如图,抛物线y=ax2+bx+c（a＜0）与x轴相交于A、B两点,与y轴的正半轴相交于点C,对称轴l与x轴的正半轴相交于点D,与抛物线相交于点F,点C关于直线l的对称点为E． （1）当a=-2,b=4,c=2时,判断四边形CDEF

Question

如图,抛物线y=ax2+bx+c（a＜0）与x轴相交于A、B两点,与y轴的正半轴相交于点C,对称轴l与x轴的正半轴相交于点D,与抛物线相交于点F,点C关于直线l的对称点为E． （1）当a=-2,b=4,c=2时,判断四边形CDEF的形状,并说明理由； （2）若四边形CDEF是正方形,且AB= 2 ,求抛物线的解析式． 这道题的第一问利用菱形的对角线互相垂直证出此图形为菱形?请您给我一个过程好吗?谢谢,急~!

Answer 1

(1)
y = -2x² + 4x + 2 = -2(x - 1)² + 4
C(0,2),F(1,4),E(2,2),D(1,0)
CD² = (0 - 1)² + (2 - 0)² = 5
DE² = (2 - 1)² + (2 - 0)² = 5
EF² = (1 - 2)² + (4 - 2)² = 5
CF² = (0 - 1)² + (2 - 4)² = 5
四边形CDEF为菱形
另法：CE(y = 2)与x轴平行,DF ( x = 1)与x轴垂直,CE与DF垂直; 交点G(1,2)等分CE,DF,四边形CDEF为菱形
(2)
设对称轴x = d,D(d,0)
AB = 2:A(d -1,0),B(d + 1,0)
y = a[x - (d - 1)][x - (d + 1)]
x = 0,y = a(d² -1)
C(0,a(d² -1))
E(2d,a(d² -1))
x = d,y = -a,F(d,-a)
CD² = DE² = d² + a²(d² - 1)²
EF² = CF² = d² + (ad²)²
四边形CDEF是正方形:CD² = CF²,d² + a²(d² - 1)² = d² + (ad²)²
(d² - 1)² = d⁴
d² - 1 = d² (无解)
或1 - d² = d²,d² = 1/2,d = ±1/√2
CD的斜率p = a(d² - 1)/(0 - d) = a(1 - d²)/d
CF的斜率q = (ad² - a + a)/(0 - d) = -ad
pq = -1,a²(1 - d²) = 1
a² = 1/(1 - d²) = 1/(1 - 1/2) = 2
a = -√2 (舍去a = √2 > 0)
抛物线的解析式 (请自己简化):
y = -√2(x - 1/√2 + 1)(x - 1/√2 -1)
或
y = -√2(x + 1/√2 + 1)(x +1/√2 -1)