If $x = at^2, y = 2at$; then $\frac{d^2y}{dx^2}$ is equal to

$-\frac{1}{2at^3}$

The correct answer is Option (4) → $-\frac{1}{2at^3}$

Given: $x = a t^{2},\; y = 2 a t$

$\frac{dx}{dt} = 2 a t,\;\;\frac{dy}{dt} = 2 a$

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

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

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

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

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