Find the coordinates of C and D, and you can get ∵ A C (3 √ 33,0 0) ∴ OA = √ 3, radius ∵ A is 2√3, connection AD∴AC=AD=AB=2√3∴CO=3√3, B0 = √. X 2+bx+c，so: ( 1)c=-3，(2)9+3√3b+c=0，so: b=-(2/3)√3，c =-3 ∴ y = ( 1/。 Then b is also the point on the parabola (2) the axis of symmetry of the parabola: the straight line x =-b/(2a) =-(-2/3) √ 3 ÷ (1/3× 2) = √ 3 makes the circumference of △PBD minimum even if: PB+PD+DB is minimum ÷BD is a fixed value. That is, the line segment and the minimal problem (refer to/czsx/jszx/bnjsc/dzkb/200805/t20080505 _ 464988.htm) find the symmetric point C (3 √ 3,0) of B about the straight line x=√3, and connect the intersection line x=√3 with P. At this time, PB+PD has a minimal CD: BC.

Then when m is on the axis of symmetry, x=√3, y=-4∴M(√3, -4) 2. Take BC as the edge.

There are two cases where the quadrilateral BCQM of m and M' ∵ is a parallelogram ∴QM=QM'=BC=3√3+√3=4√3 and ∵ axis of symmetry X = ∴ 3 ∴ qn = ∴ 3 ∴ nm = 4 √ 3-√ M' (5 √ 3, 12) So: M (√ 3, 4), (-3 √ 3, 12) or (5 √ 3, 12) I used the junior high school solution and standard format. Ask me if you don't understand anything.