MathematicsEasy128×since 2002Q5562Locus of centroid of the triangle whose vertices are (acost,asint),(bsint,−bcost)\left( {a\cos t,a\sin t} \right),\left( {b\sin t, - b\cos t} \right)(acost,asint),(bsint,−bcost) and (1,0),\left( {1,0} \right),(1,0), where ttt is a parameter, is :A(3x+1)2+(3y)2=a2−b2{\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}(3x+1)2+(3y)2=a2−b2B(3x−1)2+(3y)2=a2−b2{\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}(3x−1)2+(3y)2=a2−b2C(3x−1)2+(3y)2=a2+b2{\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}(3x−1)2+(3y)2=a2+b2D(3x+1)2+(3y)2=a2+b2{\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}(3x+1)2+(3y)2=a2+b2Check answerSkip