Question

The tangents 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. $x+y=0$
• C. $x=0$
• D. $y=0$

Answer 1

The correct option is D $y=0$

To find tangents to the curve at given points is parallel to which line we just need to find slope of the tangent,

Now Slope of the curve $=\frac{dy}{dx}=\frac{dy}{d\theta }\frac{d\theta }{dx}$

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

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

$\therefore$ Slope$=\frac{dy}{dx}=-a\sin \theta \times \frac{1}{a(1-\cos \theta )}$

Now at $\theta =(2k+1)\pi$, $\sin \theta =0$.

$\therefore \frac{dy}{dx}=0$

$\to$ Tangent at this points will be parallel to $y=0$.

Hence, $\left(D\right)$