The midpoint of MN is e,

So x 1+x2=6,

m，。 f，。 Three-point * * * line,

So | Mn | =| MF | +| fn| = x1+x2+p = 6+p = 8. p = 2。

So the parabolic equation is y 2 = 4x ①.

(2) Let the equation between P(- 1, m) and the straight line L be x=ny+t, ②.

Substitute into ①, y 2-4ny-4t = 0,

Let a (x3, y3) and b (x4, y4), then y3+y4 = 4n and y3y4 =-4t.

By ②, x3+x4 = n (y3+y4)+2t = 4n 2+2t,

x3x4=(ny3+t)(ny4+t)=n^2y3y4+nt(y3+y4)+t^2=-4n^2t+4n^2t+t^2=t^2，

x4y 3+x3y 4 = y3(ny4+t)+y4(ny3+t)= 2ny3y 4+t(y3+y4)=-8nt+4nt =-4nt，

PA slope k 1=(y3-m)/(x3+ 1),

Slope k2 of PB = (y4-m)/(x4+1),

The slope k3 of PT =-m/(t+1),

k 1+k2 =(y3-m)/(x3+ 1)+(y4-m)/(x4+ 1)

=[(y3-m)(x4+ 1)+(y4-m)(x3+ 1)]/[(x3+ 1)(x4+ 1)]

=[x4y 3+x3y 4+y3+y4-m(x3+x4+2)]/(x3x 4+x3+x4+ 1)

=[-4nt+4n-m(4n^2+2t+2)]/(t^2+4n^2+2t+ 1)，

So k1+k2-2k3 = [-4nt+4n-m (4n2+2t+2)]/(t2+4n2+2t+1)+2m/(t+1).

={(t+ 1)[-4nt+4n-m(4n^2+2t+2)]+2m(t^2+4n^2+2t+ 1)}/[(t^2+4n^2+2t+ 1)(t+ 1)]

= [4n (1-t2)+4mn2 (1-t)]/[(t2+4n 2+2t+1)] is a constant value, that is, it has nothing to do with m,

So t= 1, the fixed value is 0, and the coordinate of t is (1, 0).