Question

Locus of centroid of the triangle whose vertices are $(a\cos \theta ,a\sin \theta ),(b\sin \theta ,-b\cos \theta )$ and $(1,0)$ where $\theta$ 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 the Triangle

Centroid $=(\frac{a\cos \theta +b\sin \theta +1}{3},\frac{-a\sin \theta +b\cos \theta }{3})$

Therefore $h=\frac{a\cos \theta +b\sin \theta +1}{3}\Rightarrow 3h=a\cos \theta +b\sin \theta +1\Rightarrow 3h-1=a\cos \theta +b\sin \theta$

$k=\frac{-a\sin \theta +b\cos \theta }{3}\Rightarrow 3k=-a\sin \theta +b\cos \theta$

squaring and adding both the above equations

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

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

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

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