Q89 / 225 · JEE Main 2003 · Mathematics

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

  • A
    $(3 x+1)^2+(3 y)^2=a^2-b^2$
  • B
    $(3 x-1)^2+(3 y)^2=a^2-b^2$
  • Correct answer
    C
    $(3 x-1)^2+(3 y)^2=a^2+b^2$
  • D
    $(3 x+1)^2+(3 y)^2=a^2+b^2$

Solution

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

Hence, the correct answer is option: 'C'.

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