38解析几何解法:识破天机-中点弦问题
38:识破天机 - 中点弦问题
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对于解决圆锥曲线中以定点为中点的弦相关问题中,设出弦的两端点坐标,代入圆锥曲线方程得到两等式后再相减,从而得到弦中点坐标与所在直线的斜率的关系,使问题得以解决.此方法巧妙地将斜率公式和中点坐标公式结合起来,设而不求,代点作差,可以减少计算量,提高解题速度,优化解题过程,对解决此类问题确实具有很好的效果.这个方法我们称为
“点差法”.
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1.椭圆中点差法的应用
已知椭圆
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的方程为
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,若一条斜率为
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的直线
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交椭圆
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于
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,
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两个不同交点,且线段
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的中点为
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,
此时我们将
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,
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两点坐标分别带入椭圆
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的方程中可得:
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,
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,
将两式相减得:
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,
整理得:
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,
又
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,
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,
所以可得:
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.
2.双曲线中点差法的应用
已知双曲线
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的方程为
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,若一条斜率为
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的直线
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交双曲线
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于
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,
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两个不同交点,且线段
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的中点为
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,
则可得结论为:
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(推导过程同上)
3.抛物线中点差法的应用
已知抛物线
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的方程为
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,若一条斜率为
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的直线
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交抛物线
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于
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,
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两个不同交点,且线段
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的中点为
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,
将
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,
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两点坐标分别带入抛物线方程中可得:
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,
两式相减并整理可得:
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(2018全国Ⅱ卷理)在直角坐标系
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中,曲线
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的参数方程为
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,
(
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为参数),直线
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的参数方程为
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,(
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为参数).
(1)求
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与
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的普通方程;
(2)若曲线
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截直线
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所得线段的中点坐标为
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,求
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的方程.
【答案】见解析
【解析】
(1)曲线
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的普通方程为
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,
直线
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的普通方程为
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或
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(当
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时,其中
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),计算过程略.
(2)当直线斜率不存在时,直线方程为
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,与
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相交截得线段中点为
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,不符合题意;
当直线斜率存在时,设直线斜率为
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,直线
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与椭圆
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交于
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,
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两点,设
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,
将
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两点坐标带入
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可得,
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,
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,将两式相减并整理可得,
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,即
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,解得
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,则直线
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的方程为
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.
【点评】已知直线与椭圆交于两点且已知弦中点坐标,可利用点差法快速求得直线斜率,进而求得直线方程.
1.(改编)若直线
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交椭圆
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所得的弦
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的中点为
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,则椭圆
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的离心率为__________.
2.(改编)若点
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平分双曲线
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的一条弦,则这条弦所在直线的方程为__________.
3.(改编)直线
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与抛物线
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交于
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两点,若
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恰为线段
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的中点,则抛物线的方程为__________.