A equiv (cos theta, sin theta) and B equiv (s | vedclass

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(B) Let the coordinates of the vertices of $	riangle OAB$ be $O(0, 0)$,$A(\cos 	heta, \sin 	heta)$,and $B(\sin 	heta, -\cos 	heta)$.Let the centr

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$9x^{2} + 9y^{2} = 2$

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(B) Let the coordinates of the vertices of $	riangle OAB$ be $O(0, 0)$,$A(\cos 	heta, \sin 	heta)$,and $B(\sin 	heta, -\cos 	heta)$.Let the centroid of $	riangle OAB$ be $(h, k)$.The formula for the centroid $(h, k)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $h = \frac{x_1 + x_2 + x_3}{3}$ and $k = \frac{y_1 + y_2 + y_3}{3}$.Substituting the given coordinates:$h = \frac{0 + \cos 	heta + \sin 	heta}{3} \Rightarrow 3h = \cos 	heta + \sin 	heta$ ... $(i)$$k = \frac{0 + \sin 	heta - \cos 	heta}{3} \Rightarrow 3k = \sin 	heta - \cos 	heta$ ... (ii)To find the locus,we eliminate $	heta$ by squaring and adding equations $(i)$ and (ii):$(3h)^{2} + (3k)^{2} = (\cos 	heta + \sin 	heta)^{2} + (\sin 	heta - \cos 	heta)^{2}$$9h^{2} + 9k^{2} = (\cos^{2} 	heta + \sin^{2} 	heta + 2\sin 	heta \cos 	heta) + (\sin^{2} 	heta + \cos^{2} 	heta - 2\sin 	heta \cos 	heta)$$9h^{2} + 9k^{2} = 1 + 1 = 2$Replacing $(h, k)$ with $(x, y)$,the locus is $9x^{2} + 9y^{2} = 2$.

$A \equiv (\cos \theta, \sin \theta)$ and $B \equiv (\sin \theta, -\cos \theta)$ are two points. The locus of the centroid of $\triangle OAB$,where $O$ is the origin,is

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