ji0330
春芽
共回答了15个问题采纳率:86.7% 举报
(1) 抛物线过A,B,可表达为y = m(x + 3)(x - 1)
x = 0,y = -3m,D(0,-3m)
(2)抛物线对称轴 x = (-3+ 1)/2 = -1
x = -1,y = -4m
P(-1,-4m)
C(-1,0),圆半径r = [1 - (-3)]/2 = 2
PD的斜率:(-3m+ 4m)/(0 + 1) = m,方程:y = mx - 3m,mx - y - 3m = 0
C与PD的距离d = |-m - 0 - 3m|/√(m² + 1) = 4m/√(m² + 1) = 2
相切,d = r,m = √3/3 (舍去m = -√3/3 < 0)
不用点与直线的距离也行.先求出的斜率;然后可以求出过的垂线的斜率;然后通过二者的方程得到交点,以及三边的长.用勾股定理即可.
(3)
m = 1,P(-1,-4),D(0,-3),B(-3,0)
BP² = 20
PD² = 2
BD² = 18
BP² = PD² + BD²
BPD为直角三角形,∠BDP为直角
tan∠BPD = BD/PD = √18/√2 = 3
1年前
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