...小题满分5分,第3小题满分8分.在平面直角坐标系 中
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发布时间：2024-09-30 04:52
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时间：2024-10-26 10:51
试题分析：本题属于新定义问题，（1）我们只要利用题设定义求出 的值，若 ，则结论就可得证；（2）直线 是曲线 的分隔线，首先直线与曲线无交点，即直线方程与曲线方程联立方程组 ，方程组应无实解，方程组变形为 ，此方程就无实解，注意分类讨论，按二次项系数为0和不为0分类，然后在曲线上找到两点位于直线 的两侧．则可得到所求范围；（3）首先求出轨迹 的方程 ，化简为 ，过原点的直线中，当斜率存在时设其方程为 ，然后解方程组 ，变形为 ，这个方程有无实数解，直接判断不方便，可转化为判断函数 与 的图象有无交点，而这可利用函数图象直接判断． 是开口方向向上的二次函数， 是幂函数，其图象一定有交点，因此直线 不是 的分隔线，过原点的直线还有一条就是 ，它显然与曲线 无交点，又曲线 上两点 一定在直线 两侧，故它是分隔线，结论得证．
试题解析：（1）由题得， ，∴ 被直线4 分隔.
（2）由题得，直线 与曲线已赞过已踩过你对这个回答的评价是？评论收起 ._1uevpeq{zoom:1;background-color:#fff;border:0;margin-bottom:10px;padding:30px 0 20px 42px;position:relative}._1uevpeq.ec-1841{padding:20px 0}._1uevpeq.ec-2246{padding:20px 0 10px}.ec-1841 .y7we4hu{font-size:16px;margin-bottom:-5px}.y7we4hu{color:#7a8f9a;height:25px;line-height:25px;overflow:hidden;position:relative}.y7we4hu h2{margin:0;padding:0}.y7we4hu:after{clear:both;content:" ";display:block;height:0;visibility:hidden}a.tycfu7u{color:#666;float:right;font-size:12px;margin-left:8px;text-decoration:none}.hhhv6ex{color:#666;font-size:13px;line-height:normal;line-height:20px;margin-top:10px}.vnsdjzp{margin-top:15px;position:relative}.vnsdjzp h3{font-weight:400;padding:0}.vnsdjzp a{text-decoration:none}.vnsdjzp em{color:#d81419;font-style:normal}.ec-2246 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