A) \[a\sin t\]
B) \[2a\sin {{t}^{3}}\left( \frac{t}{2} \right)\sec \left( \frac{t}{2} \right)\]
C) \[2a\sin \left( \frac{t}{2} \right)\tan \left( \frac{t}{2} \right)\]
D) \[2a\sin \left( \frac{t}{2} \right)\]
Correct Answer: C
Solution :
\[x=a(t+\sin t),y=a(1-\cos t)\] \[\therefore \] \[\frac{dx}{dt}=a(1+\cos t),\frac{dy}{dt}=a\sin t\] \[\therefore \] \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\sin t}{a(1+\cos t)}=\frac{\sin t}{1+\cos t}\] Length of normal \[=y\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}\] \[=y\sqrt{1+\frac{{{\sin }^{2}}t}{{{(1+\cos t)}^{2}}}}\] \[=a(1-\cos t).\sqrt{1+\frac{{{\sin }^{2}}t}{{{\left( 2{{\cos }^{2}}\frac{t}{2} \right)}^{2}}}}\] \[=a(1-\cos t).\sqrt{1+\frac{{{\left( 2\sin \frac{t}{2}.\cos \frac{t}{2} \right)}^{2}}}{4{{\cos }^{4}}\frac{t}{2}}}\] \[=a(1-\cos t).\sqrt{1+\frac{4{{\sin }^{2}}\frac{t}{2}{{\cos }^{2}}\frac{t}{2}}{4{{\cos }^{4}}\frac{t}{2}}}\] \[=a(1-\cos t).\sqrt{1+{{\tan }^{2}}\frac{t}{2}}\] \[=a(1-\cos t).\sec \frac{t}{2}\] \[=2a{{\sin }^{2}}\frac{t}{2}.\sec \frac{t}{2}\] \[=2a\sin \frac{t}{2}.\tan \frac{t}{2}\]
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