The potential energy is $mgy$ . The x-component of the particle's velocity, expressed in terms of $y$ and $\dot y$ is
$\begin{align*}\dot x = \frac{d}{dt}\left(\sqrt{\frac{y}{a}} \right) = \frac{\dot y}{2 \sqrt{y a}}\end{align*}$
Therefore, the particle's kinetic energy, expressed in terms of $y$ and $\dot y$ is
$\begin{align*}\frac{1}{2}m \left(\dot y ^2 + \dot x^2 \right) = \frac{1}{2}m \dot y^2 \left(1 + \frac{1}{4 y a}\right)\end{al...$
Therefore, the Lagrangian of the particle is
$\begin{align*}L = T - U = \frac{1}{2}m \dot y^2 \left(1 + \frac{1}{4 y a}\right) - mgy\end{align*}$
Therefore, answer (A) is correct.

Solution to 1992 Problem 44