question_answer

If ST and SN are the lengths of the sub tangent and the subnormal at the point \[\theta =\frac{\pi }{2}\] on the curve \[x=a\,(\theta +\sin \theta )\], \[y=a\,(1-\cos \theta ),\,a\ne 1\], then :

A)  \[ST=SN\]                          

B)  \[Sr=2SN\]

C)  \[S{{T}^{2}}=a\,S{{N}^{3}}\]                      

D)  \[S{{T}^{3}}=a\,SN\]

Correct Answer: A

Solution :

Given that \[\Rightarrow \,\frac{dx}{d\theta }=a\,(1+\cos \theta )\] and \[\,\frac{dy}{d\theta }=a\,\sin \theta \] \[\therefore \,\,\frac{dy}{dx}=\frac{a\sin \theta }{a\,\,(1+\cos \theta )}=\frac{2\sin \,\theta /2\cos \theta /2}{2{{\cos }^{2}}\theta /2}\]                                 \[=\tan \theta /2\] Now length of sub tangent \[=\left| \frac{y}{dy/dx} \right|\] \[\therefore \] \[ST=\frac{a\,(1-\cos \theta )}{\tan \theta /2}\]                 \[=a\,\,.\,\,\frac{2{{\sin }^{2}}\theta /2}{\sin \theta /2}\,\,.\,\,\cos \theta /2=\sin \theta \]                                                                                                                           Length of sub tangent at x\[\theta =\pi /2\],                 \[ST=a\sin \frac{\pi }{2}=a\] and length of sub normal \[=\left| y\frac{dy}{dx} \right|\] \[\Rightarrow \,\,SN=a\,(1-cos\theta )\,\,.\,\,\tan \theta /2\]                 \[=a\,.\,\,2\,{{\sin }^{2}}\theta /2\tan \frac{\theta }{2}\] \[\Rightarrow \] Length of subnormal at                 \[\theta =\pi /2,\,SN=a\,.\,\,2\,.\,\frac{1}{2}=a\] Hence \[SN=ST\]