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$\hebl$

**The Beltrami Identity:**

If the Lagrangian $\:L\plr{y,y',x}\:$ of a system does not depend explicitly on $\:x$, that is
\begin{equation}
\dfrac{\partial L}{\partial x}\e 0
\tl{01}
\end{equation}
then from the Euler-Lagrange equation
\begin{equation}
\dfrac{\mr d}{\mr dx}\plr{\dfrac{\partial L}{\partial y'}}\m\dfrac{\partial L}{\partial y}\e 0
\tl{02}
\end{equation}
we have
\begin{equation}
\dfrac{\mr d}{\mr dx}\plr{y'\dfrac{\partial L}{\partial y'}\m L}\e 0
\tl{03}
\end{equation}
so
\begin{equation}
\boxed{\:\:y'\dfrac{\partial L}{\partial y'}\m L\e \texttt{constant}\:\:}\quad \texttt{(Beltrami Identity)}
\tl{04}
\end{equation}

$\hebl$

For your Lagrangian
\begin{equation}
\begin{split}
\frac{\sqrt{\dot x ^2 \p \dot y ^2}}{y}\mr dt & \e \frac{\sqrt{1\p\plr{\dfrac{\dot y}{\dot x}}^2}}{y}\dot x\,\mr dt\e\frac{\sqrt{1\p\plr{\dfrac{\mr dy/\mr dt}{\mr dx/\mr dt}}^2}}{y}\dfrac{\mr dx}{\mr dt}\,\mr dt\\
&\e\frac{\sqrt{1\p\plr{\dfrac{\mr dy}{\mr dx}}^2}}{y}\mr dx\e\frac{\sqrt{1\p y'^{2}}}{y}\mr dx\\
\end{split}
\tl{05}
\end{equation}
that is
\begin{equation}
L\plr{y,y',x}\e\frac{\sqrt{1\p y'^{2}}}{y}
\tl{06}
\end{equation}

Using the Lagrangian \eqref{06} we could find the $\:x\m$parametric representation $\:\blr{x,y\plr{x}}\:$ of the curve directly bypassing its $\:t\m$parametric representation $\:\blr{x\plr{t},y\plr{t}}$, that is the equations of the motion.

$\hebl$