(2)b=-3/2 c=4

(3) 1. Minimum perimeter 4(√5+√2)

2.s = | 2t+4 | t≦-2

When t >-2, the relation of area function is wrong! ! ! And the algorithm is a bit cumbersome. We can consider that the angle between the AC extension line and the parabola symmetry axis is 45, and the AC value is constant, so the height on AC is | t+2 | * sin 45°, and the triangle area can be obtained.

The third question and the first question can also directly use the given G point to be symmetrical about the parabola symmetry axis and the C point, so it is not necessary to extend the AB intersection parabola to the A' point.

/math/ques/detail/b 794 a 59 c-6cd 1-476 b-a 146-2f 229 c 528 16 1