早教吧作业答案频道 -->数学-->
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
题目详情
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~!
▼优质解答
答案和解析
(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)
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)
看了 如图,抛物线y=ax2+bx...的网友还看了以下:
已知抛物线y=x^2-4x+3与x轴交于点AB(A左B右)与y轴交于C点P是抛物线对称轴上一点,且 2020-05-16 …
在平面直角坐标系中,已知抛物线经过点A(0,4)B(1,0)C(5,0)抛物线对称轴与X轴交于M. 2020-05-16 …
若抛物线y=x^2-2x+c与y轴的交点坐标为(0,-3),则说法不正确的是A抛物线开口向上B抛物 2020-06-12 …
若抛物线y=x^2-2x+c与y轴的交点坐标为(0,-3),则说法不正确的是A抛物线开口向上B抛物 2020-06-12 …
如图,在平面直角坐标系xoy中,已知抛物线经过点A(0,4),B(1,0),C(5,0),抛物线对 2020-06-14 …
如图,在平面直角坐标系xoy中,已知抛物线经过点A(0,4),B(1,0),C(5,0),抛物线对 2020-06-14 …
已知在直角坐标系中,抛物线y=ax2-8ax+3(a<0)与y轴交于点A,顶点为D,其对称轴交x轴 2020-07-20 …
(2005•福州)已知:抛物线y=x2-2x-m(m>0)与y轴交于点C,C点关于抛物线对称轴的对 2020-07-21 …
(2014•北京)在平面直角坐标系xOy中,抛物线y=2x2+mx+n经过点A(0,-2),B(3 2020-07-29 …
把抛物线y=-x2+x沿x轴向右平移1个单位后,再沿x轴翻折得到抛物线C1称为第一次操作,把抛物线C 2020-12-15 …