Question

The slope of the normal to the curve $x=a(\theta -\sin \theta ),y=a(1-\cos \theta )$ at point $\theta =\frac{\pi }{2}$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (C)

$-1$

The slope of normal $x=a(\theta -\sin \theta );y=a(1-\cos \theta )at\theta =\frac{\pi }{2}$
${\left(\frac{dx}{d\theta }\right)}_{\theta =\frac{\pi }{2}}=a(1-\cos \theta )=a$
${\left(\frac{dy}{d\theta }\right)}_{\theta =\frac{\pi }{2}}=a\sin \theta =a$
$\therefore \left(\frac{dy}{dx}\right)=1$
Therefore, slope of normal at the given point is, $m=(-\frac{dx}{dy})=-1$
Hence, option 'C' is correct.