4p03004p03
因0p1所以Eξ1Eξ2时p的取值范围是0p03【点评】本小题考查二项分布分布列数学期望方差等基础知识考查同学们运用概率知识解决实际问题的能力20本小题满分14分已知点Ax1y1Bx2y2x1x2≠0是抛物线y22pxp0上的两个动点O是坐标原点向量OAOB满足OAOBOAOB设圆C的方程为
x2y2x1x2xy1y2y0
I证明线段AB是圆C的直径
fII当圆C的圆心到直线X2Y0的距离的最小值为
25时求p的值5
22
【解析】I证明1∵OAOBOAOB∴OAOBOAOB
OA2OAOBOBOA2OAOBOB
整理得OAOB0
2
2
2
2
∴x1x2y1y20
设Mxy是以线段AB为直径的圆上的任意一点则MAMB0即xx1xx2yy1yy20整理得xyx1x2xy1y2y0
22
故线段AB是圆C的直径证明2∵OAOBOAOB∴OAOBOAOB
22
OA2OAOBOBOA2OAOBOB
整理得OAOB0
2
2
2
2
∴x1x2y1y20……1
设xy是以线段AB为直径的圆上则即
yy2yy11x≠x1x≠x2xx2xx1
去分母得xx1xx2yy1yy20点x1y1x1y2x2y1x2y2满足上方程展开并将1代入得
x2y2x1x2xy1y2y0
故线段AB是圆C的直径证明3∵OAOBOAOB∴OAOBOAOB
22
OA2OAOBOBOA2OAOBOB
整理得OAOB0
2
2
2
2
∴x1x2y1y20……1
f以线段AB为直径的圆的方程为
x
x1x22yy221y1x1x22y1y22224
展开并将1代入得
x2y2x1x2xy1y2y0
故线段AB是圆C的直径II解法1设圆C的圆心为Cxy则
xxx122yy1y22
∵y122px1y222px2p0
y12y22∴x1x24p2
又因x1x2y1y20
∴x1x2y1y2
∴y1y2
y12y224p2
∵x1x2≠0∴y1y2≠0
∴y1y24p2xx1x211yyy12y22y12y222y1y21224p4p4p

12y2p2p
所以圆心的轨迹方程为y2px2p2设圆心C到直线x2y0的距离为d则
12y2p22yx2yy22py2p2pd555pyp2p25p
f当yp时d有最小值
pp25由题设得555
∴p2
解法2设圆C的圆心为Cxy则
x1x2x2yy1y22
∵y122px1y222px2p0
∴x1x2
y12y224p2
又因x1x2y1y20
∴x1x2y1y2
∴y1y2
y12y224p2
∵x1x2≠0∴y1y2≠0
∴y1y24p2xx1x211yyy12y22y12y222y1y21224p4p4p

12y2p2p
所以圆心的轨迹方程为y2px2p2
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