题目内容
已知点A(-1,0),点B (1,0),点P(x+1,y)在x轴的下方,设a=
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(1)求a、b、c关于x、y的表达式;
(2)求y关于x的函数关系式y=f(x),并求当y取得最小值时P点的坐标.
【答案】分析:(1)因为
=(-x-2,-y),
=(-x,-y),
=(x+2,y),
=(2,0),
=(x,y),
=(-2,0),由此能求出a、b、c关于x、y的表达式.
(2)因为
=0,所以3x2+y2+6x=0,由于点P(x+1,y)在x轴的下方,所以y=-
,(-2<x<0),y=-
=-
,(-2<x<0).由此能求出当y取得最小值时P点的坐标.
解答:解:(1)因为
=(-x-2,-y),
=(-x,-y),
所以a=
=x2+y2+2x,…(2分)
=(x+2,y),
=(2,0),b=
=2x+4,…(3分)
=(x,y),
=(-2,0),c=
=-2x,…(4分)
d=
=2,…(5分)
(2)因为
=0,所以2(x2+y2+2x)-(2x+4)(-2x)=0,即:3x2+y2+6x=0,…(7分)
由于点P(x+1,y)在x轴的下方,所以y=-
,(-2<x<0)
y=-
=-
,(-2<x<0)…(10分)
所以当x=-1时,ymin=-
,此时P(0,-
)…(12分)
点评:本题考查矩阵现向量乘法的意义和应用,解题时要认真审题,注意平面向量的数量积公式的灵活运用.
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(2)因为
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解答:解:(1)因为
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所以a=
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d=
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(2)因为
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由于点P(x+1,y)在x轴的下方,所以y=-
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y=-
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所以当x=-1时,ymin=-
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点评:本题考查矩阵现向量乘法的意义和应用,解题时要认真审题,注意平面向量的数量积公式的灵活运用.
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