Consider the functional $ I(y(x))=\int_{x_0}^{x_1}{f(x,y) \sqrt{1+y'^2} e^{\tan^{-1}{y'}}}dx$ where $f(x,y)\ne 0$. Let the left end of the extremal be fixed at the point $ A(x_0, y_0) $ and the right end $ B(x_1, y_1) $ be movable along the curve $y=\psi(x). $ Then the extremal $y=y(x)$ intersects the curve $ y=\psi(x)$ along which the boundary point $ B(x_1,y_1) $ slides at an angle
$ 1. \frac{\pi}{3}\\ 2.\frac{\pi}{2}\\ 3. \frac{\pi}{4}\\ 4. \frac{\pi}{6}$
My Attempt: Here $F(x,y,y')=f(x,y)\sqrt{1+y'^2} e^{\tan^{-1}{y'}} \\ \frac{\partial F}{\partial y'}=f(x,y)[\sqrt{1+y'^2}\frac{e^{\tan^{-1}{y'}}}{1+y'^2}+\frac{e^{\tan^{-1}{y'}}}{2\sqrt{1+y'^2}}] \\ \frac{\partial F}{\partial y'}=f(x,y)[\frac{e^{\tan^{-1}{y'}}}{\sqrt{1+y'^2}}+\frac{e^{\tan^{-1}{y'}}}{2\sqrt{1+y'^2}}] \\ \frac{\partial F}{\partial y'}=\frac{3}{2}f(x,y)\frac{e^{\tan^{-1}{y'}}}{\sqrt{1+y'^2}} \\ $
given that the right end $B(x_1,y_1)$ be movable along the curve $y=\psi(x) \\ \Rightarrow y_1'=\psi'(x)$
using transversality condition for variable right end $[F(x_1,y_1,y_1')+(y_1'-y')F_y']_{x=x_1}=0 \\ [f(x,y)\sqrt{1+y'^2} e^{\tan^{-1}{y'}}+ (\psi'(x)-y')(\frac{3}{2}f(x,y)\frac{e^{\tan^{-1}{y'}}}{\sqrt{1+y'^2}})]_{x=x_1}=0 \\ [f(x,y)(e^{\tan^{-1}{y'}}[\sqrt{1+y'^2}+3\frac{(\psi'(x)-y')}{2\sqrt{1+y'^2}})]_{x=x_1}=0 \\ [\frac{f(x,y)e^{\tan^{-1}{y'}}}{\sqrt{1+y'^2}}(2(1+y'^2)+3(\psi'(x)-y'))]_{x=x_1}=0 \\\frac{f(x_1,y_1)e^{\tan^{-1}{y_1'}}}{\sqrt{1+y_1'^2}}(2(1+y_1'^2)+3(y_1'-y_1'))=0 \\\frac{f(x_1,y_1)e^{\tan^{-1}{y_1'}}}{\sqrt{1+y_1'^2}}(2(1+y_1'^2)+2y_1')=0 \\(2(1+y_1'^2)+2y_1')=0 \\ since \;\; f(x,y)\ne 0 \;\; and \;\; e^{\tan^{-1}y'}\ne 0 \\ $
I am not getting how to apply Euler's equation $ \frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y'})=0 \\ as \;\;F(x,y,y')=f(x,y)\sqrt{1+y'^2} e^{\tan^{-1}{y'}} $
and also how to solve further