is symmetric relative to the normal line of $\alpha$ at $0$.
We have defined the curvature of $ \alpha $ as $ k_{\alpha}(s) = \langle T '(s), N (s) \rangle $, where $ T (s) = \alpha' (s) $ and $N (s) = J T (s) $. Here $ J: \mathbb {R}^2 \rightarrow \mathbb {R}^2 $ is the ninety degree counterclockwise rotation. We also have $ T'(s) = k_{\alpha}(s)N(s) $ and $ N'(s) = -k_{\alpha} (s) T(s) $, which are the Frenet equations of the curve $ \alpha $.
I have the following suggestion, but I don't know how to use this in the proof of this statement. I have not really been able to understand how to get to the result using the two mentioned exercises and the fundamental theorem. How can I do this?
Hint: Let $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the plane symmetry whose axis is the normal line of $\alpha$ at $\alpha(0)$. Define another curve $\beta : (-a,a) \rightarrow \mathbb{R}^2$ by $\beta(s) = \phi(\alpha(-s))$ for each $s \in (-a,a)$. Using Exercises 1.22 and 1.23, check that $k_{\beta} = k_{\alpha}$. Also, since $\beta(0) = \alpha(0)$ and $\beta'(0) = \alpha'(0)$, we complete the exercise by using the fundamental Theorema of the local theory of planes curves.
The two exercises mentioned are:
Exercise 1.22: Let $\alpha, \beta : I \rightarrow \mathbb{R}^2$ be toe plane curves p.b.a.l such that $k_{\alpha}(s)=-k_{\beta}(s)$ for every $s \in I$. Then there is an inverse rigid motion $M: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $\beta = M \circ \alpha.$
Exercise 1.23: If $\alpha : I \rightarrow \mathbb{R}^2$ is a curve p.b.a.l. defined on an open interval of $\mathbb{R}$ containing the origin and symmetric relative to it, we define another curve $\beta : I \rightarrow \mathbb{R}^2$ by $\beta(s) = \alpha(-s)$ for each $s \in I$. Then $\beta $ is p.b.a.l. and $k_{\beta}(s) = -k_{\alpha}(-s)$.
Using these two problems I have the following equality Let's see first that: $$k_{\alpha}(s)=k_{\beta}(s)$$
Let $\gamma=\phi \circ \alpha$ then by exercise 1.23, $\beta(s)=\gamma(-s)$ then $-k_{\beta}(s)=-k_{\gamma}(-s)$ but I need prove that $k_{\alpha}(s)=-k_{\beta}(s)$ for use exercise 1.22 Could you give me a suggestion or a better way to test this last equality? and then But if I try this last equality I don't know what to use
