JEE Main 202125 Jul, Morning shiftMathematicsMediumCoordinate Geometry478× since 2002The locus of the centroid of the triangle formed by any point P on the hyperbola 16x2−9y2+32x+36y−164=016{x^2} - 9{y^2} + 32x + 36y - 164 = 016x2−9y2+32x+36y−164=0, and its foci is :AOption A. 16x2−9y2+32x+36y−36=016{x^2} - 9{y^2} + 32x + 36y - 36 = 016x2−9y2+32x+36y−36=0BOption B. 9x2−16y2+36x+32y−144=09{x^2} - 16{y^2} + 36x + 32y - 144 = 09x2−16y2+36x+32y−144=0COption C. 16x2−9y2+32x+36y−144=016{x^2} - 9{y^2} + 32x + 36y - 144 = 016x2−9y2+32x+36y−144=0DOption D. 9x2−16y2+36x+32y−36=09{x^2} - 16{y^2} + 36x + 32y - 36 = 09x2−16y2+36x+32y−36=0CheckSkip · show solution
