题目内容
设P是椭圆
A.9,12
B.8,11
C.8,12
D.10,12
【答案】分析:首先将P点固定于一处,设两圆心分别为C1,C2,则r1=1,r2=c且C1,C2为椭圆的焦点,PC1≤PM+MC1,PC2≤PN+NC2,PM+PN=PM+MC1+PN+NC2-(MC1+NC2)≥PC1+PC2-(MC1+NC2)=8,所以PM+PN的最小值为8.PM+PN=PM+MC1+PN+NC2-(MC1+NC2)≤PC1+PC2+(MC1+NC2)=12.所以PM+PN的最大值为12.
解答:解:首先将P点固定于一处,设两圆心分别为C1,C2,
则r1=1,r2=c且C1,C2为椭圆的焦点,
PC1≤PM+MC1
PC2≤PN+NC2
PM+PN=PM+MC1+PN+NC2-(MC1+NC2)≥PC1+PC2-(MC1+NC2)
=2a-(r1+r2)
=10-2=8
所以,PM+PN的最小值为8.
PM+PN=PM+MC1+PN+NC2-(MC1+NC2)≤PC1+PC2+(MC1+NC2)
=2a+(r1+r2)
=10+2=12.
所以,PM+PN的最大值为12.
故选C.
点评:本题考查圆的性质和应用,解题时要注意椭圆的性质和应用.
解答:解:首先将P点固定于一处,设两圆心分别为C1,C2,
则r1=1,r2=c且C1,C2为椭圆的焦点,
PC1≤PM+MC1
PC2≤PN+NC2
PM+PN=PM+MC1+PN+NC2-(MC1+NC2)≥PC1+PC2-(MC1+NC2)
=2a-(r1+r2)
=10-2=8
所以,PM+PN的最小值为8.
PM+PN=PM+MC1+PN+NC2-(MC1+NC2)≤PC1+PC2+(MC1+NC2)
=2a+(r1+r2)
=10+2=12.
所以,PM+PN的最大值为12.
故选C.
点评:本题考查圆的性质和应用,解题时要注意椭圆的性质和应用.

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