= Solution
Choose the tangent angle $\psi=\arctan y'$ in $(-\pi/2,\pi/2)$, so $1/\sqrt{1+(y')^2}=\cos\psi$. At the maximum, the <first integral> has value $C=y_0(1-\lambda)$. Hence
$$
y\cos\psi-\lambda y=y_0(1-\lambda),\qquad
\boxed{\lambda=\frac{y_0-y\cos\psi}{y_0-y}=1+\frac{y(1-\cos\psi)}{y_0-y}.}
$$
A nonconstant regular branch below its positive maximum has $y>0$ and $\psi\ne0$, giving \b[$\lambda>1$]. For completeness, strictness follows from the differential equation rather than assuming it at every point: the <Euler-Lagrange equation> simplifies to
$$
yy''=1+(y')^2-\lambda[1+(y')^2]^{3/2}.
$$
At the maximum, $y_0y''=1-\lambda$ implies $\lambda\geq1$. If $\lambda=1$, the positive-height initial data $y=y_0$, $y'=0$ give the constant solution uniquely, contradicting the existence of the lower-height branch. Thus $\lambda=1$ is only the degenerate horizontal solution. The first integral with $\lambda>1$ also excludes $y=0$ on this regular branch.
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