A) \[ST=SN\]
B) \[ST=2\,SN\]
C) \[S{{T}^{2}}=a\,S{{N}^{3}}\]
D) \[S{{T}^{3}}=a\,SN\]
Correct Answer: A
Solution :
\[\frac{dx}{d\theta }=a(1+\cos \theta ),\,\frac{dy}{d\theta }=a\,(\sin \theta )\] \[{{\left. \frac{dy}{dx} \right|}_{\theta =\frac{\pi }{2}}}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\sin \theta }{a(1+\cos \theta )}=1,\] \[{{\left. y \right|}_{\theta =\frac{\pi }{2}}}=a\] Length of sub-tangent ST = \[\frac{y}{dy/dx}=\frac{a}{1}=a.\] and length of sub-normal SN =\[y\frac{dy}{dx}=a\,.\,1=a\] Hence \[ST=SN\].
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