9.已知橢圓方程是.橢圓左焦點(diǎn)為F1.O為坐標(biāo)原點(diǎn).A為橢圓上一點(diǎn).M在線(xiàn)段AF1上.且滿(mǎn)足.||=2.則A的橫坐標(biāo)是 查看更多

 

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 已知橢圓方程是,橢圓左焦點(diǎn)為F1,O為坐標(biāo)原點(diǎn),A為橢圓上一點(diǎn),M在線(xiàn)段AF1上,且滿(mǎn)足,||=2,則A的橫坐標(biāo)是(    )

A.          B.            C.          D.

 

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已知橢圓C:
x2
a2
+
y2
b2
=1(a>b>0)
的離心率為
2
2
,其左、右焦點(diǎn)分別為F1、F2,點(diǎn)P是橢圓上一點(diǎn),且
PF1
PF2
=0
,|OP|=1(O為坐標(biāo)原點(diǎn)).
(Ⅰ)求橢圓C的方程;
(Ⅱ)過(guò)點(diǎn)S(0,-
1
3
)
且斜率為k的動(dòng)直線(xiàn)l交
橢圓于A、B兩點(diǎn),在y軸上是否存在定點(diǎn)M,使以AB為直徑的圓恒過(guò)這個(gè)點(diǎn)?若存在,求出M的坐標(biāo),若不存在,說(shuō)明理由.

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已知橢圓C1
x2
a2
+
y2
b2
=1(a>b>0)的左、右焦點(diǎn)分別為F1、F2,其中F2也是拋物線(xiàn)C2:y2=4x的焦點(diǎn),M是C1與C2在第一象限的交點(diǎn),且|MF2|=
5
3

(1)求橢圓C1的方程;
(2)已知菱形ABCD的頂點(diǎn)A,C在橢圓C1上,對(duì)角線(xiàn)BD所在的直線(xiàn)的斜率為1.
①當(dāng)直線(xiàn)BD過(guò)點(diǎn)(0,
1
7
)時(shí),求直線(xiàn)AC的方程;
②當(dāng)∠ABC=60°時(shí),求菱形ABCD面積的最大值.

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已知橢圓
x2
a2
+
y2
b2
=1
上的點(diǎn)P到左、右兩焦點(diǎn)F1、F2的距離之和為2
2
,離心率e=
2
2

(I)求橢圓的方程;
(II)過(guò)右焦點(diǎn)F2且不垂直于坐標(biāo)軸的直線(xiàn)l交橢圓于A,B兩點(diǎn),試問(wèn):險(xiǎn)段OF2上是否存在一點(diǎn)M,使得|MA|=|MB|?請(qǐng)作出并證明.

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已知橢圓C的對(duì)稱(chēng)中心為坐標(biāo)原點(diǎn)O,焦點(diǎn)在x軸上,左右焦點(diǎn)分別為F1,F(xiàn)2,且|F1F2|=2
5
,點(diǎn)(
5
,
4
3
)
在該橢圓上.
(1)求橢圓C的方程;
(2)設(shè)橢圓C上的一點(diǎn)p在第一象限,且滿(mǎn)足PF1⊥PF2,⊙O的方程為x2+y2=4.求點(diǎn)p坐標(biāo),并判斷直線(xiàn)pF2與⊙O的位置關(guān)系;
(3)設(shè)點(diǎn)A為橢圓的左頂點(diǎn),是否存在不同于點(diǎn)A的定點(diǎn)B,對(duì)于⊙O上任意一點(diǎn)M,都有
MB
MA
為常數(shù),若存在,求所有滿(mǎn)足條件的點(diǎn)B的坐標(biāo);若不存在,說(shuō)明理由.

