= Solution
{c}
With $H>0$, set
$$
s=\sin(\phi/2),\qquad y^{-1/2}+\lambda=\frac Hs,\qquad c=H^2>0,\qquad b=\frac\lambda H.
$$
Then the <first integral> gives $1/\sqrt{1+y'^2}=s$. On the descending and ascending halves of the physical track choose respectively $0<\phi<\pi$ and $\pi<\phi<2\pi$, so that $y'=\cot(\phi/2)$. Solving for $y$ gives
$$
\boxed{y(\phi)=\frac{\sin^2(\phi/2)}{c(1-b\sin(\phi/2))^2}.}
$$
The unsquared equation requires $1-bs>0$. At $s=1$ this says \b[$b<1$]; it is not permissible to retain other values merely because the displayed formula for $y$ has a squared denominator.
For completeness, the <Euler-Lagrange equation> also gives $y''=(A'(y)/A(y))(1+y'^2)<0$. Thus $y'$ decreases through zero exactly once, so the chosen parameter runs through the whole single arch without an extra branch. Differentiating the formula for $y$ and dividing by $y'=\cot(\phi/2)$ gives, continuously also at $\phi=\pi$,
$$
\frac{dx}{d\phi}=\frac{\sin^2(\phi/2)}{c(1-b\sin(\phi/2))^3}.
$$
Using $x(0)=0$ yields
$$
\boxed{x(\phi)=\frac1c\int_0^\phi\frac{\sin^2(u/2)}{(1-b\sin(u/2))^3}\,du.}
$$
Since $\sqrt{1+y'^2}=1/\sin(\phi/2)$, the <arc length> is obtained by cancelling one factor of $\sin(\phi/2)$ in $dx/d\phi$. The conditions $x(2\pi)=1$ and $S=L$ become
$$
\boxed{\begin{aligned}
c&=\int_0^{2\pi}\frac{\sin^2(u/2)}{(1-b\sin(u/2))^3}\,du,\\
L&=\frac{\displaystyle\int_0^{2\pi}\frac{\sin(u/2)}{(1-b\sin(u/2))^3}\,du}{\displaystyle\int_0^{2\pi}\frac{\sin^2(u/2)}{(1-b\sin(u/2))^3}\,du}.
\end{aligned}}
$$
These determine a unique parameter pair for each $L>1$. To see uniqueness, regard $L$ as the <expected value> of $1/s$ under a <probability density function> proportional to $s^2/(1-bs)^3$ on $0<u<2\pi$. <Differentiation under the integral sign> gives the <covariance> identity
$$
\frac{dL}{db}=3\operatorname{Cov}\!\left(\frac1s,\frac{s}{1-bs}\right)<0.
$$
The first argument decreases with $s$ and the second increases. The strict sign follows by applying $(f(s)-f(t))(g(s)-g(t))<0$ to two independent unequal values and averaging; the integrable endpoint singularity of $1/s$ causes no problem. As $b\uparrow1$, the weight concentrates at $s=1$, so $L\downarrow1$. As $b=-k\to-\infty$, endpoint estimates give numerator $\sim2/k^2$ and denominator $\sim4\log k/k^3$, so $L\sim k/(2\log k)\to\infty$. Continuity then covers every $L>1$. At $b=0$ one recovers the <cycloid> and $L=4/\pi$.
\b[Smoothness qualification.] Near $\phi=0$,
$$
x\sim\frac{\phi^3}{12c},\qquad y\sim\frac{\phi^2}{4c},\qquad y'\to+\infty,
$$
and the other endpoint has $y'\to-\infty$. These are smooth graphs on $0<x<1$, not $C^1$ graphs on the closed interval. Indeed a finite endpoint slope would make the conserved $H$ diverge as $y\to0$. The <travel time> remains finite because
$$
T=\frac1{\sqrt{2gc}}\int_0^{2\pi}\frac{d\phi}{(1-b\sin(\phi/2))^2}<\infty\qquad(b<1).
$$
Thus the printed phrase “smooth solution” must be understood on the interior, with improper endpoint limits. The variational calculation supplies necessary stationary arches; it does not by itself prove global minimality in every larger class of tracks.
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