题目内容
已知动点P的轨迹方程为:
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-
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=1(x>2),O是坐标原点.
①若直线x-my-3=0截动点P的轨迹所得弦长为5,求实数m的值;
②设过P的轨迹上的点P的直线与该双曲线的两渐近线分别交于点P
1、P
2,且点P分有向线段
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所成的比为λ(λ>0),当λ∈[
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,
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]时,求|
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|•|
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|的最值.
【答案】
分析:①先确定直线与双曲线的右支相交,设两个交点坐标分别为D(x
D,y
D)、E(x
E,y
E),由双曲线的第二定义,求出|DF|、|EF|,从而可得|DE|,利用直线x-my-3=0截动点P的轨迹所得弦长为5,即可求得m的值;
②先确定P的坐标,进而可表示|
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|•|
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|,利用基本不等式及端点的函数值,即可求得|
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|•|
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|的最值.
解答:解:①由动点P的轨迹方程为:
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-
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=1(x>2),∴直线x-my-3=0恒过双曲线的右焦点F(3,0),于是直线与双曲线的右支相交,
设两个交点坐标分别为D(x
D,y
D)、E(x
E,y
E),
由双曲线的第二定义得
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,∴|DF|=ex
D-a
同理|EF|=ex
E-a,∴|DE|=e(x
D+x
E)-2a
∵a=2,c=3,∴e=
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,∴|DE|=
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(x
D+x
E)-4
∵若直线x-my-3=0截动点P的轨迹所得弦长为5
∴
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(x
D+x
E)-4=5
∴x
D+x
E=6
由直线过右焦点F(3,0),知x
D=x
E=3,此时直线垂直于x轴,∴m=0.
②设P(x,y),P
1(x
1,y
1),P
2(x
2,y
2),则
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∴x=
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,y=
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=
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∵点P(x,y)在双曲线:
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-
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=1上
∴
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-
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=1,化简可得
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∵
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=
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,
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=
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∴
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=
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令u=
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=λ
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+2
∵λ∈[
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,
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],∴λ=1时,λ
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+2取得最小值4
∵λ=
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时,u=
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,λ=
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时,u=
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,∴λ
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+2的最大值为
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∴|
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|•|
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|的最小值为9,最大值为
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.
点评:本题考查直线与双曲线的位置关系的综合应用,考查化归与转化、分类与整合的数学思想,培养学生的抽象概括能力、推理论证能力、运算求解能力和创新意识.
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