轨迹

问题描述:

已知圆A:(x+2)^2+y^2=1与定直线l:x=1,且动圆P和圆A外切并与直线l相切,则动圆的圆心P的轨迹方程是() A:y^2=-8x B:y^2=8x C:y^2=-4x D:y^2=4x
1个回答 分类:数学 2011-04-20

问题解答:

我来补答
解题思路: 抛物线
解题过程:
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最终答案:略
 
 
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