【答案】
分析:(Ⅰ)由題意得到F
1和F
2的坐標(biāo),設(shè)出P,Q的坐標(biāo),然后直接利用
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/0.png)
進(jìn)行求解;
(Ⅱ)①設(shè)出橢圓標(biāo)準(zhǔn)方程,利用橢圓過點(diǎn)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/1.png)
,結(jié)合a
2=b
2+1 即可求得a
2,b
2的值,則橢圓方程可求;
②當(dāng)直線斜率不存在時(shí),直接求解A,B的坐標(biāo)得到
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/2.png)
的值,當(dāng)直線斜率存在時(shí),設(shè)出直線方程,和橢圓方程聯(lián)立后,利用
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/3.png)
,消掉點(diǎn)的坐標(biāo)得到λ與k的關(guān)系,根據(jù)λ的范圍求k的范圍,然后把
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/4.png)
轉(zhuǎn)化為含有k的函數(shù)式,最后利用基本不等式求出
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/5.png)
的取值范圍.
解答:解:(Ⅰ)如圖,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/images6.png)
由題意得F
2(1,0),F(xiàn)
1(-1,0),設(shè)P(x
,y
),則Q(x
,-y
),
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/6.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/7.png)
.
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/8.png)
,
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/9.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/10.png)
①
又P(x
,y
)在拋物線上,則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/11.png)
②
聯(lián)立①、②得,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/12.png)
,解得:x
=2.
所以點(diǎn)T的橫坐標(biāo)x
=2.
(Ⅱ)(�。┰O(shè)橢圓的半焦距為c,由題意得c=1,
設(shè)橢圓C的標(biāo)準(zhǔn)方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/13.png)
,
因橢圓C過點(diǎn)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/14.png)
,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/15.png)
③
又a
2=b
2+1 ④
將④代入③,解得b
2=1或
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/16.png)
(舍去)
所以a
2=b
2+1=2.
故橢圓C的標(biāo)準(zhǔn)方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/17.png)
.
(ⅱ)1)當(dāng)直線l的斜率不存在時(shí),即λ=-1時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/19.png)
,
又T(2,0),所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/20.png)
;
2)當(dāng)直線l的斜率存在時(shí),即λ∈[-2,-1)時(shí),設(shè)直線l的方程為y=k(x-1).
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/21.png)
,得(1+2k
2)x
2-4k
2x+2k
2-2=0
設(shè)A(x
1,y
1),B(x
2,y
2),顯然y
1≠0,y
2≠0,則由根與系數(shù)的關(guān)系,
可得:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/23.png)
.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/24.png)
⑤
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/25.png)
⑥
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/26.png">,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/27.png)
,且λ<0.
將⑤式平方除以⑥式得:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/28.png)
由λ∈[-2,-1),得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/29.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/30.png)
.
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/31.png)
,解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/32.png)
.
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/33.png">,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/34.png)
,
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/35.png)
,
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/36.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/37.png)
.
令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/38.png)
,因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/39.png">,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/40.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/41.png)
,
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/42.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/43.png)
.
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/44.png)
綜上所述:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124502816258465/SYS201310251245028162584020_DA/45.png)
.
點(diǎn)評(píng):本題考查了橢圓的標(biāo)準(zhǔn)方程,考查了直線與圓錐曲線的關(guān)系,訓(xùn)練了平面向量數(shù)量積的運(yùn)算,考查了分類討論的數(shù)學(xué)解題思想,訓(xùn)練了利用基本不等式求最值,考查了學(xué)生的計(jì)算能力,是難度較大的題目.