設拋物線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617881911.png)
的焦點為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617913302.png)
,準線為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617913280.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617944531.png)
,以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617959399.png)
為圓心的圓
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617959399.png)
與
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617913280.png)
相切于點
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617991333.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617991333.png)
的縱坐標為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618022441.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618037559.png)
是圓
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617959399.png)
與
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618069266.png)
軸除
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617913302.png)
外的另一個交點.
(I)求拋物線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618100313.png)
與圓
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617959399.png)
的方程;
( II)已知直線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618131872.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618147297.png)
與
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618100313.png)
交于
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618209423.png)
兩點,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618147297.png)
與
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617913280.png)
交于點
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618349315.png)
,且
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618365573.png)
, 求
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618365557.png)
的面積.
試題分析:(I)根據(jù)拋物線的方程與準線,可得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618443826.png)
,由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617991333.png)
的縱坐標為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618022441.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617959399.png)
的縱坐標為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618022441.png)
,即
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240246185211124.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618552895.png)
,由題意可知:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618568674.png)
,則在等腰三角形中有
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240246185832096.png)
或
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618599429.png)
,由于
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618615426.png)
不重合,則
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618630421.png)
.則拋物線與圓的方程就得出.
(II)根據(jù)題意可得三角形
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618661489.png)
是直角三角形,又因
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618661524.png)
,則
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617913302.png)
是
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618708385.png)
的中點,即
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240246187082892.png)
解得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618724457.png)
.
聯(lián)立直線與拋物線方程得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618739656.png)
則由弦長公式得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618771621.png)
,又根據(jù)點到直線的距離得出
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617991333.png)
到
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618786396.png)
的距離
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618817544.png)
,從而得出
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618817674.png)
.
試題解析:(I)根據(jù)拋物線的定義:有
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240246188331102.png)
由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617991333.png)
的縱坐標為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618022441.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617959399.png)
的縱坐標為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618022441.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240246185211124.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618552895.png)
,則
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618942648.png)
,又由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618568674.png)
得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618630421.png)
則拋物線為:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618381525.png)
,圓的方程為:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240246183961142.png)
( II)由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240246190361283.png)
,
根據(jù)題意可得三角形
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618661489.png)
是直角三角形,又因
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618661524.png)
,則
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617913302.png)
是
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618708385.png)
的中點,即
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240246190982864.png)
解得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618724457.png)
.
由
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240246191291928.png)
,根據(jù)點到直線的距離得出
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024617991333.png)
到
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618786396.png)
的距離
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024618817544.png)
,從而得出
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240246191761065.png)
.
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824024230870399.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240236391361183.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639167313.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639198312.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639230280.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639167313.png)
相交于
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639292421.png)
兩點,直線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639308410.png)
、
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639323395.png)
分別交直線
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639339382.png)
于
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639354399.png)
、
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639386357.png)
兩點,線段
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639401513.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639417289.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639448350.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824023639464396.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021927179280.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021927116313.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021927210357.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021927225423.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021927241573.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824021927257458.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824030714448638.png)
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![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824020400287473.png)
與
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824020400303372.png)
有公共點,求證
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824020400318427.png)
,進而證明原點不是“C
1—C
2型點”;
(3)求證:圓
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824020400318673.png)
內(nèi)的點都不是“C
1—C
2型點”.
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