【答案】
分析:(Ⅰ)設(shè)動(dòng)點(diǎn)為M,其坐標(biāo)為(x,y),求出直線A
1、MA
2M的斜率,并且求出它們的積,即可求出點(diǎn)M軌跡方程,根據(jù)圓、橢圓、雙曲線的標(biāo)準(zhǔn)方程的形式,對(duì)m進(jìn)行討論,確定曲線的形狀;(Ⅱ)由(I)知,當(dāng)m=-1時(shí),C
1方程為x
2+y
2=a
2,當(dāng)m∈(-1,0)∪(0,+∞)時(shí),C
2的焦點(diǎn)分別為F
1(-a
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/0.png)
,0),F(xiàn)
2(a
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/1.png)
,0),假設(shè)在C
1上存在點(diǎn)N(x
,y
)(y
≠0),使得△F
1NF
2的面積S=|m|a
2,的充要條件為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/2.png)
,求出點(diǎn)N的坐標(biāo),利用數(shù)量積和三角形面積公式可以求得tanF
1NF
2的值.
解答:解:(Ⅰ)設(shè)動(dòng)點(diǎn)為M,其坐標(biāo)為(x,y),
當(dāng)x≠±a時(shí),由條件可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/3.png)
,
即mx
2-y
2=ma
2(x≠±a),
又A
1(-a,0),A
2(a,0)的坐標(biāo)滿足mx
2-y
2=ma
2.
當(dāng)m<-1時(shí),曲線C的方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/4.png)
,C是焦點(diǎn)在y軸上的橢圓;
當(dāng)m=-1時(shí),曲線C的方程為x
2+y
2=a
2,C是圓心在原點(diǎn)的圓;
當(dāng)-1<m<0時(shí),曲線C的方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/5.png)
,C是焦點(diǎn)在x軸上的橢圓;
當(dāng)m>0時(shí),曲線C的方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/6.png)
,C是焦點(diǎn)在x軸上的雙曲線;
(Ⅱ)由(I)知,當(dāng)m=-1時(shí),C
1方程為x
2+y
2=a
2,
當(dāng)m∈(-1,0)∪(0,+∞)時(shí),C
2的焦點(diǎn)分別為F
1(-a
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/7.png)
,0),F(xiàn)
2(a
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/8.png)
,0),
對(duì)于給定的m∈(-1,0)∪(0,+∞),C
1上存在點(diǎn)N(x
,y
)(y
≠0),使得△F
1NF
2的面積S=|m|a
2,
的充要條件為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/9.png)
由①得0<|y
|≤a,由②得|y
|=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/10.png)
,
當(dāng)0<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/11.png)
≤a,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/12.png)
,或
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/13.png)
時(shí),
存在點(diǎn)N,使S=|m|a
2,
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/14.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/15.png)
,或
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/16.png)
時(shí),不存在滿足條件的點(diǎn)N.
當(dāng)m∈[
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/17.png)
,0)∪(0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/18.png)
]時(shí),由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/19.png)
=(-a
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/20.png)
-x
,-y
),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/21.png)
=(a
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/22.png)
-x
,-y
),
可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/23.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/24.png)
=x
2-(1+m)a
2+y
2=-ma
2.
令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/25.png)
=r
1,|
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/26.png)
|=r
2,∠F
1NF
2=θ,
則由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/27.png)
=r
1r
2cosθ=-ma
2,可得r
1r
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/28.png)
,
從而s=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/29.png)
r
1r
2sinθ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/30.png)
=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/31.png)
,于是由S=|m|a
2,
可得-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/32.png)
=|m|a
2,即tanθ=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/33.png)
,
綜上可得:當(dāng)m∈[
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/34.png)
,0)時(shí),在C
1上存在點(diǎn)N,使得△F
1NF
2的面積S=|m|a
2,且tanθ=2;
當(dāng)m∈(0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/35.png)
]時(shí),在C
1上存在點(diǎn)N,使得△F
1NF
2的面積S=|m|a
2,且tanθ=-2;
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214306724884956/SYS201310232143067248849019_DA/36.png)
時(shí),不存在滿足條件的點(diǎn)N.
點(diǎn)評(píng):此題是個(gè)難題.考查曲線與方程、圓錐曲線等基礎(chǔ)知識(shí),同時(shí)考查推理運(yùn)算的能力,以及分類與整合和數(shù)形結(jié)合的思想.其中問題(II)是一個(gè)開放性問題,考查了同學(xué)們觀察、推理以及創(chuàng)造性地分析問題、解決問題的能力.