A particle of unit mass is moving in the \(x, y\) plane under the influence of
the potential
$$
U=-\frac{1}{r_{1}}-\frac{1}{r_{2}},
$$
where \(r_{1}\) and \(r_{2}\) are the respective distances from the points \((1,0)\)
and \((-1,0)\). Show that if new coordinates are introduced by putting
$$
x=\cosh \varphi \cos \theta, \quad y=\sinh \varphi \sin \theta,
$$
then the Lagrangian governing the motion becomes
$$
L=\frac{1}{2} \Omega\left(\dot{\theta}^{2}+\dot{\varphi}^{2}\right)+2
\Omega^{-1} \cosh \varphi,
$$
where \(\Omega=\cosh ^{2} \varphi-\cos ^{2} \theta\).
Write down the equations of motion and show that if the particle is set in
motion with \(T+U=0\), then
$$
\frac{1}{2}\left(\theta^{2}+\dot{\varphi}^{2}\right)-\frac{2 \cosh
\varphi}{\Omega^{2}}=0
$$
Deduce that
$$
\frac{\mathrm{d}^{2} \theta}{\mathrm{d} \tau^{2}}=0, \quad
\frac{\mathrm{d}^{2} \varphi}{\mathrm{d} \tau^{2}}=2 \sinh \varphi
$$
where \(\mathrm{d} t / \mathrm{d} \tau=\Omega\). Hence find the path of the
particle.