JEE Main 2005Annual, Paper 1MathematicsEasyCoordinate Geometry478× since 2002Let PPP be the point (1,0)(1, 0)(1,0) and QQQ a point on the parabola y2=8x{y^2} = 8xy2=8x. The locus of mid point of PQPQPQ is :AOption A. y2−4x+2=0{y^2} - 4x + 2 = 0y2−4x+2=0BOption B. y2+4x+2=0{y^2} + 4x + 2 = 0y2+4x+2=0COption C. x2+4y+2=0{x^2} + 4y + 2 = 0x2+4y+2=0DOption D. x2−4y+2=0{x^2} - 4y + 2 = 0x2−4y+2=0CheckSkip · show solution
