If x = 2at, y = at2, then \(\frac{d^2y}{dx^2}\) is

1 \(\frac{1}{2a}\) t 0

\(\frac{1}{2a}\)

Given, $x=at^2$ and $y=2at$ On differentiating both sides w.r.t. t, we get, $\frac{dx}{dt}=2a$ and $\frac{dy}{dt}=2at$ Therefore $\frac{dy}{dx}=\frac{2at}{2a}=t$ Now, $\frac{d^2y}{dx^2}=\frac{d}{dt}(\frac{dy}{dx})×\frac{dt}{dx}⇒\frac{d}{dt}(t)×\frac{1}{2a}=1×\frac{1}{2a}=\frac{1}{2a}$ Option B is correct.