题目内容
设直线
与抛物线C:
(
,p为常数)交于不同两点A、B,点D为抛物线准线上的一点。
(I)若t=0,且三角形ABD的面积为4,求抛物线的方程;
(II)当△ABD为正三角形时,求出点D的坐标。
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049214831500.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049221951076.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/20121005104922309662.png)
(I)若t=0,且三角形ABD的面积为4,求抛物线的方程;
(II)当△ABD为正三角形时,求出点D的坐标。
解:(I)直线
过焦点![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049225481372.png)
时,
不妨设
,则
,
又D点到直线l的距离d=p 所以
=4∴p=2
∴抛物线的方程为
(II)设
由
得
则
从而![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049240434574.png)
∴线段AB的中点为
由DM⊥AB得
,即
,
解得
从而
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049261015980.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049268244510.png)
由
得到
= ![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/20121005104927169928.png)
,
解
此时,点
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049224221620.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049225481372.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/20121005104922658536.png)
不妨设
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049227812512.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049228941298.png)
又D点到直线l的距离d=p 所以
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049233153305.png)
∴抛物线的方程为
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/20121005104923425882.png)
(II)设
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049235563340.png)
由
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049236832584.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049237951706.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049239152194.png)
从而
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049240434574.png)
∴线段AB的中点为
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049241582156.png)
由DM⊥AB得
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/20121005104924270990.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049243962810.png)
解得
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049245161514.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049255212301.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049261015980.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049268244510.png)
由
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049269452279.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049270571817.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/20121005104927169928.png)
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049272801243.png)
解
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049273921018.png)
此时,点
![](http://thumb.zyjl.cn/pic1/upload/papers/g02/20121005/201210051049277732221.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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