The slope of the tangent to a curve $C: y=y(x)$ at any point $(x, y)$ on it is $\frac{2 \mathrm{e}^{2 x}-6 \mathrm{e}^{-x}+9}{2+9 \mathrm{e}^{-2 x}}$. If $C$ passes through the points $\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)$ and $\left(\alpha, \frac{1}{2} \mathrm{e}^{2 \alpha}\right)$, then $\mathrm{e}^{\alpha}$ is equal to :

$\frac{d y}{d x}=\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}=e^{2 x}-\frac{6 e^{-x}}{2+9 e^{-2 x}}$ <br/><br/> $$ \begin{aligned} &\int d y=\int e^{2 x} d x-3 \int \underbrace{1+\left(\frac{3 e^{-x}}{\sqrt{2}}\right)^{2}}_{\text {put } e^{-x}=t} d x \\\\ &=\frac{e^{2 x}}{2}+3 \int \frac{d t}{1+\left(\frac{3 t}{\sqrt{2}}\right)^{2}} \\\\ &=\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1} \frac{3 t}{\sqrt{2}}+C \end{aligned} $$<br/><br/> $y=\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3 e^{-x}}{\sqrt{2}}\right)+C$ <br/><br/> It is given that the curve passes through <br/><br/> $\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)$ <br/><br/> $$ \begin{aligned} & \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}=\frac{1}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)+C \end{aligned} $$ <br/><br/> $\Rightarrow \quad C=\frac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)$ <br/><br/> Now if $\left(\alpha, \frac{1}{2} e^{2 \alpha}\right)$ satisfies the curve, then <br/><br/> $$ \frac{1}{2} e^{2 \alpha}=\frac{e^{2 \alpha}}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3 e^{-\alpha}}{\sqrt{2}}\right)+\frac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right) $$ <br/><br/> $\tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)-\tan ^{-1}\left(\frac{3 e^{-\alpha}}{\sqrt{2}}\right)=\frac{\pi}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{\pi}{4}$ <br/><br/> $$ \frac{\frac{3}{\sqrt{2}}-\frac{3 e^{-\alpha}}{\sqrt{2}}}{1+\frac{9}{2} e^{-\alpha}}=1 $$ <br/><br/> $\frac{3}{\sqrt{2}} e^{\alpha}-\frac{3}{\sqrt{2}}=e^{\alpha}+\frac{9}{2}$ <br/><br/> $$ e^{\alpha}=\frac{\frac{9}{2}+\frac{3}{\sqrt{2}}}{\frac{3}{\sqrt{2}}-1}=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right) $$