The slope of the normal to the curve $x=a \cos ^3 \theta, y=a \sin ^3 \theta$ at $\theta=\frac{\pi}{4}$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - 1
$x=a \cos ^3 \theta, y=a \sin ^3 \theta$
$\frac{dx}{dθ}=-3a\cos^2θ\sin θ$, $\frac{dy}{dθ}=3a\sin^2θ\cos θ$
slope of normal = $-\frac{dx}{dy}=\frac{\cos θ}{\sin θ}$
$\left.-\frac{dx}{dy}\right]_{θ=\frac{\pi}{4}}=1$