CE A.Maths (locus)

Let L be y = m(x-2)
Then [m(x-2)]^2 = 4x
m^2 (x^2 - 4x+ 4) = 4x
m^2x^2 - (4m^2 +4) x + 4m^2 = 0 ...(*)
Let A, B and P be (x1, y1), (x2, y2) and (x3, y3)
x1 and x2 are the roots of (*)
x1+x2 = (4m^2 + 4) /m^2
y1= m(x1-2) and y2 = m(x2-2)
x3 = (x1+x2)/2 = 2(m^2 + 1)/m^2 ...($)
y3 = (y1+y2)/2 = m(x1+x2)/2 - 2
y3 = mx3-2
m = (y3+2)/x3 ...(#)
Substitute (#) into ($),
x3 = 2[(y3+2)^2/x3^2 + 1]/[(y3+2)^2/x3^2]
x3 = 2[(y3+2)^2 + x3^2]/(y3+2)^2
The equation of locus of P is
x(y+2)^2 = 2(y+2)^2 + 2x^2
(x-2)(y+2)^2 = 2x^2