Why is the Brachistochrone cycloid flipped

Question

While solving the Brachistochrone problem, I setup a variable $T$, $$T=\int\dfrac{\sqrt{1+(y')^2}}{\sqrt{2gy}}$$ Then employing the Euler-Lagrange equations (Beltrami's identity): $$\dfrac{dy}{dx}=\sqrt{\dfrac{1}{c^2y}-1}$$ Whose parametric solution is $$x=\dfrac{1}{2c^2}[\theta-\sin(\theta)]+k$$ $$y=\dfrac{1}{2c^2}[1-\cos(\theta)]$$ Where $k$ is the integration constant. When applied the boundary condition $(0,0)$, the $k$ vanishes. When plotted, the curve is a cycloid. However, if I were to drop a ball, the optimum curve would be flipped, looking more like: $$x=\dfrac{1}{2c^2}[\theta-\sin(\theta)]$$ $$y=\dfrac{1}{2c^2}[\cos(\theta)-1]$$ Why is it that many papers have the cycloid as the solution although it in reality is flipped? What would be a suitable explanation? Do I need to specify any other boundary conditions?

Answer 1

There is no need for extra boundary conditions. It's just that you have implicitly assumed that the $y$-axis is pointing downward; hence, the value of the $y$ coordinate is positive downward, and negative upward. This is evident from the equation for $T$, since $2gy$ has to be positive.
If you had written the equation as $T = \int \frac{\sqrt{1+y'^2}}{\sqrt{-2gy}}$ ($y$-axis pointing up), then the result would be what you are looking for.