Question

The locus of centroid of a 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 B $(3x-1{)}^2+(3y{)}^2=a^2+b^2$

Given, vertices of triangle are $(a\cos t,a\sin t),(b\sin t,-b\cos t)$ and $(1,0),$
The centroid of the triangle is
$(x,y)=(\frac{a\cos t+b\sin t+1}{3},\frac{a\sin t-b\cos t+0}{3})\Rightarrow a\cos t+b\sin t=3x-1\Rightarrow a\sin t-b\cos t=3y$

Now,
$(3x-1{)}^2+(3y{)}^2=a^2+b^2$