A) \[\sin t=\frac{2{{a}^{2}}+{{b}^{2}}}{3ab}\]
B) \[\cos t=\frac{{{a}^{2}}+2{{b}^{2}}}{3ab}\]
C) \[\tan t=a/b\]
D) None of the above
Correct Answer: B
Solution :
We have, \[x=a\cos t+\frac{b}{2}\cos 2t\] and \[y=a\sin t+\frac{b}{2}\sin 2t\] On-differentiating both equations w.r.t.t, we get \[\frac{dx}{dt}=-a\sin t-b\sin 2t\] and \[\frac{dy}{dt}=a\cos t+b\cos 2t\] \[\therefore \] \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a\cos t+b\cos 2t}{-a\sin t-b\sin 2t}\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{(a\cos t+b\cos 2t)}{(a\sin t+b\sin 2t)}\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dt}\left( \frac{dy}{dx} \right)\cdot \frac{dt}{dx}\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dt}\left( \frac{dy}{dx} \right)\cdot \frac{dt}{dx}\] \[=\frac{\left[ \begin{align} & (a\sin t+b\sin 2t)(-a\sin t-2b\sin 2t) \\ & -(a\cos t+b\cos 2t)(a\cos t+2b\cos 2t) \\ \end{align} \right]}{{{(a\sin t+b\sin 2t)}^{2}}}\] \[\times \frac{1}{-a\sin t-b\sin 2t}\] \[=\frac{\left[ \begin{align} & {{a}^{2}}{{\sin }^{2}}t+3ab\sin t\sin 2t+2{{b}^{2}}{{\sin }^{2}}2t \\ & +{{a}^{2}}{{\cos }^{2}}t+3ab\cos t\cos 2t+2{{b}^{2}}{{\cos }^{2}}2t \\ \end{align} \right]}{{{(a\sin t+b\sin 2t)}^{3}}}\] \[=\frac{\begin{align} & {{a}^{2}}({{\sin }^{2}}t+{{\cos }^{2}}t)+{{b}^{2}}({{\sin }^{2}}2t+{{\cos }^{2}}2t) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3ab\cos (2t-t) \\ \end{align}}{{{(a\sin t+b\sin 2t)}^{3}}}\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\left[ \frac{{{a}^{2}}+2{{b}^{2}}+3ab\cos t}{{{(a\sin t+b\sin 2t)}^{3}}} \right]\] Given,\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=0\] \[\Rightarrow \] \[{{a}^{2}}+2{{b}^{2}}+3ab\cos t=0\] \[\Rightarrow \] \[\cos t=-\left( \frac{{{a}^{2}}+2{{b}^{2}}}{3ab} \right)\]
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