JEE Main 2014Session, Shift 1MathematicsMediumCoordinate Geometry478× since 2002The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6{x^2} + 3{y^2} = 6x2+3y2=6 on any tangent to it is :AOption A. (x2+y2)2=6x2+2y2\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} + 2{y^2}(x2+y2)2=6x2+2y2BOption B. (x2+y2)2=6x2−2y2\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} - 2{y^2}(x2+y2)2=6x2−2y2COption C. (x2−y2)2=6x2+2y2\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} + 2{y^2}(x2−y2)2=6x2+2y2DOption D. (x2−y2)2=6x2−2y2\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} - 2{y^2}(x2−y2)2=6x2−2y2CheckSkip · show solution
