Consider $ \bar{F} = y \sqrt{1 + (y')^2} $

To find, y = f(x) such that $ \int_{x_1}^{x_2} F \,\, dx = \int_{x_1}^{x_2} y \sqrt{1 + (y')^2} \,\, dx \hspace{0.25cm} ...(1) $ is minimum subject to the condition,

$ \int_{x_1}^{x_2} G \,\, dx = \int_{x_1}^{x_2} \sqrt{1 + (y')^2} \,\, dx = l \hspace{0.25cm} ... (2) $

To use Lagrange's method, multiply (2) by $ \lambda $ and add to (1),

$ H = F + \lambda G = \int_{x_1}^{x_2} (y + \lambda) \sqrt{1 + (y')^2} \,\, dx$

Since the integrand is free from 'x', we use the formula, $ F_1 - y' \frac{\partial F_1}{\partial y'} = c $

where, $ F_1 = (y + \lambda) \sqrt{1 + (y')^2} $

$ \frac{\partial F_1}{\partial y'} = (y + \lambda) \frac{1}{2 \times \sqrt{1 + (y')^2}} (2y') = \frac{(y + \lambda)y'}{\sqrt{1 + (y')^2}}$

$ F_1 - y' \frac{\partial F_1}{\partial y'} = c \\ \therefore (y + \lambda)[\sqrt{1 + (y')^2}] - y'[\frac{(y + \lambda)y'}{\sqrt{1 + (y')^2}}] = c \\ \therefore \frac{(y + \lambda)(1 + (y')^2) - (y + \lambda)(y')^2}{\sqrt{1 + (y')^2}} = c \\ \therefore \frac{(y + \lambda)}{\sqrt{1 + (y')^2}} = c \\ \frac{(y + \lambda)^2}{(1 + (y')^2)} = c^2 \hspace{0.5cm} [Squaring] \\ (y + \lambda)^2 = c^2(1 + (y')^2) \\ c^2 + c^2(y')^2 = (y + \lambda)^2 \\ c^2(y')^2 = (y + \lambda)^2 - c^2 \\ (y')^2 = \frac{(y + \lambda)^2 - c^2}{c^2} \\ (\frac{dy}{dx})^2 = \frac{(y + \lambda)^2 - c^2}{c^2} \\ \frac{dy}{dx} = \frac{\sqrt{(y + \lambda)^2 - c^2}}{c} \\ \frac{c}{\sqrt{(y + \lambda)^2 - c^2}} dy = dx $

Integrating, we get,

$ \int \frac{c}{\sqrt{(y + \lambda)^2 - c^2}} dy = \int dx \\ c[cosh^{-1}[\frac{y + \lambda}{c}]] = x + c' \\ cosh^{-1}(\frac{y + \lambda}{c}) = \frac{x + c'}{c} \\ \frac{y + \lambda}{c} = cosh(\frac{x + c'}{c}) \\ y + \lambda = c \times cosh(\frac{x + c'}{c})$

$ y = c \times cosh(\frac{x + c'}{c}) - \lambda $ is the required curve.