The locus of the centroid of the triangle who | vedclass

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(B) Let the centroid of the triangle be $(x, y)$.Given vertices are $(x_1, y_1) = (a \cos t, a \sin t)$,$(x_2, y_2) = (b \sin t, -b \cos t)$,and $(x_3

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the centroid of the triangle be $(x, y)$.Given vertices are $(x_1, y_1) = (a \cos t, a \sin t)$,$(x_2, y_2) = (b \sin t, -b \cos t)$,and $(x_3, y_3) = (1, 0)$.The coordinates of the centroid are given by $x = \frac{x_1 + x_2 + x_3}{3}$ and $y = \frac{y_1 + y_2 + y_3}{3}$.Thus,$3x = a \cos t + b \sin t + 1$ and $3y = a \sin t - b \cos t$.Rearranging,we get $3x - 1 = a \cos t + b \sin t$ and $3y = a \sin t - b \cos t$.Squaring and adding both equations:$(3x - 1)^2 + (3y)^2 = (a \cos t + b \sin t)^2 + (a \sin t - b \cos t)^2$.Expanding the right side:$= a^2 \cos^2 t + b^2 \sin^2 t + 2ab \sin t \cos t + a^2 \sin^2 t + b^2 \cos^2 t - 2ab \sin t \cos t$.$= a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t) = a^2 + b^2$.Therefore,the locus is $(3x - 1)^2 + (3y)^2 = a^2 + b^2$.

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

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