The equation of the locus of the centroid of | vedclass

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(A) Let the vertices of the triangle be $A = (a \cos k, a \sin k)$,$B = (b \sin k, -b \cos k)$,and $C = (1, 0)$.Let $G(x, y)$ be the centroid of the t

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$(1-3x)^2 + 9y^2 = a^2 + b^2$

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(A) Let the vertices of the triangle be $A = (a \cos k, a \sin k)$,$B = (b \sin k, -b \cos k)$,and $C = (1, 0)$.Let $G(x, y)$ be the centroid of the triangle.The coordinates of the centroid are given by $x = \frac{a \cos k + b \sin k + 1}{3}$ and $y = \frac{a \sin k - b \cos k + 0}{3}$.From these,we have $3x - 1 = a \cos k + b \sin k$ ... $(i)$ and $3y = a \sin k - b \cos k$ ... $(ii)$.Squaring and adding equations $(i)$ and $(ii)$:$(3x - 1)^2 + (3y)^2 = (a \cos k + b \sin k)^2 + (a \sin k - b \cos k)^2$.$(3x - 1)^2 + 9y^2 = a^2(\cos^2 k + \sin^2 k) + b^2(\sin^2 k + \cos^2 k) + 2ab \cos k \sin k - 2ab \sin k \cos k$.Since $\sin^2 k + \cos^2 k = 1$,we get $(3x - 1)^2 + 9y^2 = a^2 + b^2$.This is equivalent to $(1 - 3x)^2 + 9y^2 = a^2 + b^2$.

The equation of the locus of the centroid of the triangle whose vertices are $(a \cos k, a \sin k)$,$(b \sin k, -b \cos k)$ and $(1, 0)$,where $k$ is a parameter,is

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