抛物线过点A(3,0),B(-1,0),C(0,3)三点,点D是该抛物线的顶点,（1）求解析式（2）求△ACD的面积（3
1个回答
(1)
与x轴交于A, B, 则可表达为y = a(x - 3)(x + 1)
过C: 3 = a(0 -3)(0 + 1), a = -1
y = -x² + 2x + 3
(2)
对称轴x = (3 - 1)/2 = 1
顶点D(1, 4)
AC解析式: x/3 + y/3 = 1, x + y - 3 = 0
AC = √[(3 - 0)² + (0 -3)²] = 3√2
D与AC的距离h = |1 + 4 - 3|/√2 = √2
△ACD的面积 = (1/2)*3√2*√2 = 3
(3)
显然,过P的抛物线的切线与AC平行时, △PAC面积最大
AC解析式: y = 3 - x
设切线y = -x + b
-x + b = -x² + 2x + 3
x² - 3x + b - 3 = 0
∆ = 9 - 4b + 12 = 0, b = 21/4
x = 3/2
P(3/2, 15/4)
P与AC的距离h = |3/2 + 15/4 - 3|/√2 = 21√2/8
S = (1/2)*3√2*21√2/8 = 63/8
(4)
21√2/8
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