1. The intersection of parabola and y axis is a, that is, x=0, and A(0, c) is obtained; AC is a chord, so p (c/2,0), so the radius of circle p is pa 2 = op 2+OA 2 = c 2+(c/2) 2; Similarly, the radius PE 2 = PF 2+EF 2 = (C/2+1) 2+12; If the two radii are equal, the solution is c=2, so b (2,2).

Second, simplicity. The coordinates of each point can be obtained from the first solution, which is enough to prove that PM 2 = PE 2+Me 2.

Third, the straight line EF is the symmetry axis of parabola, which requires the shortest perimeter, that is, AQ+CQ is the shortest. Make point A' which is axisymmetric about straight line EF, connect point A' C', and the intersection point with EF when the perimeter is shortest is point Q. This is based on the mirror image principle, and the straight line between two points is the shortest. Find the relationship between S and T again: subtract the area of triangle CFQ from the area of trapezoid OAQC, and note that point Q and point N are not coincident. ..

Ps: Draw more pictures when you do the problem.