A) \[2a\]
B) \[a/2\]
C) \[\sqrt{2}\,a\]
D) \[a/\sqrt{2}\]
Correct Answer: C
Solution :
Length of normal = \[y\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}\] Now, \[\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{a\sin \theta }{a(1+\cos \theta )}=\frac{\sin \theta }{1+\cos \theta }=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{{\cos }^{2}}\frac{\theta }{2}}\] \\[{{\left( \frac{dy}{dx} \right)}_{\left( \theta =\frac{\pi }{2} \right)}}=\,{{\left[ \tan \frac{\theta }{2} \right]}_{\left( \theta =\frac{\pi }{2} \right)}}=1{{[y]}_{\left( \theta =\frac{\pi }{2} \right)}}=a\,\left( 1-\cos \frac{\pi }{2} \right)=a\] \ Length of normal = \[a\sqrt{1+{{(1)}^{2}}}=\sqrt{2}a\].
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