Question

The length of normal to the curve $x=a\left(\theta +\sin \left(\theta \right)\right),y=a\left(1-\cos \left(\theta \right)\right)$ at the point $\theta =\frac{\pi }{2}$ is

• A. $2a$
• B. $\frac{a}{2}$
• C. $\sqrt{2}a$
• D. $\frac{a}{\sqrt{2}}$

Answer 1

The correct option is C

$\sqrt{2}a$

Explanation for correct option:

Given, $x=a\left(\theta +\sin \left(\theta \right)\right),y=a\left(1-\cos \left(\theta \right)\right)$

The length of the normal is $y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$

Now,

$$\begin{gathered} \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}} \\ =\frac{a\sin \left(\theta \right)}{a\left(1+\cos \left(\theta \right)\right)} \\ =\frac{\sin \left(\theta \right)}{\left(1+\cos \left(\theta \right)\right)} \\ =\frac{2\sin \left({\displaystyle \frac{\theta }{2}}\right)\cos \left(\frac{\theta }{2}\right)}{2{\cos }^2\left(\frac{\theta }{2}\right)} \\ =\tan \left(\frac{\theta }{2}\right) \\ \therefore {\left[\frac{dy}{dx}\right]}_{\theta =\frac{\pi }{2}}=\tan \left(\frac{\pi }{4}\right)=1 \end{gathered}$$

Therefore the length of the normal is

${\left[y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\right]}_{\theta =\frac{\pi }{2}}=a\left(1-\cos \left(\frac{\pi }{2}\right)\right)\sqrt{1+1^2}=a\sqrt{2}$

Hence the correct option is option C.