MathematicsHard92×since 2002Q4698Let P be a variable point on the parabola y=4x2+1y = 4{x^2} + 1y=4x2+1. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is :A(3x−y)2+(x−3y)+2=0{(3x - y)^2} + (x - 3y) + 2 = 0(3x−y)2+(x−3y)+2=0B2(3x−y)2+(x−3y)+2=02{(3x - y)^2} + (x - 3y) + 2 = 02(3x−y)2+(x−3y)+2=0C(3x−y)2+2(x−3y)+2=0{(3x - y)^2} + 2(x - 3y) + 2 = 0(3x−y)2+2(x−3y)+2=0D2(x−3y)2+(3x−y)+2=02{(x - 3y)^2} + (3x - y) + 2 = 02(x−3y)2+(3x−y)+2=0Check answerSkip