JEE Main 2003Annual, Paper 1MathematicsEasyCoordinate Geometry478× since 2002Locus 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 :AOption 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−b2BOption B. (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−b2COption C. (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+b2DOption D. (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+b2CheckSkip · show solution
