人教A版选修11 2.1.2.2椭圆的简单几何性质(二) 作业.docx

2.1.2椭圆的简单几何性质(二)课时过关能力提升基础巩固1.椭圆x225+y24=1的两个焦点为f1,f2,过点f2的直线交椭圆于a,b两点.若|ab|=8,则|af1|+|bf1|的值为()a.10b.12c.16d.18解析:|ab|+|af1|+|bf1|=4a,|af1|+|bf1|=45-8=12.答案:b2.已知直线l:x+y-3=0,椭圆x24+y2=1,则直线与椭圆的位置关系是()a.相交b.相切c.相离d.相切或相交解析:将y=3-x代入x24+y2=1,得5x2-24x+32=0.=(-24)2-4532=576-640=-640),与直线方程x+3y+4=0联立,消去x,得(3m+n)y2+83my+16m-1=0,由=0,得3m+n=16mn,即3n+1m=16.又c=2,即1m-1n=4,由联立得m=17,n=13或m=1,n=15,故椭圆的长轴长为27或25.答案:c7.若直线y=x+2与椭圆x2m+y23=1有两个公共点,则m的取值范围是.解析:由x2m+y23=1,y=x+2,得(m+3)x2+4mx+m=0.直线与椭圆有两个公共点,=(4m)2-4m(m+3)=16m2-4m2-12m=12m2-12m0,解得m1或m0,且m3,m1,且m3.答案:(1,3)(3,+)8.若直线3x-y-2=0截焦点为(0,52)的椭圆所得弦中点的横坐标是12,则该椭圆的标准方程是.解析:设椭圆的标准方程为y2a2+x2b2=1(ab0),由y2a2+x2b2=1,3x-y-2=0,联立得(a2+9b2)x2-12b2x+4b2-a2b2=0,x1+x2=12b2a2+9b2=1,a2=3b2.又由焦点为(0,52)知,a2-b2=50.由,得a2=75,b2=25.故所求椭圆方程为x225+y275=1.答案:x225+y275=19.椭圆ax2+by2=1(a0,b0,且ab)与直线x+y-1=0相交于a,b两点,c是ab的中点,若|ab|=22,直线oc的斜率为22,求椭圆的方程.解:由直线方程和椭圆方程联立,得ax2+by2=1,x+y=1,则(a+b)x2-2bx+b-1=0.设a(x1,y1),b(x2,y2),则|ab|=(1+k2)(x1+x2)2-4x1x2=24b2-4(a+b)(b-1)(a+b)2.|ab|=22,a+b-aba+b=1.设c(x,y),则x=x1+x22=ba+b,y=1-x=aa+b.直线oc的斜率为22,ab=22.代入得a=13,b=23.椭圆方程为x23+2y23=1.10.如图,椭圆e:x2a2+y2b2=1(ab0)经过点a(0,-1),且离心率为22.(1)求椭圆e的方程;(2)经过点(1,1),且斜率为k的直线与椭圆e交于不同的两点p,q(均异于点a),证明:直线ap与aq的斜率之和为2.(1)解:由题设知ca=22,b=1,结合a2=b2+c2,解得a=2.所以椭圆的方程为x22+y2=1.(2)证明由题设知,直线pq的方程为y=k(x-1)+1(k2),代入x22+y2=1,得(1+2k2)x2-4k(k-1)x+2k(k-2)=0.由已知0.设p(x1,y1),q(x2,y2),x1x20,则x1+x2=4k(k-1)1+2k2,x1x2=2k(k-2)1+2k2.从而直线ap,aq的斜率之和kap+kaq=y1+1x1+y2+1x2=kx1+2-kx1+kx2+2-kx2=2k+(2-k)1x1+1x2=2k+(2-k)x1+x2x1x2=2k+(2-k)4k(k-1)2k(k-2)=2k-2(k-1)=2.能力提升1.设p,q分别为圆x2+(y-6)2=2和椭圆x210+y2=1上的点,则p,q两点间的最大距离是()a.52b.46+2c.7+2d.62解析:设q(x,y),则该点到圆心的距离d=(x-0)2+(y-6)2=x2+(y-6)2=10(1-y2)+(y-6)2=-9y2-12y+46,y-1,1,当y=-122(-9)=-23时,dmax=-9-232-12-23+46=50=52.圆上点p和椭圆上点q的距离的最大值为dmax+r=52+2=62.故选d.答案:d2.已知(4,2)是直线l被椭圆x236+y29=1所截得的线段的中点,则l的方程是()a.x-2y=0b.x+2y-4=0c.2x+3y+4=0d.x+2y-8=0解析:设l与椭圆的两交点分别为(x1,y1),(x2,y2),则有x1236+y129=1,x2236+y229=1,-,得(x1+x2)(x1-x2)36+(y1+y2)(y1-y2)9=0.由x1+x2=8,y1+y2=4,可得2(x1-x2)+4(y1-y2)=0,即y1-y2x1-x2=-12.故方程为y-2=-12(x-4),即x+2y-8=0.答案:d3.已知椭圆c:x2a2+y2b2=1(ab0)的离心率为32,过右焦点f且斜率为k(k0)的直线与c相交于a,b两点,若af=3fb,则k等于()a.1b.2c.3d.2解析:由椭圆c的离心率为32,得c=32a,b2=a24.椭圆c:x2a2+4y2a2=1.设a(xa,ya),b(xb,yb),f32a,0.af=3fb,32a-xa,-ya=3xb-32a,yb.32a-xa=3xb-32a,-ya=3yb,即xa+3xb=23a,ya+3yb=0.将点a,b的坐标代入椭圆c,得xa2a2+4ya2a2=1,xb2a2+4yb2a2=1,9-,得9xb2-xa2a2=8,(3xb+xa)(3xb-xa)a2=8,3xb-xa=433a.联立,得xa+3xb=23a,3xb-xa=433a,解得xa=33a,xb=539a.ya=-66a,yb=618a.k=yb-yaxb-xa=618a+66a539a-33a=2.答案:b4.若直线ax+by+4=0和圆x2+y2=4没有公共点,则过点(a,b)的直线与椭圆x29+y24=1的公共点个数为.解析:直线ax+by+4=0与圆x2+y2=4没有公共点,4a2+b22,a2+b21时,设切线l的方程为y=k(x-m). 由y=k(x-m),x24+y2=1,得(1+4k2)x2-8k2mx+4k2m2-4=0.设a,b两点的坐标分别为(x1,y1),(x2,y2),则x1+x2=8k2m1+4k2,x1x2=4k2m2-41+4k2.又由l与圆x2+y2=1相切,得|km|k2+1=1,即m2k2=k2+1.所以|ab|=1+k2|x1-x2|=(1+k2)(x1+x2)2-4x1x2 =(1