Question

The coordinates of the points on the curve $x=a(\theta +\sin \theta ),y=a(1-\cos \theta )$ where tangent is inclined an angle $\frac{\pi }{4}$ to the $x-$axis are -

• A. $(a,a)$
• B. $(a(\frac{\pi }{2}-1),a)$
• C. $(a(\frac{\pi }{2}+1),a)$
• D. $(a,a(\frac{\pi }{2}+1))$

Answer 1

The correct option is C $(a(\frac{\pi }{2}+1),a)$

The co-ordinates are $x=a(\theta +\sin \theta );y=a(1-\cos \theta )$
Thus, $\frac{dx}{d\theta }=a(1+\cos \theta )=a\left(2{\cos }^2\frac{\theta }{2}\right)$ and $\frac{dy}{d\theta }=a\sin \theta =2a\sin \frac{\theta }{2}\cos \frac{\theta }{2}$
$\therefore \frac{dy}{dx}=\tan \frac{\theta }{2}$
but, $\frac{dy}{dx}=\tan \frac{\pi }{4}$ (given)
$\therefore 1=\tan \frac{\theta }{2}$
$\Rightarrow \theta =\frac{\pi }{2}$
So the point is, $x=a(\theta +\sin \theta )$
$x=a(\frac{\pi }{2}+1)$
$y=a(1-0)=a$
Therefore, the required point is $(a(\frac{\pi }{2}+1),a)$
Hence, option 'C' is correct.