Question

如图,二次函数y=ax^2 bx c的图象交x轴于点A(-1,0)B()2,0

第三问谢谢高手们

Answer 1

(1) 抛物线与x轴交于A(-1, 0), (2, 0),可表达为y = a(x + 1)(x - 2)
x = 0, y = -2a = -2, a = 1
y = (x + 1)(x - 2) = x² - x - 2

(2) 设P(p,0)
PA = |p + 1|
PC = √[(p - 0)² + (0 + 2)²]
p = 3/2

(3) AC的方程: x/(-1) + y/(-2) = 1, y = -2x - 2

设M(m, m² - m - 2), H(h, -2h - 2), m > 0
AC斜率为k = (-2 - 0)/(0 + 1) = -2
HM⊥AC, MH斜率为-1/k = 1/2 = (m² - m - 2 + 2h + 2)/(m - h)
5h = 3m - 2m² (i)
两三角形相似, CH/CM = AO/AC = 1/√5
5CH² = CM²
左边 = 5[(h - 0)² + (-2h - 2 + 2)²] = 25h² = (5h)² = (3m - 2m²)²

右边 = (m - 0)² + (m² -m - 2 + 2)² = m²(m² - 2m + 2)
(3m - 2m²)² = m²(m² - 2m + 2)
3m² - 10m + 7 = 0
(m - 1)(3m - 7) = 0
m = 1或m = 7/3
M(1, -2)或M(7/3, 10/9)

AC的方程: 2x + y + 2 = 0
M与AC的距离d = r = 4√5/5 = |2m + m² - m - 2 + 2|/√(2²+1²) = |m² + m|/√5
m² + m + 4 = 0 (无解)
m² + m - 4 = 0
m = (-1 ± √17)/2
M((-1 + √17)/2, 3 - √17)或M((-1 - √17)/2, 3 + √17)

Answer 2

AC的方程可求，
二次函数的解析式可求，
第三问也就是在抛物线上求到AC的距离为半径的点的坐标。

Answer 3

这应该是高中题了吧，，具体过程就不写了
∠OAC=∠HCM 利用tan的和差化积公式可求出tan∠HCM进而求出CM斜率，然后用交轨法求M
第二问，用距离公式应该可以算出来吧，实在不行，我明天再算