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

Let $(h,k)$ be the centroid of triangle

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

$k=\frac{1}{3}(a\sin t-b\cos t+0)$

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

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

on squaring and adding eq (1) & (2) we get

$\Rightarrow (a\cos t+b\sin t{)}^2+(a\sin t-b\cos t{)}^2=(3h-1{)}^2+(3k{)}^2$

$a^2{\cos }^{2}t+b^2{\sin }^{2}t+2ab\cos t\sin t+a^2{\sin }^{2}t+b^2{\cos }^{2}t-2ab\sin t\cos t=(3h-1{)}^2+(3k{)}^2$

$\Rightarrow a^2({\cos }^{2}t+{\sin }^{2}t)+b^2({\cos }^{2}t+{\sin }^{2}t)=(3h-1{)}^2+(3k{)}^2$

$\Rightarrow a^2+b^2=(3h-1{)}^2+(3k{)}^2[\because {\cos }^{2}t+{\sin }^{2}t=1]$

$\Rightarrow$ now

locus of $(h,k)$ centroid is

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