= Solution
For the specified <functional>,
$$
F_{y'}=\frac{y'}{y\sqrt{1+y'^2}},\qquad
y'F_{y'}-F=-\frac1{y\sqrt{1+y'^2}}.
$$
Since $y>0$, the <Beltrami identity> gives $y\sqrt{1+y'^2}=R$ for some $R>0$. This suggests a <circular arc>, but it is worth deriving its equation without dividing by $y'$: such division would lose the point where the arc is horizontal.
Substitute the displayed <partial derivative> and $F_y=-\sqrt{1+y'^2}/y^2$ in the full <Euler-Lagrange equation>. Combining terms over the common denominator gives
$$
\frac d{dx}F_{y'}-F_y
=\frac{yy''+1+y'^2}{y^2(1+y'^2)^{3/2}}=0.
$$
Thus $(yy')'=-1$, so $yy'=h-x$ and a second integration gives
$$
(x-h)^2+y^2=R^2.
$$
This establishes the <semicircular extremals of the hyperbolic length functional> and excludes constant $y$, which would satisfy the first-order <Beltrami identity> for a suitable $R$ but not the full <Euler-Lagrange equation>.
The two prescribed endpoints give
$$
(-1-h)^2+4c^2=R^2=(1-h)^2+4d^2.
$$
Subtracting yields $4h=4(d^2-c^2)$. Therefore
$$
\boxed{h=d^2-c^2,\qquad R^2=(1+h)^2+4c^2,\qquad
y(x)=\sqrt{R^2-(x-h)^2}.}
$$
In particular the centre is $\boxed{(d^2-c^2,0)}$. Admissibility requires $c,d>0$, as the prescribed heights must lie in $y>0$. Under that condition the radicand is positive throughout $[-1,1]$: it is a concave <quadratic polynomial> whose two endpoint values are $4c^2$ and $4d^2$. The positive root consequently gives one smooth admissible arc, uniquely determined by the endpoints.
\Image[/past-exam-of-the-short-options-course-of-the-university-of-oxford/short-options-2015-semicircular-extremals.png]
{title=Fixed endpoint heights determine the circle centre; a free endpoint height forces a horizontal tangent}
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