已知:抛物线y=ax2-2ax+c-1的顶点A在一次函数y=-8/3x+8的图像上,该抛物线与x轴交于B,C两点（B在C
已知:抛物线y=ax2-2ax+c-1的顶点A在一次函数y=-8/3x+8的图像上,该抛物线与x轴交于B,C两点（B在C的左侧）,且过点D(0,4)
1.
2.设H为线段OC上一点,过点H作HK//BD,交AC于K,若三角形HKC的面积等于16/5,求直线的解析式
3.在（2）的基础上抛物线上是否存在一点P,过点P作PQ垂直X轴,交直线HK于Q,使A,H,P,Q点为等腰梯形的四个顶点,若存在,求P点的坐标;若不存在,说明理由
1.
2.设H为线段OC上一点,过点H作HK//BD,交AC于K,若三角形HKC的面积等于16/5,求直线的解析式
3.在（2）的基础上抛物线上是否存在一点P,过点P作PQ垂直X轴,交直线HK于Q,使A,H,P,Q点为等腰梯形的四个顶点,若存在,求P点的坐标;若不存在,说明理由
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y = a(x-1)^2 - a + c - 1,vertex is (1,-a+c-1)
when x = 1,y = -8/3 x + 8 = 8 - 8/3 = 16/3
therefore -a+c-1 = 16/3
when x = 0,y = c-1 = 4
therefore c = 5
a = 4 - 16/3 = -4/3
1.y = -4/3 (x-1)^2 + 16/3
2.B (-1,0),C(3,0)
BD slope = 4
let H(x,0),K's coordinate is x+dx,y
then y = 4dx
triangle's area HKC = (3-x)4dx /2 = 2(3-x)dx = 16/5 => dx = 8/[5(3-x)]
4dx = -4/3(x+dx-1)^2 + 16/3
32 /(5(3-x)) + 4/3(x + 8/(5(3-x))-1)^2 - 16/3 = 0
solve for x and then get equation
H' coordinate is (1.23,0)
3.
when x = 1,y = -8/3 x + 8 = 8 - 8/3 = 16/3
therefore -a+c-1 = 16/3
when x = 0,y = c-1 = 4
therefore c = 5
a = 4 - 16/3 = -4/3
1.y = -4/3 (x-1)^2 + 16/3
2.B (-1,0),C(3,0)
BD slope = 4
let H(x,0),K's coordinate is x+dx,y
then y = 4dx
triangle's area HKC = (3-x)4dx /2 = 2(3-x)dx = 16/5 => dx = 8/[5(3-x)]
4dx = -4/3(x+dx-1)^2 + 16/3
32 /(5(3-x)) + 4/3(x + 8/(5(3-x))-1)^2 - 16/3 = 0
solve for x and then get equation
H' coordinate is (1.23,0)
3.
1年前
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