The locus of the centroid of the triangle wit | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the vertices of the triangle be $A(a \cos 	heta, a \sin 	heta)$,$B(b \sin 	heta, -b \cos 	heta)$,and $C(1, 0)$.Let the centroid be $(x, y)$.Using the centroid formula,we have:$x = \frac{a \cos 	heta + b \sin 	heta + 1}{3} \implies 3x - 1 = a \cos 	heta + b \sin 	heta$$y = \frac{a \sin 	heta - b \cos 	heta + 0}{3} \implies 3y = a \sin 	heta - b \cos 	heta$Squaring and adding both equations:$(3x - 1)^2 + (3y)^2 = (a \cos 	heta + b \sin 	heta)^2 + (a \sin 	heta - b \cos 	heta)^2$$(3x - 1)^2 + 9y^2 = a^2 \cos^2 	heta + b^2 \sin^2 	heta + 2ab \sin 	heta \cos 	heta + a^2 \sin^2 	heta + b^2 \cos^2 	heta - 2ab \sin 	heta \cos 	heta$$(3x - 1)^2 + 9y^2 = a^2(\cos^2 	heta + \sin^2 	heta) + b^2(\sin^2 	heta + \cos^2 	heta)$Since $\sin^2 	heta + \cos^2 	heta = 1$,we get:$(3x - 1)^2 + 9y^2 = a^2 + b^2$.

The locus of the centroid of the triangle with vertices at $(a \cos \theta, a \sin \theta)$,$(b \sin \theta, -b \cos \theta)$ and $(1, 0)$ is (where $\theta$ is a parameter).

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