1、
y = (x-2m)²+(m-3),由于x²项的系数为正数,抛物线开口向上
当x = 2m,y = m-3
当顶点在x轴上,y = 0,0 = m-3,m = 3
该方程是y = (x-6)²
当顶点在y轴上,x = 0,m = 0
该方程是y = x²-3
当图像经过原点(0,0),则
0 = (0-2m)²+(m-3)
4m²+m-3 = 0
m = -1 or m = 3/4
该方程是y = (x+2)²-4
2、
y = x²-2x-3 = (x-1)²-4,对称轴为x = 1,设C点的坐标为(1,k)
解x²-2x-3 = 0得交点A(-1,0),B(3,0)
AB中点M为(1,0),则建立ΔBCM,面积为1/2*ΔABC = 5
BM = 2,CM = k
1/2*BM*CM = 5
2*k = 10
k = 5
∴C的坐标为(1,5)
3、(1)：
直线经过y轴,所以当x = 0,y = 3
直线经过x轴,所以当y = 0,x = 3
所以直线经过的两点为：C(0,3)、B(3,0)
而抛物线亦经过B、C两点,所以将B、C的坐标分别代入,得：
代入C点：3 = 0 + 0 + c,c = 3
代入B点：0 = -3² + 3b + (3),b = 2
∴抛物线的解析式是y = -x² + 2x + 3
(2)：由图像可见,P和C是关于x = 1对称的,所以P的坐标是(2,3)
(3)：经过AP的直线方程与直线y = -x+3垂直,斜率为-1/(-1) = 1
该方程为y - 3 = (1)(x - 2),i.e.y = x+1
两直线方程的相交点为N(1,2),PN⊥BN
考虑ΔBPN,PN = √[(2-1)²+(3-2)²] = √2,BN = √[(3-1)²+(0-2)²] = 2√2
所以tan∠PBC = tan∠PBN = PN/BN = √2/(2√2) = 1/2