 Solution

The cycloid has limiting endpoints $(0,0)$ and $(2\pi a,0)$, with its highest point $(\pi a,2a)$ at $t=\pi$. The transformation $t\mapsto 2\pi -t$ sends $x$ to $2\pi a-x$ and preserves $y$, so the arch is symmetric about $x=\pi a$. Its slope decreases from $+\infty$ through zero to $-\infty$, giving vertical limiting tangent lines at the cusps and a horizontal tangent line at the top. More explicitly,so the arch is a concave function throughout its interior. The endpoint cusps are limiting points, excluded by the open parameter interval.