解题方法
1 . “费马点”是由十七世纪法国数学家费马提出并征解的一个问题,该问题是:“在一个三角形内求作一点,使其与此三角形的三个顶点的距离之和最小”.如图1,三个内角都小于
的
内部有一点
,连接
,求
的最小值.我们称三角形内到三角形三个顶点距离之和最小的点为费马点.要解决这个问题,首先应想办法将这三条端点重合于一点的线段分离,然后再将它们连接成一条折线,并让折线的两个端点为定点,这样依据“两点之间,线段最短”,就可求出这三条线段和的最小值.某数学研究小组先后尝试了翻折、旋转、平移的方法,发现通过旋转可以解决这个问题,具体的做法如图2,将
绕点
顺时针旋转
,得到
,连接
,则
的长即为所求,此时与三个顶点连线恰好三等分费马点
的周角.同时小组成员研究教材发现:已知对任意平面向量
,把
绕其起点沿逆时针方向旋转
角得到向量
.
,把点
绕点
沿顺时针方向旋转
后得到点
,求点
的坐标;
(2)在
中,
,借助研究成果,直接写出
的最小值;
(3)已知点
,求
的费马点
的坐标.
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dad2a36927223bd70f426ba06aea4b45.png)
(2)在
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/15c0dbe3c080c4c4636c64803e5c1f76.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b11bf8ee11289d13cf5dd0ea9505e699.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c7ed53a398b1d6b7b4abbb43a9abcf1f.png)
(3)已知点
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/24a65f35281b21fdfaf7c437fbd321eb.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/15c0dbe3c080c4c4636c64803e5c1f76.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dad2a36927223bd70f426ba06aea4b45.png)
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解题方法
2 . 如图,
为坐标原点,
为抛物线
的焦点,过
的直线交抛物线于
两点,直线
交抛物线的准线于点
,设抛物线在
点处的切线为
.
与
轴的交点为
,求证:
;
(2)过点
作
的垂线与直线
交于点
,求证:
.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1dde8112e8eb968fd042418dd632759e.png)
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ab2e095298cfb23d9f47811556fc9f9a.png)
(2)过点
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7f9e8449aad35c5d840a3395ea86df6d.png)
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2024-03-13更新
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