【答案】
分析:(I)根據(jù)橢圓的特征可得當(dāng)點(diǎn)P在點(diǎn)(0,b)時(shí),△APB面積的最大,結(jié)合題中的條件可得a、b與c的關(guān)系進(jìn)而得到答案.
(II)設(shè)點(diǎn)P的坐標(biāo)為(x
,y
),由題意可設(shè)直線AP的方程為y=k(x+2),可得點(diǎn)D與BD中點(diǎn)E的坐標(biāo),聯(lián)立直線與橢圓的方程得(3+4k
2)x
2+16k
2x+16k
2-12=0,進(jìn)而表示出點(diǎn)P的坐標(biāo),結(jié)合點(diǎn)F坐標(biāo)為(1,0),再寫出直線PF的方程,根據(jù)點(diǎn)E到直線PF的距離等于直徑BD的一半,進(jìn)而得到答案.
解答:解:(Ⅰ)由題意可設(shè)橢圓C的方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/0.png)
,F(xiàn)(c,0).
由題意知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/1.png)
解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/2.png)
,c=1.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/images3.png)
故橢圓C的方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/3.png)
,離心率為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/4.png)
.
(Ⅱ)以BD為直徑的圓與直線PF相切.
證明如下:由題意可設(shè)直線AP的方程為y=k(x+2)(k≠0).
則點(diǎn)D坐標(biāo)為(2,4k),BD中點(diǎn)E的坐標(biāo)為(2,2k).
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/5.png)
得(3+4k
2)x
2+16k
2x+16k
2-12=0.
設(shè)點(diǎn)P的坐標(biāo)為(x
,y
),則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/6.png)
.
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/7.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/8.png)
.
因?yàn)辄c(diǎn)F坐標(biāo)為(1,0),
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/9.png)
時(shí),點(diǎn)P的坐標(biāo)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/10.png)
,點(diǎn)D的坐標(biāo)為(2,±2).
直線PF⊥x軸,此時(shí)以BD為直徑的圓(x-2)
2+(y±1)
2=1與直線PF相切.
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/11.png)
時(shí),則直線PF的斜率
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/12.png)
.
所以直線PF的方程為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/13.png)
.
點(diǎn)E到直線PF的距離
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/14.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/15.png)
.
又因?yàn)閨BD|=4|k|,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184252738755610/SYS201310241842527387556021_DA/16.png)
.
故以BD為直徑的圓與直線PF相切.
綜上得,當(dāng)直線AP繞點(diǎn)A轉(zhuǎn)動時(shí),以BD為直徑的圓與直線PF相切.
點(diǎn)評:解決此類問題的關(guān)鍵是熟練掌握橢圓中有關(guān)數(shù)值的關(guān)系,以及橢圓與直線的位置關(guān)系、圓與直線的位置關(guān)系.