试题:
有两点M(-1,0),N(1,0),点P(x,y)使
MP
MN
PM
PN
NM
NP
成公差小于零的等差数列;
1)求x,y满足的关系式;2)若P横坐标x0=
2
,记 θ为
PM
PN
夹角,求tanθ值.

答案:

我来补答
(1)由已知,得到:
MP
=(x+1,y)  
MN
=(2,0)
NP
=(x-1,y)
 
PM
=-
MP
=(-x-1,-y),
PN
=-
NP
=(-x+1,-y),
MP
MN
=2x+2,
PM
PN
=x2-1+y2, 
NM
NP
=-2x+2

MP
MN
PM
PN
NM
NP
成公差小于零的等差数列,
∴2x+2+(-2x+2)=2(x2-1+y2),且2x+2>-2x+2 整理得出x2+y2=3(x>0).
(2)
PM
=-
MP
=(-x-1,-y),
PN
=-
NP
=(-x+1,-y),
PM
PN
=x2+y2-1=2,
|PM|
•|
PN|
=
(-x+1)2+(-y)2
(-x-1)2+(-y)2
=
x2+y2+2x+1 
x2+y2-2x+1
4+2x
4-2x
=
16-4x2

若P横坐标x0=
2
,则
|PM|
•|
PN|
=
16-8
=2
2
,cos<
PM
PN
>=
PM
PN
|
PM|
 |
PN
|
=
2
2

θ=45°tanθ=1.
 
 
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