题目内容
已知过抛物线y2 =2px(p>0)的焦点F的直线x-my+m=0与抛物线交于A,B两点,且△OAB(O为坐标原点)的面积为2
,则m6+ m4的值为( )
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824011844354344.png)
A.1 | B. 2 | C.3 | D.4 |
B
试题分析:由题意,可知该抛物线的焦点为
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824011844385572.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824011844401656.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824011844416577.png)
∴直线方程变为:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824011844447691.png)
A,B两点是直线与抛物线的交点,
∴它们的坐标都满足这两个方程.
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824011844463840.png)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240118444791269.png)
∴方程的解
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240118444941313.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240118445101316.png)
代入直线方程,可知:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240118445411362.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240118445571410.png)
△OAB的面积可分为△OAP与△OBP的面积之和,
而△OAP与△OBP若以OP为公共底,
则其高即为A,B两点的y轴坐标的绝对值,
∴△OAP与△OBP的面积之和为:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240118445721647.png)
求得p=2,
∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824011844588570.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824011844603452.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824011844619571.png)
故答案为:B
点评:本题主要考查了椭圆的简单性质,直线,抛物线与椭圆的关系.考查了学生综合分析问题和基本的运算能力.
![](http://thumb.zyjl.cn/images/loading.gif)
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