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圆C的方程为(x-2)2+y2=4,圆M的方程为(x-2-5sinθ)2+(y-5cosθ)2=1(θ∈R),过圆C上任意一点P作圆M的两条切线PE、PF,切点分别为E、F,则PE•PF的最小值是()A.6B.569C.7D.659
题目详情
圆C的方程为(x-2)2+y2=4,圆M的方程为(x-2-5sinθ)2+(y-5cosθ)2=1(θ∈R),过圆C上任意一点P作圆M的两条切线PE、PF,切点分别为E、F,则
•
的最小值是( )
A.6
B.
C.7
D.
2222
•
的最小值是( )
A.6
B.
C.7
D.
PE PE
PF PF
C.7
D.
56 56 9 9
65 65 9 9
PE |
PF |
A.6
B.
56 |
9 |
C.7
D.
65 |
9 |
PE |
PF |
A.6
B.
56 |
9 |
C.7
D.
65 |
9 |
PE |
PF |
56 |
9 |
C.7
D.
65 |
9 |
56 |
9 |
65 |
9 |
65 |
9 |
▼优质解答
答案和解析
(x-2)22+y22=4的圆心C(2,0),半径等于2,圆M (x-2-5sinθ)22+(y-5cosθ)22=1,
圆心M(2+5sinθ,5cosθ),半径等于1.∵|CM|=
=5>2+1,故两圆相离.
∵
•
=|
|•
•cos∠EPF,要使
•
最小,需|
| 和
最小,且∠EPF 最大,
如图所示,设直线CM 和圆C 交于H、G两点,则
•
的最小值是
•
.
|H M|=|CM|-2=5-2=3,|H E|=
=
=2
,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
(5sinθ)2+(5cosθ)2 (5sinθ)2+(5cosθ)2 (5sinθ)2+(5cosθ)22+(5cosθ)22=5>2+1,故两圆相离.
∵
•
=|
|•
•cos∠EPF,要使
•
最小,需|
| 和
最小,且∠EPF 最大,
如图所示,设直线CM 和圆C 交于H、G两点,则
•
的最小值是
•
.
|H M|=|CM|-2=5-2=3,|H E|=
=
=2
,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
PE PE PE•
PF PF PF=|
|•
•cos∠EPF,要使
•
最小,需|
| 和
最小,且∠EPF 最大,
如图所示,设直线CM 和圆C 交于H、G两点,则
•
的最小值是
•
.
|H M|=|CM|-2=5-2=3,|H E|=
=
=2
,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
|
PE PE PE|•
|PF| |PF| |PF|•cos∠EPF,要使
•
最小,需|
| 和
最小,且∠EPF 最大,
如图所示,设直线CM 和圆C 交于H、G两点,则
•
的最小值是
•
.
|H M|=|CM|-2=5-2=3,|H E|=
=
=2
,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
PE PE PE•
PF PF PF 最小,需|
| 和
最小,且∠EPF 最大,
如图所示,设直线CM 和圆C 交于H、G两点,则
•
的最小值是
•
.
|H M|=|CM|-2=5-2=3,|H E|=
=
=2
,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
|
PE PE PE| 和
|PF| |PF| |PF| 最小,且∠EPF 最大,
如图所示,设直线CM 和圆C 交于H、G两点,则
•
的最小值是
•
.
|H M|=|CM|-2=5-2=3,|H E|=
=
=2
,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
PE PE PE•
PF PF PF的最小值是
•
.
|H M|=|CM|-2=5-2=3,|H E|=
=
=2
,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
HE HE HE•
HF HF HF.
|H M|=|CM|-2=5-2=3,|H E|=
=
=2
,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
|HM|2−|ME|2 |HM|2−|ME|2 |HM|2−|ME|22−|ME|22=
=2
,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
9−1 9−1 9−1=2
,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
2 2 2,sin∠MHE=
=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
|ME| |ME| |ME||MH| |MH| |MH|=
,
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
1 1 13 3 3,
∴cos∠EHF=cos2∠MHE=1-2sin22∠MHE=
,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
7 7 79 9 9,
∴
•
=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
HE HE HE•
HF HF HF=|H E|•|H E|•cos∠EHF=2
×2
×
=
,故选 B.
2 2 2×2
×
=
,故选 B.
2 2 2×
=
,故选 B.
7 7 79 9 9=
,故选 B.
56 56 569 9 9,故选 B.
圆心M(2+5sinθ,5cosθ),半径等于1.∵|CM|=
(5sinθ)2+(5cosθ)2 |
∵
PE |
PF |
PE |
|PF| |
PE |
PF |
PE |
|PF| |
如图所示,设直线CM 和圆C 交于H、G两点,则
PE |
PF |
HE |
HF |
|H M|=|CM|-2=5-2=3,|H E|=
|HM|2−|ME|2 |
9−1 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
(5sinθ)2+(5cosθ)2 |
∵
PE |
PF |
PE |
|PF| |
PE |
PF |
PE |
|PF| |
如图所示,设直线CM 和圆C 交于H、G两点,则
PE |
PF |
HE |
HF |
|H M|=|CM|-2=5-2=3,|H E|=
|HM|2−|ME|2 |
9−1 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
PE |
PF |
PE |
|PF| |
PE |
PF |
PE |
|PF| |
如图所示,设直线CM 和圆C 交于H、G两点,则
PE |
PF |
HE |
HF |
|H M|=|CM|-2=5-2=3,|H E|=
|HM|2−|ME|2 |
9−1 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
|
PE |
|PF| |
PE |
PF |
PE |
|PF| |
如图所示,设直线CM 和圆C 交于H、G两点,则
PE |
PF |
HE |
HF |
|H M|=|CM|-2=5-2=3,|H E|=
|HM|2−|ME|2 |
9−1 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
PE |
PF |
PE |
|PF| |
如图所示,设直线CM 和圆C 交于H、G两点,则
PE |
PF |
HE |
HF |
|H M|=|CM|-2=5-2=3,|H E|=
|HM|2−|ME|2 |
9−1 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
|
PE |
|PF| |
如图所示,设直线CM 和圆C 交于H、G两点,则
PE |
PF |
HE |
HF |
|H M|=|CM|-2=5-2=3,|H E|=
|HM|2−|ME|2 |
9−1 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
PE |
PF |
HE |
HF |
|H M|=|CM|-2=5-2=3,|H E|=
|HM|2−|ME|2 |
9−1 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
HE |
HF |
|H M|=|CM|-2=5-2=3,|H E|=
|HM|2−|ME|2 |
9−1 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
|HM|2−|ME|2 |
9−1 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
9−1 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
2 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
|ME| |
|MH| |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin2∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
1 |
3 |
∴cos∠EHF=cos2∠MHE=1-2sin22∠MHE=
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
7 |
9 |
∴
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
HE |
HF |
2 |
2 |
7 |
9 |
56 |
9 |
2 |
2 |
7 |
9 |
56 |
9 |
2 |
7 |
9 |
56 |
9 |
7 |
9 |
56 |
9 |
56 |
9 |
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