Question

如图所示,已知抛物线y=x2-1与x轴交于A、B两点,与y轴交于点C．

Answer 1

（1）抛物线y = x2– 1,令y = 0,可得 x2– 1 = 0 => x2 = 1 => x =±1,所以点A(-1,0),点B(1,0),令x = 0,可得y = -1,所以点C(0,-1) ；
（2）因为点B(1,0),点C(0,-1),所以kBC= 1/1 = 1,lBC：y = x – 1,过点A且与CB平行的直线的方程为y = x + 1,与抛物线y = x2– 1联立可得,x + 1 = x2– 1 => x2 – x – 2 = 0 => (x + 1)(x – 2) = 0 => x = -1或者2,所以点P(2,3).因为AP//CB,所以如图所示,四边形ACBP是梯形,而∠PAC = 90°,即AP⊥AC,所以四边形ACBP是直角梯形,S四边形ACBP = (1/2)*(AP + BC)*AC = (1/2)* [√(32 + 32) + √(12 + 12)]*√(12+ 12) = 4 ；
（3）假设线段AP上存在点M,使得CΔMBC 最小,由题意,点C关于直线AP的对称点C’ 与点B的连线C’B与直线AP的交点即为点M.
易求得点C’ (-2,1),这样kC’B= (1 – 0)/(-2 – 1) = -1/3,所以lC’B ：y = (-1/3)(x – 1),与lAP：y = x + 1联立可得(-1/3)(x – 1)= x + 1 => 1 – x = 3x + 3 => 4x = -2 => x = -1/2,即点M(-1/2,1/2) .