新编高中数学 2.3柱面与平面的截面课件 北师大版选修41

1、-1 1-3柱面与平面的截面北 师 大 版 数 学 课 件精 品 资 料 整 理 -* *-3 3柱面与平面的截面柱面与平面的截面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.了解柱面、旋转面的形成过程.2.掌握柱面、旋转面、焦球、椭圆的概念.3.知道柱面的垂直截面和一般截面的形状.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123zhis

2、hi 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】 过圆柱面的轴的平面与圆柱面的交线是 ().a.两条相交线段b.两条平行线段c.两条异面线段d.平行四边形答案:bzhishi shuli知识梳理zhongnan jvjiao重难聚焦suitangyanlian随堂演练dianli tou

3、xi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 shulizhishi shulizhishi shulizhishi shuli知识梳理知识梳理知识梳理知识梳理zhongnan jvjiaozhongnan jvjiaozh

4、ongnan 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123【做一做3】f1和f2是椭圆的焦点,p,q是椭圆上的两点,且pf1=3,pf2=2,qf1=d1,

5、qf2=d2,则().a.d1+d2=5b.d1-d2=5c.d1d2=5d.=5解析:椭圆上的点到两焦点的距离之和为常数,d1+d2=pf1+pf2=3+2=5.答案:azhishi 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 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

6、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)当点p在椭圆的任意位置时,过p作l1的垂线,垂足为q,过p作平面的垂线,垂足为k1,连接k1q,得rtpk1q,则qpk1=.从而有=cos=定值.(2)椭圆上任

7、意一点到焦点f1的距离与到直线l1的距离之比为定值cos.我们把直线l1叫做椭圆的一条准线.(3)椭圆上任意一点到焦点f2的距离与到直线l2的距离之比也为定值cos,所以l2是椭圆的另一条准线.(4)记e=cos,我们把e叫做椭圆的离心率.(5)e的几何意义是:椭圆上一点到焦点的距离和它到准线的距离的比.当e越接近于1时,c越接近于a,从而b越小,因此椭圆越扁;反之,e越接近于0,从而b越接近于a,椭圆越接近于圆.当e=0时,c=0,a=b,两焦点重合,图形就是圆了.可见离心率是刻画椭圆圆扁程度的量.zhishi shuli知识梳理zhongnan jvjiao重难聚焦suitangyanli

8、an随堂演练dianli touxi典例透析mubiaodaohang目标导航题型一题型二题型三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 touxidian

9、li 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题型三解:如图所示.abcd为边长为5的正方形,连接oc,od.则ocd为等边三角形,设cd的中点为e,连接oe,则oecd,且oe=.ad垂直上底面,adoe.又oecd,且adcd=d,oe平面abcd.oe为轴到截面的距离.故轴到截面的距离为.zhishi shuli知识梳理zhongnan jvjiao重难聚焦suitangyanlian随堂演练dianli touxi典例透析mubiaoda

10、ohang目标导航题型一题型二题型三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mubiao

11、daohangmubiaodaohangmubiaodaohangmubiaodaohang目标导航目标导航目标导航目标导航题型一题型二题型三解:(1)如图所示,设过点a的母线为aa,则ooaa,且ooaa为矩形,易知oba是等腰直角三角形,且ab=.aa=2,oo与ab所成的角为baa,tanbaa=.(2)所求截面为矩形aabb,s矩形aabb=2=2(cm2).(3)过点o作ocab,垂足为c.点o到截面的距离即oo到截面的距离,也是点o到截面的距离.故d=oc=cm.zhishi shuli知识梳理zhongnan jvjiao重难聚焦suitangyanlian随堂演练dianli

12、touxi典例透析mubiaodaohang目标导航题型一题型二题型三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题型三解:如图所示,连接ma,设圆o的半径为r.l是aq的垂直平分线,ma=mq.又om+mq=oq=r,mo+ma=r=常数.roa,动点m的轨迹是以o,a为焦点的椭圆.zhishi shulizhishi shulizhishi shulizhishi shuli知识梳理知识梳理知识梳理知识梳理zhongnan jvjiaozhongnan

13、 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题型三【变式训练2】设f1,f2为椭圆=1的两个焦点,p为椭圆上的一点,已

14、知p,f1,f2是一个直角三角形的三个顶点,且|pf1|pf2|,求的值.解:由已知|pf1|+|pf2|=6,|f1f2|=2,根据直角的不同位置,分两种情况:若pf2f1为直角,则|pf1|2=|pf2|2+|f1f2|2,即|pf1|2=(6-|pf1|)2+20,得|pf1|=,|pf2|=,故;若f1pf2为直角,则|f1f2|2=|pf1|2+|pf2|2,即20=|pf1|2+(6-|pf1|)2,解得|pf1|=4或|pf1|=2.因为|pf1|pf2|,所以|pf1|=4,|pf2|=2,故=2.zhishi shuli知识梳理zhongnan jvjiao重难聚焦suita

15、ngyanlian随堂演练dianli touxi典例透析mubiaodaohang目标导航题型一题型二题型三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 to

16、uxidianli 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题型三证明:如图所示,过g1作g1hbg2,h为垂足,则四边形abhg1是矩形.g1h=ab.q1,q2分别是p1,p2的平行射影,p1q1p2q2.p1q1q2p2是平行四边形.q1q2=p1p2,即q1q2等于底面直径.g1h=ab=q1q2=2b.又由切线长定理,得g1a=g1f1=g2f2,g2f1=g2b,g2f1-g2f2=g2b-g1a.又g1a=bh,g2f1-g2f2=

17、g2b-bh.f1f2=g2h.在rtg1g2h中,g2h=2,故f1f2=2.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题型三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suitangyanliansuitangyanl

18、iansuitangyanliansuitangyanlian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题型三【变式训练3】已知一圆柱面的半径为3,圆柱面的一截面的两焦点球的球心距为12,求截面截圆柱所得的椭圆的长轴长、短轴长、两焦点间的距离和截面与母线所夹的角.解:易知长轴长2a=12,短轴长2b=6,两焦点间的距离2c=6.a=

19、6,c=3,椭圆的离心率e=.设截面与母线所夹的角为,则cos=.=30.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 2 3 4 51用一个平面去截一个圆柱面,其交线是().a.圆b.椭圆c.两条平行线段d.以上均可能解析:当平面垂直于圆柱面的轴时,交线为圆;当平面与圆柱面的轴平行时,交线为两条平行线段;当平面与圆柱面的轴不平行也不垂直时,交线为椭圆,故选d.答案:dzhishi shuli知识梳理zhongnan jvjiao重难聚焦suitangyanlian随堂演练dianli t