Question

已知抛物线y=x2-x-2交X轴点A（-1,0）交Y轴点C（0,-2）在对称轴上是否存在点P使△PAC为直角三角形,求点

Answer 1

y = (x + 1)(x - 2)

对称轴x = (-1 + 2)/2 = 1/2

P(1/2,p)

AC² = (-1 - 0)² + (0 + 2)² = 5

AP² = (-1 - 1/2)² + (0 - p)² = p² + 9/4

PC² = (1/2 - 0)² + (p + 2)² = p² + 4p + 17/4

(1) AC² + AP² = PC²

5 + p² + 9/4 = p² + 4p + 17/4

p = 3/4,P(1/2,3/4)

(2) AP² + PC² = AC²

p² + 9/4 + p² + 4p + 17/4 = 5

4p² + 8p + 3 = 0

(2p + 3)(2p + 1) = 0

p = -3/2,P(1/2,-3/2)

p = -1/2,P(1/2,-1/2)

(3)AC² + PC² = AP²

5 + p² + 4p + 17/4 = p² + 9/4

p = -7/4

P(1/2,-7/4)