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一、選擇題

1.D  2.A  3.C  4.D  5.B  6.C  7.D  8.B  9.A  10.A

二、填空題

11.148  12.-4  13.  14.-6  15.①②③④

三、解答題

16.解:⑴

                                                                                                                 3分

=1+1+2cos2x

=2+2cos2x

=4cos2x

∵x∈[0,]  ∴cosx≥0

=2cosx                                                                                                    6分

⑵ f (x)=cos2x-?2cosx?sinx

      =cos2x-sin2x

      =2cos(2x+)                                                                                           8分

∵0≤x≤  ∴

  ∴

,當(dāng)x=時(shí)取得該最小值

 ,當(dāng)x=0時(shí)取得該最大值                                                                  12分

17.由題意知,在甲盒中放一球概率為,在乙盒放一球的概率為                    3分

①當(dāng)n=3時(shí),x=3,y=0的概率為                                              6分

②|x-y|=2時(shí),有x=3,y=1或x=1,y=3

它的概率為                                                                12分

18.解:⑴證明:在正方形ABCD中,AB⊥BC

又∵PB⊥BC  ∴BC⊥面PAB  ∴BC⊥PA

同理CD⊥PA  ∴PA⊥面ABCD    4分

⑵在AD上取一點(diǎn)O使AO=AD,連接E,O,

則EO∥PA,∴EO⊥面ABCD 過(guò)點(diǎn)O做

OH⊥AC交AC于H點(diǎn),連接EH,則EH⊥AC,

從而∠EHO為二面角E-AC-D的平面角                                                             6分

在△PAD中,EO=AP=在△AHO中∠HAO=45°,

∴HO=AOsin45°=,∴tan∠EHO=,

∴二面角E-AC-D等于arctan                                                                   8分

⑶當(dāng)F為BC中點(diǎn)時(shí),PF∥面EAC,理由如下:

∵AD∥2FC,∴,又由已知有,∴PF∥ES

∵PF面EAC,EC面EAC  ∴PF∥面EAC,

即當(dāng)F為BC中點(diǎn)時(shí),PF∥面EAC                                                                         12分

19.⑴f '(x)=3x2+2bx+c,由題知f '(1)=03+2b+c=0,

f (1)=-11+b+c+2=-1

∴b=1,c=-5                                                                                                    3分

f (x)=x3+x2-5x+2,f '(x)=3x2+2x-5

f (x)在[-,1]為減函數(shù),f (x)在(1,+∞)為增函數(shù)

∴b=1,c=-5符合題意                                                                                      5分

⑵即方程:恰有三個(gè)不同的實(shí)解:

x3+x2-5x+2=k(x≠0)

即當(dāng)x≠0時(shí),f (x)的圖象與直線(xiàn)y=k恰有三個(gè)不同的交點(diǎn),

由⑴知f (x)在為增函數(shù),

f (x)在為減函數(shù),f (x)在(1,+∞)為增函數(shù),

,f (1)=-1,f (2)=2

且k≠2                                                                                               12分

20.⑴∵

                                                                                         3分

∴{an-3n}是以首項(xiàng)為a1-3=2,公比為-2的等比數(shù)列

∴an-3n=2?(-2)n1

∴an=3n+2?(-2)n1=3n-(-2)n                                                                        6分

⑵由3nbn=n?(3n-an)=n?[3n-3n+(-2)n]=n?(-2)n

∴bn=n?(-)n                                                                                                    8分

<6

∴m≥6                                                                                                                   13分

21.⑴設(shè)M(x0,y0),則N(x0,-y0),P(x,y)

AM:y=   ①

BN:y=  、

聯(lián)立①②  ∴                                                                                      4分

∵點(diǎn)M(xo,yo)在圓⊙O上,代入圓的方程:

整理:y2=-2(x+1)  (x<-1)                                                                             6分

⑵由

設(shè)S(x1、y1),T(x2、y2),ST的中點(diǎn)坐標(biāo)(x0、y0)

則x1+x2=-(3+)

x1x2                                                                                                          8分

中點(diǎn)到直線(xiàn)的距離

故圓與x=-總相切.                                                                                        14分

⑵另解:∵y2=-2(x+1)知焦點(diǎn)坐標(biāo)為(-,0)                                                  2分

頂點(diǎn)(-1,0),故準(zhǔn)線(xiàn)x=-                                                                              4分

設(shè)S、T到準(zhǔn)線(xiàn)的距離為d1,d2,ST的中點(diǎn)O',O'到x=-的距離為

又由拋物線(xiàn)定義:d1+d2=|ST|,∴

故以ST為直徑的圓與x=-總相切                                                                      8分

 


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