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If $(p, q), (a \cos α, b \sin α), (b \cos α, a \sin α)$ are the vertices of a triangle, where $α$ is a parameter then the locus of the centroid of the triangle

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${(3 x - p)}^{2} + {(3 y - q)}^{2} = {(a + b)}^{2}$

$x^{2} + y^{2} = 9 {(a + b)}^{2}$

${(3 x + p)}^{2} + {(3 y + q)}^{2} = {(a + b)}^{2}$

$x (x - a) + y (y + b) = 0$

answer is A.

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Detailed Solution

centroid $= (x, y)$ $= (\frac{p + (a + b) \cos α}{3}, \frac{q + (a + b) \sin α}{3})$

$\cos α = \frac{3 x - p}{a + b}, \sin α = \frac{3 y - 2}{a + b}$

$\Rightarrow {(3 x - p)}^{2} + {(3 y - q)}^{2} = {(a + b)}^{2}$

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