∫DM∑on, D (3 3,0), ∴ n (-3,2), then the solution is ∴;

(2) Connect the Y axis of AC with G, where ∫m is the midpoint of BC, ∴AO=BM=MC, AB=BC=2, ∴AG=GC, that is, G (0, 1).

∫∠ABC = 90°, ∴BG⊥AC, that is, BG is the middle vertical line of AC. Let PA=PC, that is, point p is on the middle vertical line of AC, so p is on the straight line BG.

Point p is the intersection of straight line BG and parabola,

Let the analytical formula of straight line BG be, then the solution is, Ⅷ.

Do something,

∴ point p () or p (),

③ ∫∴ axis of symmetry,

Make, solve, ∴E (,0),

Therefore, e and d are symmetrical about a straight line, ∴QE=QD, ∴|QE-QC|=|QD-QC|,

Maximize |QE-QC| and extend the intersection of DC and Q, that is, Q is the intersection of DC and straight line.

Since m is the midpoint of BC, ∴ c (1, 2), let the analytical formula of straight CD be y = kx+b.

So, the solution is,

When,,

Therefore, when q is in the position of (), |QE-QC| is the largest.

If the intersection C is the CF⊥x axis and the vertical foot is F, then CD =.