Let a curve y = y x be given by the solution of the differential equation cos 1 2 cos - 1 e - x d x = e 2 x…

We have,

$\cos \left(\frac{1}{2}{\cos }^{-1}\left(e^{-x}\right)\right)dx=\left(\sqrt{e^{2x}-1}\right)dy ...\left(i\right)$

Put ${\cos }^{-1}\left({\mathrm{e}}^{-\mathrm{x}}\right)=\theta , \theta \in \left[0,\pi \right]$

$\Rightarrow \cos \theta =e^{-x}$

$\Rightarrow 2{\cos }^2\frac{\theta }{2}-1=e^{-x}$

$\Rightarrow \cos \left(\frac{\theta }{2}\right)=\sqrt{\frac{e^{-x}+1}{2}}=\sqrt{\frac{e^x+1}{2e^x}}$

Hence, by $\left(i\right)$, we have

$\cos \left(\frac{\theta }{2}\right)dx=\left(\sqrt{e^{2x}-1}\right)dy$

$\Rightarrow \sqrt{\frac{e^x+1}{2e^x}}dx=\sqrt{e^{2x}-1}dy$

$\Rightarrow \left(\sqrt{\frac{e^x+1}{2e^x}}\right)dx=\left(\sqrt{e^x-1}\right)\left(\sqrt{e^x+1}\right)dy$

$\Rightarrow \frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{e^x}\sqrt{e^x-1}}=\int dy$

Put  $e^x=t\Rightarrow dt=e^{x}dx$

$\frac{1}{\sqrt{2}}\int \frac{dt}{e^x\sqrt{e^x}\sqrt{e^x-1}}=\int dy$

$\int \frac{dt}{t\sqrt{t^2-t}}=\sqrt{2}\int dy$

Put $t=\frac{1}{z}\Rightarrow \frac{dt}{dz}=-\frac{1}{z^2}$

$\int \frac{-\frac{dz}{z^2}}{\frac{1}{z}\sqrt{\frac{1}{z^2}-\frac{1}{z}}}=\sqrt{2}\int dy$

$\Rightarrow -\int \frac{dz}{\sqrt{1-z}}=\sqrt{2}\int dy$

$\Rightarrow \frac{-2(1-z{)}^{1/2}}{-1}=\sqrt{2}y+c$

$\Rightarrow 2{\left(1-\frac{1}{t}\right)}^{1/2}=\sqrt{2}y+c$

$\Rightarrow 2{\left(1-e^{-x}\right)}^{1/2}=\sqrt{2}y+c$

Now, it meets $y-\text{axis}$ at $\left(0,-1\right)$, hence

$0=-\sqrt{2}+c\Rightarrow c=\sqrt{2}$

Hence,

$2{\left(1-e^{-x}\right)}^{1/2}=\sqrt{2}\left(y+1\right)$

It passes through $\left(\alpha , 0\right)$

$2{\left(1-e^{-\alpha }\right)}^{1/2}=\sqrt{2}$

$\Rightarrow \sqrt{1-e^{-\alpha }}=\frac{1}{\sqrt{2}}$

$\Rightarrow 1-e^{-\alpha }=\frac{1}{2}$

$\Rightarrow e^{-\alpha }=\frac{1}{2}\Rightarrow e^{\alpha }=2$