Question

Find the locus of the centroid of a triangle whose vertices are $(acost,asint),(bsint,-bcost)$ and $(1,0)$, where T is the parameter.

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

Answer 1

Answer: (D)

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

given points are $A(acost,asint),B(bsint,-bcost),C(1,0)$

therefore the centroid of the triangle is $(x,y)=(\frac{acost+bsint+1}{3},\frac{asint-bcost+0}{3})$

so $x=\frac{acost+bsint+1}{3},y=\frac{asint-bcost+0}{3}$

$\Rightarrow 3x-1=acost+bsint,3y=asint-bcost$

now squaring and adding both equations we get

${(3x-1)}^2+{\left(3y\right)}^2={(acost+bsint)}^2+{(asint-bcost)}^2$

${(acost+bsint)}^2+{(asint-bcost)}^2=a^2({\left(cost\right)}^2+{\left(sint\right)}^2)+b^2({\left(cost\right)}^2+{\left(sint\right)}^2)+2ab.cost.sint-2ab.cost.sint$

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

this is the locus of the centroid