【解析】
(1)设椭圆的焦距为2c,由椭圆右顶点为圆心可得a值,进而由离心率可得c值,根据平方关系可得b值;
(2)由点G在线段AB上,且|AG|=|BH|及对称性知点H不在线段AB上,所以要使|AG|=|BH|,只要|AB|=|GH|,设A(x1,y1),B(x2,y2),联立直线方程与椭圆方程消掉y得x的二次方程,利用韦达定理及弦长公式可得|AB|,在圆中利用弦心距及勾股定理可得|GH|,根据|AB|=|GH|得r,k的方程,分离出r后按k是否为0进行讨论,借助基本函数的范围即可求得r范围;
【答案】
解:(1)设椭圆的焦距为2c,由椭圆右顶点为圆M的圆心( 2,0),得a= 2,又 ca= 22,所以c=1,b=1.所以椭圆C的方程为: x22+y2=1.
(2)设A(x1,y1),B(x2,y2),由直线l与椭圆C交于两点A,B,则{y=kxx2+2y2−2=0,所以(1+2k2)x2−2=0,则x1+x2=0,x1x2=− 21+2k2,所以|AB|= (1+k2)81+2k2= 8(1+k2)1+2k2,点M( 2,0)到直线l的距离d= |||2k||| 1+k2,则|GH|=2 r2− 2k21+k2,显然,若点H也在线段AB上,则由对称性可知,直线y=kx就是y轴,矛盾,所以要使|AG|=|BH|,只要|AB|=|GH|,所以 8(1+k2)1+2k2=4(r2? 2k21+k2),r2= 2k21+k2+ 2(1+k2)1+2k2= 2(3k4+3k2+1)2k4+3k2+1=2(1+ k42k4+3k2+1),当k=0时,r= 2,当k≠0时,r2=2(1+ 1 1k4+ 3k2+2)<2(1+ 12)=3,又显然r2=2(1+ 1 1k4+ 3k2+2)>2,所以 2<r< 3,
综上, 2≤r< 3.
故答案为:
(1) x22+y2=1;
(2) 2≤r< 3
追答![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/faf2b2119313b07e80537d1708d7912396dd8cc6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/bf096b63f6246b6027189e6beff81a4c510fa29e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/7af40ad162d9f2d35e2008f0adec8a136327ccff?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
答案可以采纳吗
追问为什么 根据对称性
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/f636afc379310a556ec7f2ebb34543a9832610c9?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
有这个
追答解析第二点不是有吗
追问OK了!
追答嗯嗯。第一张图就是解析。