Show that in polar coordinates the functional
$$
S[y]=\int_a^b d x \sqrt{x^2+y^2} \sqrt{1+y^{\prime}(x)^2} \text { becomes } S[r]=\int_{\theta_a}^{\theta_b} d \theta r \sqrt{r^2+r^{\prime}(\theta)^2}
$$
and that the resulting Euler-Lagrange equation is
$$
\frac{d^2 r}{d \theta^2}-\frac{3}{r}\left(\frac{d r}{d \theta}\right)^2-2 r=0 \quad \text { which can be written as } \frac{d^2}{d \theta^2}\left(\frac{1}{r^2}\right)+\frac{4}{r^2}=0
$$
Hence show that equations for the stationary paths are
$$
\frac{1}{r^2}=A \cos 2 \theta+B \sin 2 \theta \quad \text { or } \quad A\left(x^2-y^2\right)+2 B x y=1
$$
where $A$ and $B$ are constants and $0 \leq \theta<\pi$.