It is known that the vertex of parabola y = (X-2) ^ 2-m ^ 2 (constant M greater than 0) is p Q: if the two intersections of the parabola and the x-axis are a and B from left to right, and the angle APB is 90 degrees, try to find the perimeter of the triangle APB
The vertex P (2, - m ^ 2) of parabola y = (X-2) ^ 2-m ^ 2 (constant M greater than 0),
A(2-m,0),B(2+m,0),
The midpoint of AB is C (2,0), CP is the symmetry axis of parabola, CP ⊥ ab
Angle APB = 90 degrees, triangle ACP, BCP are isosceles right triangle,
CP=|-m^2|=m^2,AC=BC=m,
M ^ 2 = m, M = 0 (rounding off), M = 1,
AP=BP=√2,
The perimeter of triangle APB = 2 + 2 √ 2
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