Locus of a centriod 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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Aicharya · Accepted Answer

$x=\frac{\cos t+b \sin t+1}{3} \Rightarrow a \cos t+b \sin t=3 x-1$ $y=\frac{a \sin t-b \cos t}{3} \Rightarrow a \sin t-b \sin t=3 y$ Squaring \& adding, $(3 \mathrm{x}-1)^2+(3 \mathrm{y})^2=\mathrm{a}^2+\mathrm{b}^2$

Locus of a centriod of the triangle whose vertices are $(a \cos t, a \sin t),(b \sin t,-b \cos t)$ and $(1…

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