= Solution
{c}
A constant displacement proportional to $(1,-3)$ leaves the <velocities> unchanged and changes $3x+y$ by zero. It is therefore a continuous symmetry of the <Lagrangian>. The <generalized momentum> components are
$$
p_x=\frac{\partial\mathcal L}{\partial\dot x}=m(6\dot x+\dot y),\qquad
p_y=\frac{\partial\mathcal L}{\partial\dot y}=m(\dot x+4\dot y).
$$
<Noether's theorem> gives the conserved quantity
$$
\boxed{P=p_x-3p_y=m(3\dot x-11\dot y)=\text{constant}.}
$$
A direct check using the <Euler-Lagrange equations> is
$$
6\ddot x+\ddot y=-3g,\qquad \ddot x+4\ddot y=-g
\quad\Longrightarrow\quad 3\ddot x-11\ddot y=0.
$$
The normalization of a symmetry generator is arbitrary. One can take the transformation parameter to have <length> units and $(K_x,K_y)=(1,-3)$ dimensionless, giving $P$ the units of <momentum>. Alternatively, for the paper's convention with dimensionless parameter and <length>-valued $K_i$, use $(K_x,K_y)=(\ell,-3\ell)$; the conserved quantity is then $\ell P$. The conserved content is identical.
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