Help needed in finding the integral curves given by orbits of one-parameter groups Equip $\mathbb{R}^2$ with standard symplectic structure and inner product. Consider a Hamiltonian $$H=(x,y)A(x,y)^t.$$  I have to determine orbits of one-parameter groups acting by isometries of $\mathbb{R}^2$. 
I understand the following: 

Let $M$ be a smooth manifold. If $\theta: \mathbb{R} \times M \to M $
   is a global flow, then we immediately have following maps: 
$\forall t \in \mathbb{R}, \quad \theta_ t: M \to M$ is diffeomorphism. 
$\forall \ p \in M,$ we have smooth curve $\theta ^ {(p)}: \mathbb{R} \to M$ with image as the orbit of $p$ under the group action. 
And lastly, $\mathbb{R} \to Diff(M)$ is a group homomorphism. 

But I don't know how to use this understanding to solve my problem . Please help! 
 A: The phase space has points $x^{i}\in V_{2}$. The standard inner product is $x^{i}\delta_{ij}x^{j}$ with $\delta_{ij}$ an invariant tensor under the action of the isometry group SO(2). If $R^{i}_{j}$ are the group matrices,
\begin{equation}
\delta_{ij}=[R^{-T}]_{i}^{\ k}[R^{-T}]_{j}^{\ l}\delta_{kl} \ .
\end{equation}
Set up a one-parameter subroup $R=\exp(\theta J)$. Differentiating the previous equation by $\theta$ and setting $\theta=0$ gives the following condition on the Lie algebra elements $J$.
\begin{equation}
J^{i}_{\ j}+J^{\!j}_{\;i}=0 \ .
\end{equation}
These one-parameter subgroups give orbits,
\begin{equation}
x^{i}(\theta)=[\exp(\theta J)]^{i}_{\ j}x^{j}(0)
\end{equation}
so the ODE is,
\begin{equation}
\frac{dx^{i}}{d\theta}=J^{i}_{\ j}x^{j} \ .
\end{equation}
The symplectic space has invariant tensor $\epsilon_{ij}$. The one-parameter subgroups of the symplectic group Sp(2,R) give orbits given by,
\begin{equation}
\frac{dx^{i}}{d\theta}=\epsilon^{jk}\frac{\partial x^{i}}{\partial x^{j}}\frac{\partial H}{\partial x^{k}}=\epsilon^{ik}\frac{\partial H}{\partial x^{k}}
\end{equation}
Equating the last two equations gives the condition for the Hamiltonian to generate isometries.
\begin{equation}
J^{i}_{\ j}x^{j}=\epsilon^{ij}\frac{\partial H}{\partial x^{j}}
\end{equation}
Multiply by $\epsilon_{ik}$,
\begin{equation}
\epsilon_{ik}J^{i}_{\ j}x^{j}=\frac{\partial H}{\partial x^{k}}
\end{equation}
and integrate,
\begin{equation}
H=x^{k}\epsilon_{ik}J^{i}_{\ j}x^{j}=-x^{i}\epsilon_{ik}J^{k}_{\ j}x^{j}
\end{equation}
so that, symmetrizing,
\begin{equation}
A_{ij}=-\frac{1}{2}(\epsilon_{ik}J^{k}_{\ j}+J^{i}_{\ k}\epsilon_{kj})
\end{equation}
so, using the parameters of the A matrix in the question, $\alpha=\delta=J^{1}_{\ 2}$, $\beta=-(J^{1}_{\ 1}+J^{2}_{\ 2})/2$.
