Lagrangian first integral I want to extremize $$\int dt \frac{\sqrt{\dot x ^2 + \dot y ^2}}{y}.$$
I have thought that, since the Lagrangian $L(y, \dot y, \dot x)$ is $t$ dependent only implicitly, that i could use the fact that $$L(z,z') \implies  L - z' \partial L / \partial z' = c.$$
So $$L - y' \partial L / \partial y' = c_1,$$
$$L - x' \partial L / \partial x' = c_2$$
But these two equations, when we substitute the values and arrange it, give us
$$dy/dx = c_3 \implies y = c_3 x +b.$$
This is certainly wrong, the answer is supposed to be a circle equation. Even so we can solve it another way, i am still confused: Why did we got the wrong answer using the above two equation? If, for example, the Lagrangian was $\int dt \sqrt{\dot x ^2 + \dot y ^2}$, we could use the above approach to get the answer (in this case, a line is the right answer).
 A: Hint: Noether's theorem yields that
$$\begin{align} L\text{ has no }&x\text{-dependence} \cr
 \quad& \Downarrow&\quad\cr \text{momentum } &p_x \text{ is conserved}, \end{align} $$
and
$$\begin{align} L\text{ has no explicit }&t\text{-dependence} \cr
 \quad& \Downarrow&\quad\cr \text{energy } p_x\dot{x}+p_y\dot{y}&-L\text{ is conserved}. \end{align} $$
A: with
$$L=\frac{\sqrt{\dot x^2+\dot y^2}}{y}$$
and because L is not a function of x you obtain that
$$\frac{\partial L}{\partial \dot x}=\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}\,y}=\text{constant}$$
from here
$$\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}\,y}\mapsto 
\frac{1}{\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,y(x)}=\text{constant}$$
or
$$\sqrt{1+\frac{dy}{dx}^2}\,y(x)=k^2$$
A: $\newcommand{\bl}[1]{\boldsymbol{#1}} 
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\newcommand{\hebl}{\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}}$
$\hebl$
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.
$\hebl$
Hint for the Solution
Insert the Lagrangian \eqref{06} in the Beltrami Identity \eqref{04} to find
\begin{equation}
f\plr{y,y'\e\dfrac{\mr dy}{\mr dx}}\e a\e \texttt{positive constant}
\tl{H-01}
\end{equation}
Solve equation \eqref{H-01} with respect to $\:\mr dx$ to find
\begin{equation}
\mr dx\e g\plr{y}\mr dy
\tl{H-02}
\end{equation}
In equation \eqref{H-02} make a proper convenient change from the variable $\:y\:$ to an angle variable  $\:\theta\:$
\begin{equation}
y\e h\plr{\theta}
\tl{H-03}
\end{equation}
Convert equation \eqref{H-02} to something like that
\begin{equation}
\mr dx\e q\plr{\theta}\mr d\theta
\tl{H-04}
\end{equation}
Integrate equation \eqref{H-04} to have
\begin{equation}
x\e u\plr{\theta}
\tl{H-05}
\end{equation}
Equations \eqref{H-03} and \eqref{H-05} give a $\:\theta\m$parametric representation of the motion orbit.
