Question

If $\theta$ is parameter, $A=(a\cos \theta ,a\sin \theta )$ and $B=(b\sin \theta ,-b\cos \theta )$ , $C=(1,0)$ then the locus of the centroid of $\Delta ABC$ is

• A. $(3x+1{)}^2+9y^2=a^2+b^2$
• B. $(3x-1{)}^2+9y^2=a^2-b^2$
• C. $(3x-1{)}^2+9y^2=a^2+b^2$
• D. $(3x+1{)}^2+9y^2=a^2-b^2$

Answer 1

The correct option is C $(3x-1{)}^2+9y^2=a^2+b^2$

Centroid of triangle

$x=\frac{a\cos \theta +b\sin \theta +1}{3};y=\frac{a\sin \theta -b\cos \theta +0}{3}$

$3x-1=a\cos \theta +b\sin \theta ;3y=a\sin \theta -b\cos \theta$

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

${(3x-1)}^2+9{\left(y\right)}^2=a^2+b^2$