Solve this

Given values are:

$x=a(2 \theta-\sin 2 \theta)$

and

$y=a(1-\cos 2 \theta)$

Applying parametric differentiation

$\frac{d x}{d \theta}=2 a-2 a \cos 2 \theta$

$\frac{d y}{d \theta}=0+2 \operatorname{asin} 2 \theta$

$\frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}=\frac{\sin 2 \theta}{1-\cos 2 \theta}$

Now putting the value of $\theta=\frac{\pi}{3}$

$\left.\frac{d y}{d x}\right|_{\theta=\frac{\pi}{3}}=\frac{\sin 2\left(\frac{\pi}{3}\right)}{1-\cos 2\left(\frac{\pi}{3}\right)}$

$=\frac{\frac{\sqrt{3}}{2}}{1+\frac{1}{2}}$

$=\frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}=\frac{1}{\sqrt{3}}$

So, $\frac{d y}{d x}$ is $\frac{1}{\sqrt{3}}$ at $\theta=\frac{\pi}{3}$.