1.Let PPP be a variable point on the parabola y=4x2+1y = 4x^2 + 1y=4x2+1. Then, the locus of the mid-point of the point PPP and the foot of the perpendicular drawn from the point PPP to the line y=xy = xy=x is:a.(3x−y)2+(x−3y)+2=0(3x - y)^2 + (x - 3y) + 2 = 0(3x−y)2+(x−3y)+2=0b.2(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=0d.2(x−3y)2+(3x−y)+2=02(x - 3y)^2 + (3x - y) + 2 = 02(x−3y)2+(3x−y)+2=0Login to continueOnly logged in users canattempt or see the solution.