Question

The tangent to the curve $x=a(\theta -\sin \theta );y=a(1+\cos \theta )$ at the points $\theta =(2k+1)\pi ,k\in Z$ are parallel to

• A. $y=x$
• B. $y=-x$
• C. $y=0$
• D. $x=0$

Answer 1

The correct option is C $y=0$

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

$\therefore$ Slope of tangent $=\frac{-\sin \theta }{1-\cos \theta }$

at $\theta =(2k+1)\pi$

Slope of tangent $=\frac{-\sin (2k+1)\pi }{1-\cos (2k+1)\pi }=0$

Since the slope of tangent $=0$, it will be parallel to $x-$ axis

Therefore $y=0$ is a possibility.