The angle between the curve "  2y= e^-x/2  " and the y-axis??​

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ 2\mathrm{y}={\mathrm{e}}^{-\mathrm{x}/2} \\ \mathrm{For} \mathrm{y}-\mathrm{axis}, \mathrm{x}=0 \\ 2\mathrm{y}={\mathrm{e}}^0 \\ \Rightarrow \mathrm{y}=\frac{1}{2} \\ \mathrm{Hence} \mathrm{y}-\mathrm{axis} \mathrm{and} {\mathrm{e}}^{-\mathrm{x}/2} \mathrm{intersect} \mathrm{at} \left(0,\frac{1}{2}\right) \\ 2\mathrm{y}={\mathrm{e}}^{-\mathrm{x}/2} \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ 2\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2}{\mathrm{e}}^{-\frac{\mathrm{x}}{2}} \\ \mathrm{Put} \left(0,\frac{1}{2}\right) \mathrm{to} \mathrm{get} \mathrm{slope} \mathrm{of} \mathrm{tangent} \\ \Rightarrow 2\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2}{\mathrm{e}}^0 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{4}=\mathrm{tan\theta } \\ \Rightarrow \mathrm{\theta }={\tan }^{-1}\left(-\frac{1}{4}\right) \\ \mathrm{Angle} \mathrm{made} \mathrm{with} \mathrm{y}-\mathrm{axis}=\frac{\mathrm{\pi }}{2}+ \mathrm{\theta } \\ =\frac{\mathrm{\pi }}{2}+{\tan }^{-1}\left(-\frac{1}{4}\right) \\ =\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(\frac{1}{4}\right) \\ ={\cot }^{-1}\left(\frac{1}{4}\right) \\ ={\tan }^{-1}\left(4\right) \end{gathered}$$

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