A curve is given by the equations \[x=a\cos \theta +\frac{1}{2}b\cos 2\theta ,\] \[y=a\sin \theta +\frac{1}{2}b\,\sin \,2\theta \], then the points for which \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=0,\] is given by [Kurukshetra CEE 2002]

\[\cos \theta =\frac{\left( {{a}^{2}}-2{{b}^{2}} \right)}{3ab}\]

\[\sin \theta =\frac{2{{a}^{2}}+{{b}^{2}}}{5ab}\]

\[\tan \theta =\frac{3{{a}^{2}}+2{{b}^{2}}}{4ab}\]

\[\cos \theta =\frac{-\left( {{a}^{2}}+2{{b}^{2}} \right)}{3ab}\]

A curve is given by the equations \[x=a\cos \theta +\frac{1}{2}b\cos

1.	A curve is given by the equations \[x=a\cos \theta +\frac{1}{2}b\cos 2\theta ,\] \[y=a\sin \theta +\frac{1}{2}b\,\sin \,2\theta \], then the points for which \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=0,\] is given by [Kurukshetra CEE 2002]
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.	\[\cos \theta =\frac{\left( {{a}^{2}}-2{{b}^{2}} \right)}{3ab}\]
Answer» D. \[\cos \theta =\frac{\left( {{a}^{2}}-2{{b}^{2}} \right)}{3ab}\]

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A curve is given by the equations \[x=a\cos \theta +\frac{1}{2}b\cos 2\theta ,\] \[y=a\sin \theta +\frac{1}{2}b\,\sin \,2\theta \], then the points for which \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=0,\] is given by [Kurukshetra CEE 2002]

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