JEE Math Practice Question
Let P be a variable point on the parabola $$y = 4{x^2} + 1$$. 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.$${(3x - y)^2} + (x - 3y) + 2 = 0$$
- B.$$2{(3x - y)^2} + (x - 3y) + 2 = 0$$
- C.$${(3x - y)^2} + 2(x - 3y) + 2 = 0$$
- D.$$2{(x - 3y)^2} + (3x - y) + 2 = 0$$
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