A) \[\sin \theta =\frac{2{{a}^{2}}+{{b}^{2}}}{5ab}\]
B) \[\tan \theta =\frac{3{{a}^{2}}+2{{b}^{2}}}{4ab}\]
C) \[\cos \theta =\frac{-\left( {{a}^{2}}+2{{b}^{2}} \right)}{3ab}\]
D) \[\cos \theta =\frac{\left( {{a}^{2}}-2{{b}^{2}} \right)}{3ab}\]
Correct Answer: C
Solution :
\[x=a\cos \theta +\frac{1}{2}b\cos 2\theta \], \[y=a\sin \theta +\frac{1}{2}b\sin 2\theta \] \[\frac{dy}{d\theta }=a\cos \theta +b\cos 2\theta \], \[\frac{dx}{d\theta }=-a\sin \theta -b\sin 2\theta \] \ \[\therefore \] \[\frac{d}{dx}\,\left( \frac{dy}{dx} \right)\,=\,\frac{d}{d\theta }\,.\,\left( \frac{dy}{dx} \right)\,.\,\frac{d\theta }{dx}\] = \[\left[ \frac{(a\sin \theta +b\sin 2\theta )(a\sin \theta +2b\sin 2\theta )}{{{(a\sin \theta +b\sin 2\theta )}^{2}}} \right.\] \[\left. +\frac{(a\cos \theta +b\cos 2\theta )(a\cos \theta +2b\cos 2\theta )}{{{(a\sin \theta +b\sin 2\theta )}^{2}}} \right]\,.\,\frac{d\theta }{dx}\]but \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=0\] Þ \[{{a}^{2}}+2{{b}^{2}}+3ab[\sin 2\theta \,\sin \theta +\cos 2\theta \,.\,\cos \theta ]=0\] Þ \[{{a}^{2}}+2{{b}^{2}}=-3ab\cos (2\theta -\theta )\] \ \[\cos \theta =-\left( \frac{{{a}^{2}}+2{{b}^{2}}}{3ab} \right)\].
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