Question

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

Answer: (D)

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

$x=a(\theta +\sin \theta )$

$y=a(1+\cos \theta )$

If tangent to curve at $\theta =\pi /3$, makes angle $\alpha$ with x-axis,

$\Rightarrow$ slope of tangent = $\tan \alpha$

$\Rightarrow$ ${\begin{array}{c}\tan \alpha =\frac{dy/d\theta }{dx/d\theta }\end{array}|}_{\theta =\pi /3}$

$=\frac{-a\sin \theta }{a(1+\cos \theta )}$

${\begin{array}{c}\frac{-\sin \theta }{1+\cos \theta }\end{array}|}_{\theta =\pi /3}$

$=-\frac{\sqrt{3}/2}{1+1/2}$

$\tan \alpha =-\frac{1}{\sqrt{3}}$

Ans. $\alpha =5\pi /6$