From x2=2py, y=
1
4p
X2, the derivative is y'=
1
2p
x.
These two tangent equations are y-y 1=
1
2p
x 1(x-x 1)，①
y-y2=
1
2p
X2(x-x2), ②…2 points.
For equation ①, if we substitute point M(m, -p), we get -p-y 1=
1
2p
x 1(m-x 1)，
Y 1=
1
4p
x 12，
∴-p-
1
4p
x 12=
1
2p
x 1(m-x 1)，
Finishing: x 12-2mx 1-4p2=0,
Similarly, x22-2mx2-4p2 of Equation ② = 0,
That is x 1, and x2 is two of the equations x2-2mx-4p2=0.
∴x 1+x2=2m, x 1x2=-4p2, ③…4 points.
Let the slope of straight line AB be k and k=
y2-y 1
x2-x 1
=
x22-x 12
4p(x2-x 1)
=
1
4p
(x 1+x2)，
∴ The equation of straight line AB is y-
x 12
4p
=
1
4p
(x 1+x2)(x-x 1)，
Expand: y=
1
4p
(x 1+x2)x-
x 1x2
4p
Substitute ③ to get: y=
m
2p
X+p, ∴ A straight line passes through the fixed point (0, p) p). ... six points.
(2) Proof: Let M(m, -p), A(x 1, y 1) and B(x2, y2) be the conclusion of (i).
And x 1+x2=2m, x 1x2=-4p2,
∴kMA=
y 1+p
x 1-m
，kMB=
y2+p
x2-m
∴
1
kMA
+
1
Jiuba
=
x 1-m
y 1+p
+
x2-m
y2+p
=
x 1-m
x 12
4p
+p
+
x2-m
x22
4p
+p
=
4p(x 1 m)
x 12+4p2
+
4p(x2-m)
x22+4p2
=
4p(x 1 m)
x 12-x 1x2
+
4p(x2-m)
x22-x 1 x2
=
4p(x 1-m)x2-4p(x2-m)x 1
x 1 x2(x 1 x2)
=
4 p.m.
x 1x2
=
4 p.m.
-4p2
=-
m
p
It's also VIII
1
kMP
=
m
-p-p
=-
m
2p
∴
1
kMA
+
1
Jiuba
=
2
kMP
.
That is to say, the slopes of straight lines MA, MF and MB are the reciprocal of arithmetic series ... 13 points.