山花诗酒
幼苗
共回答了14个问题采纳率:85.7% 举报
(1)
![](https://img.yulucn.com/upload/5/cc/5ccc245e613aa353bd6a03a4fddea34f_thumb.jpg)
(2)见解析
学生错(1)曲线C是焦点在x轴上的椭圆,当且仅当
![](https://img.yulucn.com/upload/1/d4/1d4c0f6612194dc3d7675ee122028b7e_thumb.jpg)
解得2<m<5,所以m的取值范围是(2,5).
(2)当m=4时,曲线C的方程为x
2 +2y
2 =8,点A,B的坐标分别为(0,2),(0,-2).
由
![](https://img.yulucn.com/upload/3/9b/39bb32ad01fadb91c13ae700ba37807b_thumb.jpg)
得(1+2k
2 )x
2 +16kx+24=0.
设点M,N的坐标分别为(x
1 ,y
1 ),(x
2 ,y
2 ),则y
1 =kx
1 +4,y
2 =kx
2 +4,x
1 +x
2 =
![](https://img.yulucn.com/upload/9/f4/9f4597cec931a15f22e4bf7dae3f6320_thumb.jpg)
,x
1 x
2 =
![](https://img.yulucn.com/upload/4/85/4859eb1f4db0d2557df836646ebb76ea_thumb.jpg)
.直线BM的方程为y+2=
![](https://img.yulucn.com/upload/3/03/303d1f7802e106e47fd554c54b8fe047_thumb.jpg)
x,点G的坐标为
![](https://img.yulucn.com/upload/4/7b/47bf686777e539d988a5c5c45a6be655_thumb.jpg)
.
因为直线AN和直线AG的斜率分别为k
AN =
![](https://img.yulucn.com/upload/1/bd/1bd755ae58d81632f5351034894dbe35_thumb.jpg)
,k
AG =-
![](https://img.yulucn.com/upload/5/af/5af9f617fffb15e0261e88311692c238_thumb.jpg)
,所以k
AN -k
AG =
![](https://img.yulucn.com/upload/b/b6/bb6227b2d3f3f924bbd2ecfdfb972002_thumb.jpg)
=
![](https://img.yulucn.com/upload/3/bb/3bba82ad13b9eb06ab1ca4c2f904e098_thumb.jpg)
=
![](https://img.yulucn.com/upload/4/cd/4cd75b8338ae9676342b1941f6147391_thumb.jpg)
=
![](https://img.yulucn.com/upload/a/63/a63395dbb756c0152382e1d88fa4d83e_thumb.jpg)
=0.
即k
AN =k
AG .故A,G,N三点共线.
审题引导:(1)方程的曲线是焦点在x轴上的椭圆;
(2)证明三点共线的常用方法.
规范(1)曲线C是焦点在x轴上的椭圆,当且仅当
![](https://img.yulucn.com/upload/6/00/60040878936bdb5d7c074a418c1c2ed0_thumb.jpg)
(3分)
解得
![](https://img.yulucn.com/upload/b/48/b4803eb6993981d3dd62614b9f9112e3_thumb.jpg)
<m<5,所以m的取值范围是
![](https://img.yulucn.com/upload/5/cc/5ccc245e613aa353bd6a03a4fddea34f_thumb.jpg)
.(4分)
(2)当m=4时,曲线C的方程为x
2 +2y
2 =8,点A,B的坐标分别为(0,2),(0,-2).(5分)
由
![](https://img.yulucn.com/upload/3/9b/39bb32ad01fadb91c13ae700ba37807b_thumb.jpg)
得(1+2k
2 )x
2 +16kx+24=0.(6分)
因为直线与曲线C交于不同的两点,所以Δ=(16k)
2 -4(1+2k
2 )×24>0,即k
2 >
![](https://img.yulucn.com/upload/8/5d/85d7671dd15e343cb5e7fb59911282ef_thumb.jpg)
.(7分)
设点M,N的坐标分别为(x
1 ,y
1 ),(x
2 ,y
2 ),则y
1 =kx
1 +4,y
2 =kx
2 +4,
x
1 +x
2 =
![](https://img.yulucn.com/upload/9/f4/9f4597cec931a15f22e4bf7dae3f6320_thumb.jpg)
,x
1 x
2 =
![](https://img.yulucn.com/upload/4/85/4859eb1f4db0d2557df836646ebb76ea_thumb.jpg)
.(8分)
直线BM的方程为y+2=
![](https://img.yulucn.com/upload/3/03/303d1f7802e106e47fd554c54b8fe047_thumb.jpg)
x,点G的坐标为
![](https://img.yulucn.com/upload/4/7b/47bf686777e539d988a5c5c45a6be655_thumb.jpg)
.(9分)
因为直线AN和直线AG的斜率分别为k
AN =
![](https://img.yulucn.com/upload/1/bd/1bd755ae58d81632f5351034894dbe35_thumb.jpg)
,k
AG =-
![](https://img.yulucn.com/upload/5/af/5af9f617fffb15e0261e88311692c238_thumb.jpg)
,(11分)
所以k
AN -k
AG =
![](https://img.yulucn.com/upload/6/3d/63d08261414d322cf37dfdaf0f93f135_thumb.jpg)
=0.
即k
AN =k
AG .(13分)故A,G,N三点共线.(14分)
错因分析:易忽视焦点在x轴上,漏掉
![](https://img.yulucn.com/upload/3/02/302aea4c54c041e3e82dba10e7619c40_thumb.jpg)
这一条件,从而失误.联立消元后易忽视Δ>0这一前提条件.
1年前
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