Given,


Differentiate Eq. (i) w.r.t. $t$, we get
$$
\frac{d x}{d t}=a(1+\cos t)
$$

Differentiate Eq. (ii) w.r.t. $t$, we get
$$
\begin{aligned}
& \frac{d y}{d x}=a \sin t \\
& \therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a \sin t}{a(1+\cos t)}=\frac{\sin t}{(1+\cos t)} \\
& \text { Now, } \quad \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d t}\left(\frac{d y}{d x}\right) \times \frac{d t}{d x} \\
& =\frac{d}{d t}\left(\frac{\sin t}{1+\cos t}\right) \times \frac{1}{a(1+\cos t)} \\
& =\frac{(1+\cos t) \cos t-\sin t(-\sin t)}{(1+\cos t)^2} \times \frac{1}{a(1+\cos t)} \\
& \frac{d^2 y}{d x^2}=\frac{(\cos t+1)}{a(1+\cos t)^3}=\frac{1}{a(1+\cos t)^2} \\
& \text { At } t=2 \pi / 3 \\
& \frac{d^2 y}{d x^2}=\frac{1}{a\left(\frac{1}{2}\right)^2}=\frac{4}{a} \\
&
\end{aligned}
$$