Question

Locus of centroid of triangle with vertices $(a\cos t,a\sin t),(b\sin t,-b\cos t)$ and $(1,0)$ where $t$ is 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 A $(3x-1{)}^2+(3y{)}^2=a^2+b^2$

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

we know that if $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are vertices of a triangle then

Centroid $(h,k)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

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

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

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

$3k=a\sin t-b\cos t$

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

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

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

Replacing $h$ with $x$ and $k$ with $y$

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