We fix $q$ the quadratic form on $\mathbb{R}^2$ given by $q\left(x_1, x_2\right)=x_1 x_2$. Let
$$\mathrm{O}(q)=\left\{f \in \mathrm{GL}_2(\mathbb{R}) \mid q(x)=q(f(x)) \text { pour tout } x \in \mathbb{R}^2\right\}$$
Show that $\mathrm{O}(q)$ is the union of a set of vector symmetries and the set of applications $\gamma_a$ defined by $\gamma_a\left(e_1\right)=a e_1$ and $\gamma_a\left(e_2\right)=a^{-1} e_2$ (for $a \in \mathbb{R}^*$ ).
I know that the matrix of $q$ in the canonical basis is $$M_q = \left(\begin{array}{ll} 0 & 1/2 \\ 1/2 & 0 \end{array}\right)$$ and the relation $q(x) = q(f(x))$ gives that if $P$ is the matrix of $f \in \mathrm{GL}_2(\mathbb{R}) $ then $$ M_q = P^t M_q P. $$ What do I have to do next?
Let $ P= \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} $ be an element of $O(q)$,then $\forall x_{1},x_{2}\in \mathbb{R},acx_{1}+(ad+bc)x_{1}x_{2}+bd x_{2} =x_{1}x_{2}$ so,we must have $ac=bd=0$ and $ad+bc=1$,if $a=0$, then $c\neq 0 ,b\neq 0$, so $d=0$ and $bc=1$, in particular, $P$ must be of the form: $ P_{b}= \begin{pmatrix} 0 & b \\ b^{-1} & 0 \\ \end{pmatrix} $ if $a\neq 0$, then necesarly $c=b=0,ad=1$,So $P$ must be of the form: $ P= \begin{pmatrix} a & 0\\ 0 & a^{-1} \\ \end{pmatrix} $ which correspond to $\gamma_{\alpha}$,finally remark that $P_{b}=R\circ\gamma_{b^{-1}}$ where $R$ is a reflection along the line having equation $x_{2}=x_{1}$