If $x=a t^2, y=2 a t$, then $\frac{d^2 y}{d x^2}$ is equal to:

$-\frac{1}{2 a t^3}$

The correct answer is Option (2) → $-\frac{1}{2 a t^3}$

$x = at^2,\quad y = 2at$

$\frac{dx}{dt} = 2at,\quad \frac{dy}{dt} = 2a$

$\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$

$\frac{d}{dt}\left(\frac{dy}{dx}\right) = -\frac{1}{t^2}$

$\frac{d^2y}{dx^2} = \frac{-\frac{1}{t^2}}{2at} = -\frac{1}{2at^3}$

$\frac{d^2y}{dx^2} = -\frac{1}{2at^3}$