Question

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

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

Answer 1

The correct option is C $\sqrt{2}a$

Length of normal =$y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$
Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{asin\theta }{a(1+cos\theta )}=\frac{sin\theta }{1+cos\theta }=\frac{2sin\frac{\theta }{2}cos\frac{\theta }{2}}{2co{s}^2\frac{\theta }{2}}$
$\therefore {\left(\frac{dy}{dx}\right)}_{(\theta =\frac{\pi }{2})}={\left[tan\frac{\theta }{2}\right]}_{(\theta =\frac{\pi }{2})}=1[y{]}_{(\theta =\frac{\pi }{2})}=a(1-cos\frac{\pi }{2})=a$
$\therefore$ Length of normal =$a\sqrt{1+(1{)}^2}=\sqrt{2}a$.