Question

The inclination of the tangent at$\theta =\frac{\pi }{3}onthecurvex=a(\theta +sin\theta ),y=a(1+cos\theta )is$

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

Answer 1

Answer: (C)

$\frac{2\pi }{3}$

$\begin{array}{cc}x& =a(\theta +\sin \theta )\\ \frac{dx}{d\theta }& =a(1+\cos \theta )\\ y& =a(1+\cos \theta )\\ \frac{dy}{d\theta }& =a(0-\sin \theta )=-a\sin \theta \end{array}$

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

at $\theta =\frac{\pi }{3}$

$\frac{dy}{dx}=\frac{-\sin \frac{\pi }{3}}{1+\cos \frac{\pi }{3}}=\frac{-\frac{\sqrt{3}}{2}}{1+\frac{1}{2}}=\frac{-\sqrt{3}}{3}=\frac{-1}{\sqrt{3}}$

$\Rightarrow \mathrm{Tan}\theta =-\frac{1}{\sqrt{3}}$

$\Rightarrow \theta =\frac{2\pi }{3}$

Hence, (c) is the correct option