Solution: (1) Let the resolution function of the line where OA is located be y=kx, ∫ a (2,4), ∴2k=4, ∴k=2, and the resolution function of the line where OA is located be y = 2x. (2 points) (2)①∫. 2m)∴ The parabolic analytic function is y=(x-m)2+2m∴ When x=2, y=(2-m)2+2m=m2-2m+4(0≤m≤2)∴ The coordinate of point P is (2, m2-2m+4. Then PM = 2；; ; ①PM=PN= 2, then n1(2,3+2), N2 (2 2,3-2); (2) PM = Mn, which is known from the properties of isosceles triangle with three lines: n3 (2,1); ③PN=PM, where ∠PMN4=∠N4PM=∠PM3M, then: △PMN4∽△PN3M, so: PM2=PN4? 6? 1PN3, that is, PN4=PM2÷PN3= 1, so N4 (2,1); To sum up, the coordinates of the point N that meets the requirements are: n1(2,3+2); N2(2，3-2)； N3(2 1)； N4 (2， 1)。 (4 points) (4) When the line segment PB is the shortest, the analytical formula of parabola at this time is y=(x- 1)2+2, ① If the straight line passes through P, it is l∨OA, and if the straight line L: Y = 2x+H, it is: 4+H. ∴ Straight line L: Y = (2) The intercept above point A is AD=AP, that is, d (2 2,5); Let the straight line L '∨OA pass through d, and let the straight line L': y = 2x+h', then there is: 4+h'=5, h' =1;' ∴ straight line l': y = 2x+ 1, the analytical formula of simultaneous parabola is: {y = 2x+1y = (x-1) 2+2, and the solution is {x=2+2y=5+22, {x