What is the locus of the centroid of the tria | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the vertices of the triangle be $A(a \cos t, a \sin t)$,$B(b \sin t, -b \cos t)$,and $C(1, 0)$.Let the centroid be $(x, y)$.The coordinates of the centroid are given by:$x = \frac{a \cos t + b \sin t + 1}{3}$ and $y = \frac{a \sin t - b \cos t + 0}{3}$.Rearranging the terms,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:$(3x - 1)^2 + (3y)^2 = a^2 \cos^2 t + b^2 \sin^2 t + 2ab \sin t \cos t + a^2 \sin^2 t + b^2 \cos^2 t - 2ab \sin t \cos t$.Simplifying:$(3x - 1)^2 + (3y)^2 = a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t)$.Since $\cos^2 t + \sin^2 t = 1$,we get:$(3x - 1)^2 + (3y)^2 = a^2 + b^2$.

What is the locus of the centroid of the triangle whose vertices are $(a \cos t, a \sin t)$,$(b \sin t, -b \cos t)$,and $(1, 0)$?

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