MHT CET · Maths · Application of Derivatives
The length of normal at any point to the curve, \(y=c \cosh \left(\frac{x}{c}\right)\) is
- A fixed
- B \(\frac{y^{2}}{c^{2}}\)
- C \(\frac{y^{2}}{c}\)
- D \(\frac{y}{c^{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{y^{2}}{c}\)
Step-by-step Solution
Detailed explanation
Given, \(y=c \cosh \left(\frac{x}{c}\right)\)...(i)
\(
\frac{d y}{d x}=c \cdot \frac{1}{c} \cdot \sinh \left(\frac{x}{c}\right)=\sinh \left(\frac{x}{c}\right)
\)
Now, length of normal
\(
=y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}
\)
\(=c \cosh \left(\frac{x}{c}\right) \sqrt{1+\sin ^{2} \mathrm{~h}\left(\frac{x}{c}\right)}\)
\(=c \cosh \left(\frac{x}{c}\right) \sqrt{\cosh ^{2}\left(\frac{x}{c}\right)}\)
\(=c\left[\cosh \left(\frac{x}{c}\right)\right]^{2}\)
\(=c\left(\frac{y}{c}\right)^{2} \quad[\) from Eq. (i)]
\(=\frac{y^{2}}{c}\)
\(
\frac{d y}{d x}=c \cdot \frac{1}{c} \cdot \sinh \left(\frac{x}{c}\right)=\sinh \left(\frac{x}{c}\right)
\)
Now, length of normal
\(
=y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}
\)
\(=c \cosh \left(\frac{x}{c}\right) \sqrt{1+\sin ^{2} \mathrm{~h}\left(\frac{x}{c}\right)}\)
\(=c \cosh \left(\frac{x}{c}\right) \sqrt{\cosh ^{2}\left(\frac{x}{c}\right)}\)
\(=c\left[\cosh \left(\frac{x}{c}\right)\right]^{2}\)
\(=c\left(\frac{y}{c}\right)^{2} \quad[\) from Eq. (i)]
\(=\frac{y^{2}}{c}\)
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