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1、1.3简单曲线的极坐标方程,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,教学目标,1、认识几种圆的极坐标方程，比较它与直角坐标方程的异同。 2、掌握求圆的极坐标方程的方法。 3、能应用极坐标方程解决圆与直线的关系。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2
2、011 Aspose Pty Ltd.,教学重点： 求圆的极坐标方程的方法与步骤 教学难点： 极坐标方程是涉及长度与角度的问题，列方程实质是解直角或斜三角形问题，要使用旧的三角知识。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,答：与直角坐标系里的情况一样，求曲线的极坐标方程就是建系设点（点与坐标的对应）列式（方程与坐标的对应）化简得方程f(,)=0 说明,求曲线的极坐标方程的步骤：,Evaluation
3、only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,新课引入：,热身训练：在平面直角坐标系中,1、圆心坐标为(3,0)且半径为3的圆方程为,（x-3）2+y2=9,2、圆心坐标为（0,3）且半径为3的圆线方程为_ 3、圆心在原点半径为3的圆方程为_,X2+(y-3)2=9,X2+y2=9,变式：将以上三个方程化为极坐标方程并画出对应的图形。,Evaluation only. Created with Aspose.Slides for
4、.NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,（x-3）2+y2=9,X2+(y-3)2=9,X2+y2=9,6cos ,6sin ,3,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,曲线的极坐标方程,定义：如果曲线上的点与方程f(,)=0有如下关系 ()曲线上任一点的坐标(所有坐标中至少有一个)符合方程f(,)=0
5、； ()方程f(,)=0的所有解为坐标的点都在曲线上。 则曲线的方程是f(,)=0 。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,题组练习1,求下列圆的极坐标方程 ()中心在极点，半径为r； ()中心在(a,0)，半径为a； ()中心在(a,/2)，半径为a； ()中心在(0,)，半径为r。,r,2acos ,2asin ,2+ 0 2 -2 0 cos( - )= r2,Evaluation only.
6、 Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,你可以用极坐标方程直接来求吗？,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 Pr
7、ofile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,练习,以极坐标系中的点(1,1)为圆心，1为半径的圆的方程是,C,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,题组练习2,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 20
8、04-2011 Aspose Pty Ltd.,（ ）,A、双曲线 B、椭圆 C、抛物线 D、圆,D,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.,（ ）,C,Evaluatio
9、n 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
10、.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.,直线的极坐标方程,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,答：与直角坐标系里的情况一样，
11、求曲线的极坐标方程就是找出曲线上动点的坐标与之间的关系，然后列出方程(,)=0 ，再化简并讨论。,怎样求曲线的极坐标方程？,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例题1：求过极点，倾角为 的射线的极坐标方程。,分析：,如图，所求的射线上任一点的极角都是 ，其,极径可以取任意的非负数。故所求,直线的极坐标方程为,新课讲授,Evaluation only. Created with Aspose.Slid
12、es for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,1、求过极点，倾角为 的射线的极坐标方程。,易得,思考：,2、求过极点，倾角为 的直线的极坐标方程。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,和前面的直角坐标系里直线方程的表示形式比较起来，极坐标系里的直线表示起来很不方便，要用两条射线组合而成。原因在哪
13、？,为了弥补这个不足，可以考虑允许极径可以取全体实数。则上面的直线的极坐标方程可以表示为,或,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例题2、求过点A(a,0)(a0)，且垂直于极轴的直线L的极坐标方程。,解：如图，设点,为直线L上除点A外的任意一点，连接OM,在 中有,即,可以验证，点A的坐标也满足上式。,Evaluation only. Created with Aspose.Slides for
14、.NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,求直线的极坐标方程步骤,1、根据题意画出草图；,2、设点 是直线上任意一点；,3、连接MO；,4、根据几何条件建立关于 的方 程，并化简；,5、检验并确认所得的方程即为所求。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,练习：设点P的极坐标为A ，直线 过点P且与极轴所成
15、的角为 ,求直线 的极坐标方程。,解：如图，设点,为直线 上异于的点,连接OM，,在 中有,即,显然A点也满足上方程。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,例题3设点P的极坐标为 ，直线 过点P且与极轴所成的角为 ,求直线 的极坐标方程。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.
16、Copyright 2004-2011 Aspose Pty Ltd.,则 由点P的极坐标知,由正弦定理得,显然点P的坐标也是它的解。,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,O,H,M,A,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.,A、两条相交的直线,B、两条射线,C、一条直线,D、一
