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

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

$x=\frac{a\cos t+b\sin t+1}{3}$ and $y=\frac{a\sin t-b\cos t}{3}$
$\therefore 3x-1=a\cos t+b\sin t$ --(1) and

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