A) \[-\frac{1}{a}\]
B) \[\frac{1}{a}\]
C) \[-1\]
D) \[-2\]
E) \[-\frac{2}{a}\]
Correct Answer: A
Solution :
Given that, \[x=a(1+\cos \theta )\] ...(i) and \[y=a(\theta +\sin \theta )\] ...(ii) Differentiating Eqs. (i) and (ii) w.r.t.\[\theta ,\]we get \[\frac{dx}{d\theta }=-a\sin \theta \] ...(iii) and \[\frac{dy}{d\theta }=a+a\cos \theta \] ...(iv) Dividing Eq. (iii) by Eq. (iv), we get \[{\frac{dy}{d\theta }}/{\frac{dx}{d\theta }}\;=\frac{a(1+\cos \theta )}{-a\sin \theta }\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1-\cos \theta }{-\sin \theta }\] \[=-\frac{1+2{{\cos }^{2}}\frac{\theta }{2}-1}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}=-\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}\] \[=-\cot \frac{\theta }{2}\] \[\therefore \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\cos e{{c}^{2}}\frac{\theta }{2}.\frac{1}{2}\frac{d\theta }{dx}\] \[=-\frac{1}{2}\frac{1}{a}\frac{\cos e{{c}^{2}}\frac{\theta }{2}}{\sin \theta }\] \[\therefore \] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{\theta =\frac{\pi }{2}}}=-\frac{1}{a}\]
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