Question

Locus of centroid 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. $(3x +1)^2 + (3y)^2 = a^2 - b^2$
• B. $(3x -1)^2 + (3y)^2 = a^2 - b^2$
• C. $(3x -1)^2 + (3y)^2 = a^2 + b^2$
• D. $(3x +1)^2 + (3y)^2 = a^2 + b^2$

Answer 1

Answer: (C)

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

Answer 2

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