If $x = a\sin 2t(1 + \cos 2t)$ and $y = b\cos 2t(1 − \cos 2t)$, then $\left(\frac{dy}{dx}\right)_{at\,t=\frac{\pi}{4}}$ is equal to

$\frac{b}{a}$

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

$x = a \sin 2t (1 + \cos 2t)$

$y = b \cos 2t (1 - \cos 2t)$

$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$x = a \sin 2t + a \sin 2t \cos 2t$

$\frac{dx}{dt} = a(2\cos 2t + 2\cos^2 2t - 2\sin^2 2t)$

$\frac{dx}{dt} = a(2\cos 2t - 2 + 4\cos^2 2t)$

$y = b \cos 2t - b \cos^2 2t$

$\frac{dy}{dt} = b(-2\sin 2t + 4\sin 2t \cos 2t)$

$\frac{dy}{dt} = -2b \sin 2t (1 - 2\cos 2t)$

At $t = \frac{\pi}{4}$:

$\sin 2t = 1, \ \cos 2t = 0$

$\frac{dx}{dt} = -2a, \quad \frac{dy}{dt} = -2b$

$\displaystyle \left( \frac{dy}{dx} \right)_{t=\pi/4} = \frac{-2b}{-2a} = \frac{b}{a}$

Final Answer: $\frac{b}{a}$