Question

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

• A. $\sqrt{2}|a|$ units
• B. $\sqrt{3}|a|$ units
• C. $2|a|$ units
• D. $3|a|$ units

Answer 1

The correct option is A $\sqrt{2}|a|$ units

Here, $\frac{dx}{d\theta }=a(1+\cos \theta )$ and $\frac{dy}{d\theta }=a\sin \theta$
$\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\sin \theta }{a(1+\cos \theta )}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\theta =\frac{\pi }{2}}=\frac{\sin \frac{\pi }{2}}{1+\cos \frac{\pi }{2}}=1$
Length of normal is $={\left|y_1\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\right|}_{\theta =\frac{\pi }{2}}$
$=\left|a(1-\cos \frac{\pi }{2})\sqrt{1+1^2}\right|=\sqrt{2}|a|$ units.