∴A(0,-2).
The midpoint of MN is p, ∫(BM+MP)? MN =0，
∴BP？ MN = 0, ∴PB vertically bisects the line segment MN,
Let MN be: y=kx-2, which is the same as x2=8y, so.
x2-8kx+ 16=0。
xM+xN=8k，xMxN= 16。
From △ > 0? 64k2-4× 16>0？ k2> 1。
The coordinates of point P are XP = XM+XN2 = 8K2 = 4K, YP = KXP? 2=4k2？ 2.
The linear PB equation is: y? 4k2+2=？ 1k (x？ 4k)。
Let x=0, y = 2+4k2 > 6, and the range of ∴|OB | is (6, +∞);
(2) The existence point B (0, 10) is the demand.
In fact, if there is a point B, then △BMN is an isosceles right triangle, and ∠ B = 90.
Because it is known from (1) that PB bisects the line segment MN vertically,
So | BP | =| Mn | 2,
By b (0 0,2+4k2), P(4k, 4k2-2),
∴|BP|=(4k)2+(4k2？ 2? 2? 4k2)2=4k2+ 1。
12 | MN | = 12 1+k2(xM+xN)2？ 4xMxN
= 12 1+k264k2？ 64=4k4？ 1.
∴4k2+ 1=4k4？ 1.
The solution is k2=2,
Point B (0, 10) is what you want.
I hope I can help you. Oh, my God, I'm exhausted from typing! ! ! ! ! ! ! !