The length of the normal at any point to the | vedclass

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(C) Given the curve equation: $y=c \cosh \left(\frac{x}{c}\right)$ ... $(i)$Differentiating with respect to $x$:$\frac{dy}{dx} = c \cdot \frac{1}{c} \

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$\frac{y^{2}}{c}$

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(C) Given the curve equation: $y=c \cosh \left(\frac{x}{c}ight)$ ... $(i)$Differentiating with respect to $x$:$\frac{dy}{dx} = c \cdot \frac{1}{c} \cdot \sinh \left(\frac{x}{c}ight) = \sinh \left(\frac{x}{c}ight)$The formula for the length of the normal is given by:$L = |y| \sqrt{1 + \left(\frac{dy}{dx}ight)^{2}}$Substituting the values:$L = y \sqrt{1 + \sinh^{2} \left(\frac{x}{c}ight)}$Using the identity $\cosh^{2} 	heta - \sinh^{2} 	heta = 1$,we have $1 + \sinh^{2} 	heta = \cosh^{2} 	heta$:$L = y \sqrt{\cosh^{2} \left(\frac{x}{c}ight)}$$L = y \cosh \left(\frac{x}{c}ight)$From equation $(i)$,we know $\cosh \left(\frac{x}{c}ight) = \frac{y}{c}$:$L = y \cdot \left(\frac{y}{c}ight) = \frac{y^{2}}{c}$Thus,the length of the normal is $\frac{y^{2}}{c}$.

The length of the normal at any point to the curve $y=c \cosh \left(\frac{x}{c}\right)$ is

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