2011年高考数学理一轮复习 8-4直线与圆锥曲线的位置关系 精品课件.ppt

Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 备考例题3　　设椭圆ax2＋by2＝1与直线x＋y－1＝0相交于A、B两点，点C是AB的中点，若|AB|＝2 ，OC的斜率为 ，求椭圆的方程． Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 第四节　直线与圆锥曲线的位置关系 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 知识自主·梳理 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 必威体育精装版考纲 高考热点 了解圆锥曲线的初步应用． 以解答题的形式考查直线与圆锥曲线相交的交点、弦长等问题，解决直线与圆锥曲线位置关系中的定值、定点以及参数的范围问题. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 1.直线与圆锥曲线的位置关系． 要解决直线与圆锥曲线的位置关系问题，可把直线方程与圆锥曲线方程联立，消去y(或消去x)得到关于x(或关于y)的一元二次方程．如联立后得到以下方程： Ax2＋Bx＋C＝0(A≠0)，Δ＝B2－4AC 若Δ＜0，则直线与圆锥曲线 ； 若Δ＝0，则直线与圆锥曲线 ； 若Δ＞0，则直线与圆锥曲线 ． 没有公共点 有且只有一个公共点 有两个不同的公共点 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2．弦长公式． 直线与圆锥曲线相交时，常常借助根与系数的关系解决弦长问题．直线方程与圆锥曲线方程联立，消去y后得到关于x的一元二次方程．当Δ＞0时，直线与圆锥曲线相交，设交点为A(x1，y1)，B(x2，y2)，直线AB的斜率为k，则直线被圆锥