The locus of the centroid of a triangle whose | vedclass

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(B) The centroid $(x, y)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $x = \frac{x_1 + x_2 + x_3}{3}$ and $y = \frac{

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$a^2 + b^2 - 1$

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(B) The centroid $(x, y)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $x = \frac{x_1 + x_2 + x_3}{3}$ and $y = \frac{y_1 + y_2 + y_3}{3}$.Given vertices are $(1, 0)$,$(a \cos t, a \sin t)$,and $(b \sin t, -b \cos t)$.So,$x = \frac{1 + a \cos t + b \sin t}{3} \Rightarrow 3x - 1 = a \cos t + b \sin t$ ...$(i)$And $y = \frac{0 + a \sin t - b \cos t}{3} \Rightarrow 3y = a \sin t - b \cos t$ ...(ii)Squaring and adding equations $(i)$ and (ii):$(3x - 1)^2 + (3y)^2 = (a \cos t + b \sin t)^2 + (a \sin t - b \cos t)^2$$9x^2 - 6x + 1 + 9y^2 = a^2 \cos^2 t + b^2 \sin^2 t + 2ab \sin t \cos t + a^2 \sin^2 t + b^2 \cos^2 t - 2ab \sin t \cos t$$9x^2 + 9y^2 - 6x + 1 = a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t)$$9x^2 + 9y^2 - 6x + 1 = a^2 + b^2$$9x^2 + 9y^2 - 6x = a^2 + b^2 - 1$Comparing this with the given equation $9x^2 + 9y^2 - 6x = k$,we get $k = a^2 + b^2 - 1$.

The locus of the centroid of a triangle whose vertices are $(1, 0)$,$(a \cos t, a \sin t)$,and $(b \sin t, -b \cos t)$ is $9x^2 + 9y^2 - 6x = k$. Then,the value of $k$ is equal to

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