If $\mathrm{x}=\mathrm{a}(\mathrm{\theta }+\sin \mathrm{\theta }),\mathrm{y}=\mathrm{a}(1-\cos \mathrm{\theta })$, then ${\mathrm{y}}_2$ at $\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$, is

$$\begin{gathered} \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{a\sin \theta }{a(1+\cos \theta )}=\frac{2\sin {\displaystyle \frac{\theta }{2}}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}=\tan \frac{\theta }{2} \\ \Rightarrow \frac{dy}{dx}=\tan \frac{\theta }{2} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\frac{d\theta }{dx}=\frac{d}{dx}\left(\tan \frac{\theta }{2}\right)\frac{1}{a(1+\cos \theta )} \\ \Rightarrow y_2=\left(\frac{1}{2}se{c}^2\frac{\theta }{2}\right)\frac{1}{a(1+\cos \theta )} \\ \Rightarrow y_2=\frac{1}{a{(1+\cos \theta )}^2} \\ substitute \theta =\frac{\mathrm{\pi }}{4}\Rightarrow y_2=\frac{1}{a{(1+{\displaystyle \frac{1}{\sqrt{2}}})}^2}=\frac{2\left(3-2\sqrt{2}\right)}{a} \end{gathered}$$