Question

Locus of centroid of the triangle whose vertices are (a cos t, sin t), (b sin t, -b cost 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 C ${(3x-1)}^2+(3y{)}^2=a^2+b^2$

Triangle vertices are

$A(x_1,y_1)=(a\cos t,asint)$

$B(x_2,y_2)=(b\sin t,-b\cos t)$

$C(x_3,y_3)=(1,0)$

Let the centroid point is $(x,y)$.

We know that,

$(x,y)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

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

$3x=a\cos t+b\sin t+1,3y=a\sin t-b\cos t$

$3x-1=a\cos t+b\sin t,3y=a\sin t-b\cos t$

On squaring and adding both side we get,

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

$\Rightarrow {(3x-1)}^2+9y^2=a^2{\cos }^{2}t+b^2{\sin }^{2}t+2ab\sin t\cos t+a^2{\sin }^{2}t+b^2{\cos }^{2}t-2ab\sin t\cos t$

$\Rightarrow {(3x-1)}^2+9y^2=a^2{\cos }^{2}t+b^2{\sin }^{2}t+a^2{\sin }^{2}t+b^2{\cos }^{2}t$

$\Rightarrow {(3x-1)}^2+9y^2=a^2{\cos }^{2}t+a^2{\sin }^{2}t+b^2{\sin }^{2}t+b^2{\cos }^{2}t$

$\Rightarrow {(3x-1)}^2+9y^2=a^2({\cos }^{2}t+{\sin }^{2}t)+b^2({\sin }^{2}t+{\cos }^{2}t)$

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

Hence, this is the answer.