Question

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

• A. $2\left(a\right|$
• B. $\sqrt{2}\left(a\right|$
• C. $\frac{\left(}{a}$
• D. $\frac{\left(}{a}$

Answer 1

The correct option is B $\sqrt{2}\left(a\right|$

Here,
$\frac{\mathrm{dx}}{\mathrm{d\theta }}=a\left(1+\cos \theta \right)\mathrm{and}\frac{\mathrm{dy}}{\mathrm{d\theta }}=a\left(\sin \theta \right)\mathrm{or},\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d\theta }}}{\frac{\mathrm{dx}}{\mathrm{d\theta }}}=\frac{a\left(\sin \theta \right)}{a\left(1+\cos \theta \right)}\mathrm{or},{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\theta =\frac{\pi }{2}}=\frac{\sin \frac{\pi }{2}}{1+\cos \frac{\pi }{2}}=1\therefore \mathrm{Length}\mathrm{of}\mathrm{normal}\mathrm{is}{\left(\left(y\right|\sqrt{1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}\right\}}_{\theta =\frac{\pi }{2}}=\left(a\left(1-\cos \frac{\pi }{2}\right)\right|\sqrt{1+1^2}=\sqrt{2}\left(a\right|.$