Question

Length of the normal to the curve at any point on the curve $y=\frac{a(e^{x/a}+e^{-x/a})}{2}$ varies as

• A. $x$
• B. $x^2$
• C. $y$
• D. $y^2$

Answer 1

The correct option is D $y^2$

Given,

$y=\frac{a(e^{\frac{x}{a}}+e^{-\frac{x}{a}})}{2}$

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

$\frac{dy}{dx}=a\sin h\left(\frac{x}{a}\right)\frac{1}{a}$

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

Length of normal $=y{[1+{\left(\frac{dy}{dx}\right)}^2]}^{\frac{1}{2}}$

$=a\cos h\left(\frac{x}{a}\right){[1+{\sin }^{2}h\left(\frac{x}{a}\right)]}^{\frac{1}{2}}$

$=a\cos h\left(\frac{x}{a}\right){\left[{\cos }^{2}h\left(\frac{x}{a}\right)\right]}^{\frac{1}{2}}$

$=a{\cos }^{2}h\left(\frac{x}{a}\right)$

$=a\times \frac{y^2}{a^2}$

$=\frac{y^2}{a}$