The slope of the tangent to a curve C: y = y (x) at any point [ x,...

Given,

$\dfrac{dy}{dx} = \dfrac{2{e}^{2x} - 6{e}^{- x} + 9}{2 + 9{e}^{- 2x}}$

On rearranging we get,

$\dfrac{dy}{dx} = {e}^{2x} - \dfrac{6{e}^{x}}{2{e}^{2x} + 9}$

Integrating both side we get,

$y = \dfrac{{e}^{2x}}{2} - \sqrt{2}\tan^{- 1}\left(\dfrac{\sqrt{2}{e}^{x}}{3}\right) + c$

If the curve passes through the point \left(0, \dfrac{1}{2} + \dfrac{\pi}{2\sqrt{2}}\right)$</p><p>Then$c = \sqrt{2}\left(\dfrac{\pi}{4} + \tan^{- 1}\dfrac{\sqrt{2}}{3}\right)$</p><p>So, curve will be$y = \dfrac{{e}^{2x}}{2} - \sqrt{2}\left(\tan^{- 1}\left(\dfrac{\sqrt{2}{e}^{x}}{3}\right) - \dfrac{\pi}{4} - \tan^{- 1}\dfrac{\sqrt{2}}{3}\right)$</p><p>Again curve passes through the point</p><p>\left(\alpha, \dfrac{1}{2}{e}^{2\alpha}\right)$

Putting the value in curve equation we get,

$\dfrac{{e}^{2\alpha}}{2} = \dfrac{{e}^{2\alpha}}{2} - \sqrt{2}\left(\tan^{- 1}\left(\dfrac{\sqrt{2}{e}^{\alpha}}{3}\right) - \dfrac{\pi}{4} - \tan^{- 1}\dfrac{\sqrt{2}}{3}\right)$

$⇒ \tan^{- 1}\left(\dfrac{\sqrt{2}{e}^{\alpha}}{3}\right) = \dfrac{\pi}{4} + \tan^{- 1}\dfrac{\sqrt{2}}{3}$

$⇒ \dfrac{\sqrt{2}{e}^{\alpha}}{3} = \tan \left(\dfrac{\pi}{4} + \tan^{- 1}\dfrac{\sqrt{2}}{3}\right)$

$⇒ \dfrac{\sqrt{2}{e}^{\alpha}}{3} = \left(\dfrac{tan\dfrac{\pi}{4} + {tantan}^{- 1}\dfrac{\sqrt{2}}{3}}{1 - tan\dfrac{\pi}{4} \times {tantan}^{- 1}\dfrac{\sqrt{2}}{3}}\right)$

$⇒ \dfrac{\sqrt{2}{e}^{\alpha}}{3} =$\left(\dfrac{1 + \dfrac{\sqrt{2}}{3}}{1 - \dfrac{\sqrt{2}}{3}}\right)$</p><p>$⇒ {e}^{\alpha} = \dfrac{3}{\sqrt{2}}\left(\dfrac{3 + \sqrt{2}}{3 - \sqrt{2}}\right)