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Domain and Range of Trigonometric Ratios

The length of...

Question

The length of normal at any point to the curve, $y = c cos h (\frac{x}{c})$ is

Fixed

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

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

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

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Solution

The correct option is C $\frac{y^{2}}{c}$
Given, $y = c cos h (\frac{x}{c})$ ...... (i)
$\frac{d y}{d c} = c \cdot \frac{1}{c} \cdot sin h (\frac{x}{c}) = sin h (\frac{x}{c})$
Now, length of normal
$= y \sqrt{1 + {(\frac{d y}{d x})}^{2}}$
$= c cos h (\frac{x}{c}) \sqrt{1 + {sin}^{2} h (\frac{x}{c})}$
$= c cos h (\frac{x}{c}) \sqrt{cos h^{2} (\frac{x}{c})}$
$= c {[cos h (\frac{x}{c})]}^{2}$
$= c {(\frac{y}{c})}^{2}$ [from Eq. (i)]
$= \frac{y^{2}}{c}$

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The length of normal at any point to the curve, y=ccosh(xc) is

The correct option is C y2cGiven, y=ccosh(xc) ...... (i)dydc=c⋅1c⋅sinh(xc)=sinh(xc)Now, length of normal=y√1+(dydx)2=ccosh(xc)√1+sin2h(xc)=ccosh(xc)√cosh2(xc)=c[cosh(xc)]2=c(yc)2 [from Eq. (i)]=y2c

The length of normal at any point to the curve, $y = c cos h (\frac{x}{c})$ is