The locus of the centroid of the triangle wit | vedclass

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Question

(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 $(x, y)$.By the centroid formula,$x = \frac{a \cos 	heta + b \sin 	heta + 1}{3}$ and $y = \frac{a \sin 	heta - b \cos 	heta}{3}$.Rearranging,we get $a \cos 	heta + b \sin 	heta = 3x - 1$ and $a \sin 	heta - b \cos 	heta = 3y$.Squaring both equations and adding them:$(a \cos 	heta + b \sin 	heta)^2 + (a \sin 	heta - b \cos 	heta)^2 = (3x - 1)^2 + (3y)^2$.Expanding the left side:$a^2 \cos^2 	heta + b^2 \sin^2 	heta + 2ab \sin 	heta \cos 	heta + a^2 \sin^2 	heta + b^2 \cos^2 	heta - 2ab \sin 	heta \cos 	heta = (3x - 1)^2 + 9y^2$.$a^2(\cos^2 	heta + \sin^2 	heta) + b^2(\sin^2 	heta + \cos^2 	heta) = (3x - 1)^2 + 9y^2$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we get $a^2 + b^2 = (3x - 1)^2 + 9y^2$.

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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