Question

The length of the normal to the curve $y=a\cos h\left(\frac{x}{a}\right)$ at any point varies as

• A. ordinate
• B. abscissa
• C. Square of the abscissa
• D. square of the ordinate

Answer 1

The correct option is D

square of the ordinate

Find the length of the normal to the given curve

Given, $y=a\cos h\left(\frac{x}{a}\right)$

Partially differentiating with respect to $x$

$\frac{\partial y}{\partial x}=\frac{\partial a}{\partial x}\cos h\left(\frac{x}{a}\right)$

$\frac{\partial y}{\partial x}=a\times \sin h\left(\frac{x}{a}\right)\times \frac{1}{a}$

$\therefore$ $\frac{\partial y}{\partial x}=\sin h\left(\frac{x}{a}\right)$

Length of the normal is given as

$PN=\left|y\right|\sqrt{1+{\left(\frac{\partial y}{\partial x}\right)}^2}$

$=y\sqrt{1+\sin h^2\left(\frac{x}{a}\right)}$

$\therefore$ $PN=y\cos h\left(\frac{x}{a}\right)$ $...\left(\because 1+sin{h}^{2}x=cos{h}^{2}x\right)$

$\therefore$ $PN=y\times \frac{y}{a}=\frac{y^2}{a}$ …$(\because \cos h\left(\frac{x}{a}\right)=\frac{y}{a})$

$\therefore$ $PN\propto y^2$

So, length of the normal to the curve varies as the square of the ordinate.

Hence, option $\left(D\right)$ is the correct answer.