试题:
已知动点P(x,y)满足,
x2+y2-4x+6y+13
+
x2+y2+6x+4y+13
=
26
,则
y-1
x-3
取值范围(  )
A.(-∞,
1
2
]∪[4,+∞)
B.(-∞,
1
4
]∪[2+∞)
C.[
1
2
,4]
D.[
1
4
,2]
圆锥曲线综合 2016-05-26

答案:

我来补答
由于动点P(x,y)满足,
x2+y2-4x+6y+13
+
x2+y2+6x+4y+13
=
26
,化为
(x-2)2+(y+3)2
+
(x+3)2+(y+2)2
=
26

设A(2,-3),B(-3,-2),则|AB|=
(-3-2)2+(-2+3)2
=
26

∴动点P(x,y)在相等AB上,
设k=
y-1
x-3
,则k表示动点P(x,y)与M(3,1)连线的斜率.
又kMA=
-3-1
2-3
=4,kMB=
-2-1
-3-3
=
1
2

1
2
≤k≤4

y-1
x-3
∈[
1
2
,4]

故选C.
 
 
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