《高中数学 2.5圆锥曲线的几何性质课件 北师大版选修41》由会员分享，可在线阅读，更多相关《高中数学 2.5圆锥曲线的几何性质课件 北师大版选修41（28页珍藏版）》请在金锄头文库上搜索。
1、-1 1-5圆锥曲线的几何性质数 学 精 品 课 件北 师 大 版-* *-5 5圆锥曲线的几何性质圆锥曲线的几何性质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.掌握椭圆、双曲线的离心率的定义.2.掌握圆锥曲线的统一定义.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12ZHISHI SHULIZHISHI SHULIZHISHI SHUL
2、IZHISHI 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
3、航目标导航目标导航12【做一做1】下列数据可能是椭圆离心率的是().A.B.1C.2D.4解析:由于椭圆的离心率e的范围是(0,1),仅有选项A中的数据(0,1).故选A.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122.圆锥曲线的统一定义抛物线、椭圆、双曲线都是平面上到定点的距离与到定直线的距离之比为常数e(离心率)的动点的轨迹,此时定点称为焦点,定直线称为准线.当e=1时,轨迹为抛物线;当0e1时,轨迹为双曲线.这就是圆锥曲线的统一定义.ZHISHI S
4、HULIZHISHI 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MUBIAODAOHANGMUBIAODAOHANGMUBIAO
5、DAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航12【做一做2】平面与圆锥的轴线平行,圆锥母线与轴线夹角为60,则平面与圆锥交线的离心率是().A.2B.C.D.2解析:设平面与轴线夹角为,母线与轴线夹角为,由题意,知=0,=60,故所求离心率e=2.答案:AZHISHI 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SUITANGYANLIANSUIT
6、ANGYANLIANSUITANGYANLIANSUITANGYANLIAN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椭圆、双曲线的两条准线间的距离剖析:椭圆的长轴长为2a,焦距为2c,则两条准线间距离为;双曲线的实轴长为2a,焦距为2c,则两条准线间距离为.如图所示,l1,l2是双曲线的准线,F1,F2是焦点,A1,A2是顶点,O为中心.由离
7、心率定义,A1H1=A1F1.A1F1=OF1-OA1=c-a,A1H1=.OH1=OA1-A1H1=a-.由对称性,得OH2=,H1H2=.即双曲线的两条准线间距离为,同理可证椭圆的两条准线间距离为.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ZHISHI SH
8、ULI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 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 J
9、VJIAO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解析:如图所示为截面的轴面,则AB=8,SB=6,SA=10.由勾股定理的逆定理,知SBA=90,则cosASB=.设圆锥的母线和轴
10、所成的角为,截面和轴所成的角为.cos2=,即2cos2-1=,cos=.+=90,cos=sin=.椭圆的离心率e=.答案:CZHISHI 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 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 SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANL
11、IAN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 TOUXIDIA
12、NLI 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】本例2中,仅把已知条件“ABF2是正三角形”改为“ABF2是直角三角形”,其他不变,重新求解.解:因为ABF2是直角三角形,所以AF2F1=,所以AF1=F1F2,AF2=F1F2.又F1F2=2c,AF2-AF1=2a,所以(2c)-(2c)=2a,整理得+1.即双曲线的离心率e=+1.ZHISHI SHULI知识梳理ZHONGNAN JVJIAO重难聚焦SUITANGYANL
13、IAN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 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 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
14、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题型三【变式训练3】已
15、知点A(1,2)在椭圆=1内,点F的坐标为(2,0),在椭圆上求一点P使|PA|+2|PF|最小.解:如图所示.a2=16,b2=12,c2=4,c=2.F为椭圆的右焦点,并且离心率e=.设点P到右准线的距离为d,则|PF|=d,d=2|PF|.|PA|+2|PF|=|PA|+d.由几何性质可知,当点P的纵坐标(横坐标大于零)与点A的纵坐标相同时,|PA|+d最小.把y=2代入=1,得x=.即点P为所求.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 5Z
16、HISHI 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MUBIAODAOHANGMUBIAODAOHAN
17、GMUBIAODAOHANGMUBIAODAOHANG目标导航目标导航目标导航目标导航1 2 3 4 5解析:在PKF2中,PKF2=90,PK=c,F2K=-c,F2P=.因为F1F2=F2P,所以2c=,化简得a2-2c2=0,解得,即椭圆的离心率e=.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
18、 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:在PF1F2中,PF1F2=90
19、,F1PF2=60,所以PF1=F1F2,则PF2=F1F2.又F1F2=2c,PF2+PF1=2a,所以2c+2c=2a,整理得,即双曲线的离心率e=.答案:BZHISHI 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
20、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 53双曲线的两个焦点为F1,F2,P是双曲线上一点,且有PF1=2PF2,则双曲线离心率的取值范围为().A.(-,3B.(1,3C.(3,+)D.3,+)解析:设双曲线的实轴长是2a,焦距为2c,则PF1-PF2=2PF2-PF2=PF2=2a,所以PF1=2PF2=4a.又PF1+PF2F1F2,F1F2=2c,所以4a+2a2
21、c,整理得3,所以双曲线离心率的取值范围为1e3.答案:BZHISHI 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
22、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分别是双曲线的左、右焦点,M是虚轴上的上边的端点,O是F1F2的中点,且MF1F2=30,则双曲线的离心率e=.解析:在RtF1OM中,F1OM=90,MF1O=30,则OF1=c,OM=b,=tan30,所以c=b,所以c2=3b2,所以c2=3(c2-a2),整理得3a2=2c2,所以双曲线的离心率e=.答案:ZHISHI SHULIZHISHI SHULIZHISHI SHULIZHISHI SHULI知识梳理知
23、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 55已知F1,F2为椭圆的焦点,长轴长2a=4,短轴长2b=2,P为椭圆上任一点,求:(1)PF1PF2的最大值;(2)P+P的最小值.解:由于P为椭圆上任一点,则PF1+PF2=2a=4,(1)PF1PF2=4,当且仅当PF1=PF2=2时取等号,所以PF1PF2的最大值是4.(2)P+P=(PF1+PF2)2-2PF1PF2=16-2PF1PF2.由(1)知PF1PF24,所以P+P=16-2PF1PF216-24=8,当且仅当PF1=PF2=2时取等号,所以P+P的最小值是8.
