If $x=k(t-\sin t),y=k(1-\cos t)(k\ne 0)$ then $\frac{d^{2}y}{d{x}^2}$ at $t=\frac{\pi }{2}$

$\frac{dx}{dt}=k(1-\cos t),\frac{dy}{dt}=k\sin t$

so $\frac{dy}{dx}=\frac{k\sin t}{k(1-\cos t)}=\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{2{\sin }^2\frac{t}{2}}=\cot \frac{t}{2}$

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}\\ =\frac{1}{2}\left(-{\mathrm{cosec}}^2\left(\frac{t}{2}\right)\right)\frac{1}{k\left(2{\sin }^2\left(\frac{t}{2}\right)\right)}\\ =-\frac{1}{4k}{\mathrm{cosec}}^4\frac{t}{2}\end{array}$

${\left.\frac{d^{2}y}{d{x}^2}\right|}_{t=\pi /2}=-\frac{1}{4k}(\sqrt{2}{)}^4=-\frac{1}{k}$