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 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

Answer: (A)

${(3x-1)}^2+{\left(3y\right)}^2=a^2+b^2$

Let $(h,k)$ be the co-ordinate of centroid
$\therefore h=\frac{a\cos t+b\sin t+1}{3}$
and $k=\frac{a\sin t-b\cos t+0}{3}$
$\Rightarrow 3h-1=a\cos t+b\sin t...\left(i\right)3k=a\sin t-b\cos t......\left(ii\right)$
By squaring (i) and (ii) then adding we get,
$(3h-1{)}^2+(3k-0{)}^2=a^2({\cos }^{2}t+{\sin }^{2}t)+b^2({\cos }^{2}t+{\sin }^{2}t)=a^2+b^2$
replacing $(h,k)$ by $(x,y)$ we get
choice (A) is correct.