If $x=a(\theta +\sin \theta ),y=a(1-\cos \theta )$, then $\frac{d^{2}y}{d{x}^2}$ at $\theta =\frac{\pi }{4}$ is

$x=a(\theta +\sin \theta );y=a(1-\cos \theta )$

$\frac{dx}{d\theta }=a(1+\cos \theta ),\frac{dy}{d\theta }=a\sin \theta$

$\frac{dy}{dx}=\frac{a\sin \theta }{a(1+\cos \theta )}\times \frac{1-\cos \theta }{1-\cos \theta }$

$\frac{dy}{dx}=\frac{\sin \theta (1-\cos \theta )}{{\sin }^2\theta }=\frac{1-\cos \theta }{\sin \theta }$

$\frac{dy}{dx}=\cos ec\theta -\cot \theta$

$\frac{d^{2}y}{d{x}^2}=(-\cos ec\theta .\cot \theta +\cos e{c}^2\theta )\frac{d\theta }{dx}$

$=\cos ec\theta (\cos ec\theta -\cot \theta )\times \frac{1}{a(1+\cos \theta )}$

${\left(\frac{d^{2}y}{dx}\right)}_{at\theta =\frac{\pi }{4}}=\sqrt{2}(\sqrt{2}-1)\times \frac{1}{a(1+\frac{1}{\sqrt{2}})}$

$=\frac{2(3-2\sqrt{2})}{a}$