The velocity of light above a hot surface decreases with the height from that surface. The velocity is given by $v={v}_{0}(\frac{1-y}{\alpha})$ where $y$ is the vertical distance above the starting position. Show that the path taken by light from (0,0) to $(2X,0)$ is the arc of a circle centeblack at $(X,\alpha )$. Note that the origin is chosen to be well above the ground, so that $y<0$ is allowable.

Attempt at a solution: After working through from the velocity relation given in the question, we get the following ODE: $\frac{1}{{v}_{0}(\frac{1+y}{\alpha})\sqrt{1+{y}^{2}}}}=-C$ where $C$ is a constant.

Attempt at a solution: After working through from the velocity relation given in the question, we get the following ODE: $\frac{1}{{v}_{0}(\frac{1+y}{\alpha})\sqrt{1+{y}^{2}}}}=-C$ where $C$ is a constant.