I am trying to compute the optimal path length for getting from Sydney to Hong Kong via a tunnel and using only the force of gravity. (See figure.) By using calculus of variations and the Euler-Lagrange I determine that the time taken is given by $$t=\int_{x_A}^{x_B}\sqrt{\frac{R}{g}}\frac{\sqrt{1+y'^2}}{\sqrt{R^2-x^2-y^2}}dx,$$ and from this I obtain using the Euler-Lagrange: $$\sqrt{\frac{R}{g}}\Big[\frac{x\sqrt{1+y'^2}}{(R^2-y^2-x^2)^{3/2}}-\frac{d}{dx}\Big(\frac{y'(1+y'^2)^{-1/2}}{\sqrt{R^2-y^2-x^2}}\Big)\Big]=0.$$ I know what I need to obtain, but have absolutely no clue how to get there. I am guessing it would make my life a lot easier if I change to polar coordinates? The answer is a set of parametric equations and is $$x(\theta)=R\Big[(1-b)\cos{\theta}+b\cos{\frac{1-b}{b}\theta}\Big],$$ $$y(\theta)=R\Big[(1-b)\sin{\theta}-b\sin{\frac{1-b}{b}\theta} \Big],$$ with $b \in [0,1]$.
Can someone please show how to obtain this?