Question

The locus of the centroid of the triangle with vertices at (a cos $\theta$, a sin $\theta$), (b sin $\theta$, -b cos $\theta$) and (1, 0) is (Here $\theta$ is a parameter)

• 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 $(3x-1{)}^2+9y^2=a^2+b^2$

The co-ordinates of the centroid are
$(x,y)=(\frac{a\cos \theta +b\sin \theta +1}{3},\frac{a\sin \theta -b\cos \theta }{3})$
Hence, $x=\frac{a\cos \theta +b\sin \theta +1}{3}$
$3x-1=a\cos \theta +b\sin \theta$ ...........(i)
And
$y=\frac{a\sin \theta -b\cos \theta }{3}$
$3y=a\sin \theta -b\cos \theta$ ...........(ii)
Hence, $(3x-1{)}^2+9y^2=(a\cos \theta +b\sin \theta {)}^2+(a\sin \theta -b\cos \theta {)}^2$
$=a^2({\cos }^2\theta +{\sin }^2\theta )+b^2({\cos }^2\theta +{\sin }^2\theta )+2ab(\sin \theta \cos \theta -\cos \theta \sin \theta )$
$(3x-1{)}^2+9y^2=a^2+b^2$