A) a sin 2t
B) \[\frac{a}{2}\sin 2t\]
C) 2a sin 2t
D) 2a
Correct Answer: B
Solution :
[b] \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3a\,\,{{\sin }^{2}}t\cos t}{-3a\,\,{{\cos }^{2}}t\sin t}=-\tan \,\,t\] \[\therefore \] equation of the tangent at ?t? is \[y-a\,\,{{\sin }^{3}}t=-\tan t(x-a\,\,{{\cos }^{3}}t)\] \[\Rightarrow x\tan t+y-a({{\sin }^{3}}t+\sin \,t.{{\cos }^{2}}t)=0\] \[\Rightarrow x\tan t+y-a\sin t=0\] \[\therefore \] distance from the origin to this tangent \[=\frac{\left| -a\,\,\sin \,\,t \right|}{\sqrt{{{\tan }^{2}}t+1}}=\frac{a\sin t}{\sec t}=\frac{a}{2}\sin 2t\]
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