Question

If the tangents to the curve x = a$(\theta +sin\theta ),y=a(1+cos\theta )at\theta =\frac{\pi }{3}$ makes an angle$\alpha (0\le \alpha <)$with x-axis , then $\alpha$

• A. $\frac{\pi }{3}$
• B. $\frac{2\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{5\pi }{6}$

Answer 1

The correct option is D $\frac{5\pi }{6}$

Given equation of the curve is

$x=a(\theta +\sin \theta )$; $y=a(1+\cos \theta )$

Now

$\frac{dx}{d\theta }=a(1+\cos \theta )$ and $\frac{dy}{d\theta }=(-a\sin \theta )$

$\therefore \frac{dy}{dx}=\frac{\left(\frac{dy}{d\theta }\right)}{\left(\frac{dx}{d\theta }\right)}=\frac{(-a\sin \theta )}{a(1+\cos \theta )}$

$=\frac{-2a\sin \frac{\theta }{2}.\cos \frac{\theta }{2}}{a.2{\cos }^2\frac{\theta }{2}}=-\left(\tan \frac{\theta }{2}\right)$

Now

${\left|\frac{dy}{dx}\right|}_{\theta =\frac{\pi }{3}}=-\frac{1}{\sqrt{3}}$

Now. angle made by the curve at $\theta =\frac{\pi }{3}$ is

${\tan }^{-1}(-\frac{1}{\sqrt{3}})=\pi -{\tan }^{-1}(-\frac{1}{\sqrt{3}})$

$=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$