Question

The length of the tangent to the curve $x=a(\theta +\sin \theta ),y=a(1-\cos \theta )$ at $\theta$ points is

• A. $2a\sin \frac{\theta }{2}$
• B. $a\sin \theta$
• C. $2a\sin \theta$
• D. $a\cos \theta$

Answer 1

The correct option is A $2a\sin \frac{\theta }{2}$

Given, $x=a(\theta +\sin \theta ),y=a(1-\cos \theta )$
$\Rightarrow \frac{dx}{d\theta }=a(1+\cos \theta )=2a{\cos }^2\frac{\theta }{2},\frac{dy}{d\theta }=a\sin \theta =2a\sin \frac{\theta }{2}\cos \frac{\theta }{2}$
Thus slope at ${}^{\prime }{\theta }^{\prime }$ is, $m=\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\tan \frac{\theta }{2}$
Hence length of tangent at ${}^{\prime }{\theta }^{\prime }$ is $=\frac{y\sqrt{1+m^2}}{m}$
$=\frac{a(1-\cos \theta )\sqrt{1+{\tan }^2\frac{\theta }{2}}}{\tan \frac{\theta }{2}}=2a\sin \frac{\theta }{2}$