JEE MAIN - Mathematics (2003 - No. 47)
Locus of centroid of the triangle whose vertices are $$\left( {a\cos t,a\sin t} \right),\left( {b\sin t, - b\cos t} \right)$$ and $$\left( {1,0} \right),$$ where $$t$$ is a parameter, is :
$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}$$
$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}$$
$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$
$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$
Explanation
$$x = {{a\cos t + b\sin t + 1} \over 3}$$
$$ \Rightarrow a\cos t + b\sin t = 3x - 1$$
$$y = {{a\sin t - b\cos t} \over 3}$$
$$ \Rightarrow a\sin t - b\cos t = 3y$$
$$ \Rightarrow a\cos t + b\sin t = 3x - 1$$
$$y = {{a\sin t - b\cos t} \over 3}$$
$$ \Rightarrow a\sin t - b\cos t = 3y$$
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