Question

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

• A. $2a$
• B. $a$
• C. $\sqrt{2}a$
• D. $2\sqrt{2}a$

Answer 1

Answer: (C)

$\sqrt{2}a$

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

$\frac{dx}{d\theta }=a(1+\cos \theta )$

$\Rightarrow {\left(\frac{dx}{d\theta }\right)}_{\theta =\frac{\pi }{2}}=a$

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

$\frac{dy}{d\theta }=a\sin \theta$

$\Rightarrow {\left(\frac{dy}{d\theta }\right)}_{\theta =\frac{\pi }{2}}=a$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\theta =\frac{\pi }{2}}=1$

Length of normal $=y_1\sqrt{1+m^2}$

$=a\sqrt{2}$