如图,已知在平面直角坐标系xoy中,抛物线y=1⼀4x눀+bx+c与x轴交于点A、B(点A在点B右侧)与y轴交于点c

2025-03-10 04:35:52
推荐回答(1个)
回答1:

∵y=1/4x²+bx+c过点C(0,-3)
∴0+0+c=-3
∴c=-3
∴y=1/4x²+bx-3

OC=3
OA=2OC=6
抛物线与x轴交于点A、B(点A在点B右侧),C点在y轴上,∴A在x正半轴上
∴A(6,0)
将(6,0)代入y=1/4x²+bx-3:
1/4*6²+6b-3=0
b=-1
y=1/4x²-x-3 = 1/4(x-2)²-4
顶点M(2,-4)

MA²=(2-6)²+(-4-0)²=32
MC²=(2-0)²+(-4+3)²=5
AC²=(6-0)²+(0+3)²=45
cos∠MAC=(MA²+AC²-MC²)/(2MA*AC) = (32+45-5)/{2*√32*√45) = 3/√15>0
tan∠MAC=√(1-cos²∠MAC)/cos∠MAC = √(1-9/15)/(3/√15) = √6/3

D点在对称轴x=2上,令D(2,t)
AD²=(2-6)²+(t-0)²=16+t²
AC²=45
CD²=(2-0)²+(t+3)²=t²+6t+13
CD² = AD²+AC²-2AD*AC*cos∠CAD = AD²+AC²-2AD*AC*cos45°
t²+6t+13 = 16+t² + 45 - 2*√45*√(16+t²)*√2/2
√10*√(16+t²) = 16-2t

160+10t² = 4t²-64t+256
3t²+32t-48=0
(t+12)(3t-4)=0
t1=-12,t2=4/3
D(2,-12),或D(2,4/3)