If $x=at^2, y -\frac{a}{t^2}, $ then the value of $\frac{d^2y}{dx^2}$ at t =1 is :

$\frac{2}{a}$

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

$x = at^2$, $y = \frac{a}{t^2}$

$\frac{dx}{dt} = 2at$, $\frac{dy}{dt} = -\frac{2a}{t^3}$

$\frac{dy}{dx} = \frac{-\frac{2a}{t^3}}{2at} = -\frac{1}{t^4}$

$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( -\frac{1}{t^4} \right) = \frac{\frac{d}{dt} \left( -\frac{1}{t^4} \right)}{\frac{dx}{dt}}$

$\frac{d}{dt} \left( -\frac{1}{t^4} \right) = 4 \cdot \frac{1}{t^5}$

$\frac{d^2y}{dx^2} = \frac{4 \cdot \frac{1}{t^5}}{2at} = \frac{2}{at^6}$

we substitute $t = 1$ into the expression for $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} \bigg|_{t=1} = \frac{2}{a \cdot 1^6} = \frac{2}{a}$