If x=acos^3θ , y = a sin^3θ . The value of (d^2y)/(dx^2) at θ =(π)/6…

The correct answer is Option (2) → $\frac{32}{27a}$

$x=a\cos^3\theta , y = a \sin^3\theta$

$\frac{dx}{dθ}=-3a\cos^2θ\sin θ$

$\frac{dy}{dx}=3a\sin^2θ\cos θ$

$⇒\frac{dy}{dx}=-\frac{\sin θ}{\cos θ}$

so $\frac{dy}{dx}=-\tan θ$

so $\frac{d^2y}{dx^2}=-\sec^2 θ\frac{dθ}{dx}$

$=\frac{-\sec^2θ}{-3a\cos^2θ\sin θ}$ at $θ=\frac{π}{6}$ is $\frac{2^5}{3a×{\sqrt{3}}^4}$

$=\frac{32}{27a}$