Question

The locus of the 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. ${(3x-1)}^2+{\left(3y\right)}^2=a^2-b^2$
• B. ${(3x-1)}^2+{\left(3y\right)}^2=a^2+b^2$
• C. ${(3x+1)}^2+{\left(3y\right)}^2=a^2+b^2$
• D. ${(3x+1)}^2+{\left(3y\right)}^2=a^2-b^2$

Answer 1

The correct option is B ${(3x-1)}^2+{\left(3y\right)}^2=a^2+b^2$

$(a\cos t,a\sin t),(b\sin t,-b\cos t),(1,0)$

Let locus of centroid be $(h,k)$

$h=\frac{(a\cos t+b\sin t+1)}{3}$

$3h-1=a\cos t+b\sin t$ ------(1)

$k=\frac{(a\sin t-b\cos t+0)}{3}$

$3k=a\sin t-b\cos t$ -------(2)

Squaring and adding (1) and (2) we get

$(3h-1{)}^2+(3k{)}^2=a^2+b^2$

Hence the locus is $(3x-1{)}^2+(3y{)}^2=a^2+b^2$