If $x=acos \theta , y = bsin \theta , $ then $\frac{d^2y}{dx^2}$ at $\theta =\frac{\pi }{2}$ is :

$-\frac{b}{a^2}$

The correct answer is option (2) → $-\frac{b}{a^2}$

$x=a\cos\theta, y = b\sin \theta$

$\frac{dx}{dθ}=-a\sin θ$, $\frac{dy}{dθ}=b\cos θ$

$\frac{dy}{dx}=-\frac{b}{a}\cot θ⇒\frac{d^2y}{dx^2}=+\frac{b}{a}×(cosec^2θ)\frac{dθ}{dx}$

$⇒-\frac{b}{a}×\sin^2θ-a\sin θ$

so at $\frac{π}{2}$ → $-\frac{b}{a^2}$