-1 1-4平面截圆锥面数 学 精 品 课 件北 师 大 版-* *-4 4平面截圆锥面平面截圆锥面ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1.掌握圆锥面、双曲线、抛物线的定义.2.掌握垂直截面、一般截面与圆锥面的交线形状.ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航123ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航123【做一做1】 直线l1与l2相交于点P,则l2绕l1旋转一周得到的是().A.圆柱面B.圆锥面C.平面D.圆锥面或平面答案:DZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航123ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航123ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航123ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航123【做一做2】 在圆锥内部嵌入两个焦球,一个位于平面的上方,一个位于平面的下方,并且与平面及圆锥均相切,若平面与两个球的切点不重合,则平面与圆锥面的交线是().A.圆B.椭圆C.双曲线D.抛物线解析:由于平面与两个球的切点不重合,则平面与圆锥母线不平行,且只与圆锥的一侧相交,即,则交线是椭圆.答案:BZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航(1)当平面与圆锥面的母线平行,且不过圆锥顶点时,交线为抛物线.(2)当平面与圆锥面的母线平行,且过圆锥顶点时,交线退化为一条直线.(3)当平面只与圆锥面一侧相交,且不过圆锥顶点时,交线为椭圆.(4)当平面只与圆锥面一侧相交,且不过圆锥顶点,并与圆锥面的对称轴垂直时,交线为圆.(5)当平面只与圆锥面一侧相交,且过圆锥顶点时,交线退化为一个点.(6)当平面与圆锥面两侧都相交,且不过圆锥顶点时,交线为双曲线的一支(另一支为此圆锥面的对顶圆锥面与平面的交线).(7)当平面与圆锥面两侧都相交,且过圆锥顶点时,交线为两条相交直线.ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航题型一题型二解:由双曲线的定义,得AF2+BF2-(AF1+BF1)=4a.AF1+BF1=AB=m,AF2+BF2=4a+m,ABF2的周长为AF2+BF2+AB=(4a+m)+m=4a+2m.ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航题型一题型二解:如图所示,由双曲线的定义得(BF1-AF1)+(AF2-BF2)=4a.又BF1-AF1=AB=m,AF2-BF2=4a-m,AF2=4a-m+BF2,ABF2的周长为AF2+BF2+AB=(4a-m+BF2)+BF2+m=4a+2BF2,即ABF2的周长不是定值.ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航题型一题型二解:如图所示,设直线l与直线l的距离为1,且直线l与点C位于直线l的两侧,动圆M的半径为r,直线l与圆M相切于点A,连接MC,MA并延长MA交l于点B,则MBl.动圆M与圆C外切,MC=1+r.动圆M与直线l相切,MA=r,MB=1+r,MC=MB,即动点M到定点C的距离和到定直线l的距离相等,动圆圆心M的轨迹是以C为焦点,以直线l为准线的抛物线.ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航题型一题型二【变式训练2】已知抛物线y2=2px上有三点A(x1,y1),B(x2,y2),C(x3,y3),且x1x2PF2,所以PF1-PF2=2a=6,所以PF1=2a+PF2=6+10=16.答案:DZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航1 2 3 4 5解析:如图所示,连接MF,则MF=MA=3,则在MFK中,MF2+MK2=32+42=25=FK2,所以MKMF,所以MO=FK=.答案:ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知识梳理知识梳理知识梳理ZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAOZHONGNAN JVJIAO重难聚焦重难聚焦重难聚焦重难聚焦SUITANGYANLIANSUITANGYANLIANSUITANGYANLIANSUITANGYANLIAN随堂演练随堂演练随堂演练随堂演练DIANLI TOUXIDIANLI TOUXIDIANLI TOUXIDIANLI TOUXI典例透析典例透析典例透析典例透析MUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航1 2 3 4 54设F1,F2是双曲线的左、右两个焦点,实轴长2a=4,虚轴长2b=6,P是双曲线左支上的点,若PF1,PF2,F1F2成等差数列,且公差大于0,则F1PF2=.解析:F1F2=2=14,由于PF1,PF2,F1F2成等差数列,且公差大于0,则PF1+F1F2=2PF2,PF2PF1,所以PF2-PF1=4,PF1+14=2PF2,解得PF1=6,PF2=10,在F1PF2中,由余弦定理可得cosF1PF2=-,故F1PF2=120.答案:120ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANLIAN随堂演练DIANLI TOUXI典例透析MUBIAODAOHANG目标导航1 2 3 4 5