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

$\begin{array}{l}\frac{dy}{dx}=\frac{2e^{2x}-6e^{-x}+9}{2+9e^{-2x}}\\ \frac{dy}{dx}=e^{2x}-\frac{6e^x}{2e^{2x}+9}\\ y=\frac{e^{2x}}{2}-{\tan }^{-1}\left(\frac{\sqrt{2}{e}^x}{3}\right)+c\end{array}$

If C passes through the point $\left(0,\frac{1}{2}+\frac{\pi }{2\sqrt{2}}\right)$

$c=-\frac{\pi }{4}-{\tan }^{-1}\frac{\sqrt{2}}{3}$

Again C passes through the point $\left(\alpha ,\frac{1}{2}{\mathrm{e}}^{2\alpha }\right)$ then ${\mathrm{e}}^{\alpha }=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$