鑫颖
幼苗
共回答了25个问题采纳率:92% 举报
(1)设 D ( x , y ), A ( a , a ), B ( b ,- b ),
∵ D 是 AB 的中点, ∴ x =
![](https://img.yulucn.com/upload/d/29/d29b63d38b7d9a5595a9b73336cec2dc_thumb.jpg)
, y =
![](https://img.yulucn.com/upload/4/06/4066fd6263452e631994ea28ff8a5023_thumb.jpg)
,
∵ | AB |=2
![](https://img.yulucn.com/upload/9/d3/9d33f50e2cd31a585714ce698a736fe3_thumb.jpg)
,∴( a - b )
2 +( a + b )
2 =12,
∴(2 y )
2 +(2 x )
2 =12,∴点 D 的轨迹 C 的方程为 x
2 + y
2 =3.
(2) ①当直线 l 与 x 轴垂直时, P (1,
![](https://img.yulucn.com/upload/6/0c/60c6306b4ebd99307c9f72b25684cfda_thumb.jpg)
), Q (1,-
![](https://img.yulucn.com/upload/6/0c/60c6306b4ebd99307c9f72b25684cfda_thumb.jpg)
),
此时| PQ |=2
![](https://img.yulucn.com/upload/6/0c/60c6306b4ebd99307c9f72b25684cfda_thumb.jpg)
,不符合题意;
当直线 l 与 x 轴不垂直时,设直线 l 的方程为 y = k ( x -1),
由于| PQ |=3,所以圆心 C 到直线 l 的距离为
![](https://img.yulucn.com/upload/7/89/7896c03a9c10a688a3ef4263b633a105_thumb.jpg)
,
由
![](https://img.yulucn.com/upload/f/4f/f4fa7ba1dbe2d74ec1f5d82faa2203d2_thumb.jpg)
=
![](https://img.yulucn.com/upload/7/89/7896c03a9c10a688a3ef4263b633a105_thumb.jpg)
,解得 k =
![](https://img.yulucn.com/upload/c/85/c85b108fbdeffc53501ba73204ec40c4_thumb.jpg)
.故直线 l 的方程为 y =
![](https://img.yulucn.com/upload/c/85/c85b108fbdeffc53501ba73204ec40c4_thumb.jpg)
( x -1).
②当直线 l 的斜率存在时,设其斜率为 k ,则 l 的方程为 y = k ( x -1),
由消去 y 得( k
2 +1) x
2 -2 k
2 x + k
2 -3=0,
设 P ( x
1 , y
1 ), Q ( x
2 , y
2 )则由韦达定理得 x
1 + x
2 =
![](https://img.yulucn.com/upload/c/68/c68f62b52af60ddef7ff19f7bdca4b17_thumb.jpg)
, x
1 x
2 =
![](https://img.yulucn.com/upload/d/99/d99a6d4f3e59a98f47fbd994e17b2ec3_thumb.jpg)
,
则
![](https://img.yulucn.com/upload/2/27/2277b58236018b5df38eda2d342c1424_thumb.jpg)
=( m - x
1 ,- y
1 ),
![](https://img.yulucn.com/upload/d/46/d467c48c90330e30cdb600243bb839e7_thumb.jpg)
=( m - x
2 ,- y
2 ),
∴
![](https://img.yulucn.com/upload/2/27/2277b58236018b5df38eda2d342c1424_thumb.jpg)
·
![](https://img.yulucn.com/upload/d/46/d467c48c90330e30cdb600243bb839e7_thumb.jpg)
=( m - x
1 )( m - x
2 )+ y
1 y
2 = m
2 - m ( x
1 + x
2 )+ x
1 x
2 + y
1 y
2 = m
2 - m ( x
1 + x
2 )+ x
1 x
2 + k
2 ( x
1 -1)( x
2 -1)
= m
2 -
![](https://img.yulucn.com/upload/0/3d/03d801bf0a40ec5aa659cac33b26b6bf_thumb.jpg)
+
![](https://img.yulucn.com/upload/d/99/d99a6d4f3e59a98f47fbd994e17b2ec3_thumb.jpg)
+ k
2 (
![](https://img.yulucn.com/upload/d/99/d99a6d4f3e59a98f47fbd994e17b2ec3_thumb.jpg)
-
![](https://img.yulucn.com/upload/c/68/c68f62b52af60ddef7ff19f7bdca4b17_thumb.jpg)
+1)=
要使上式为定值须
![](https://img.yulucn.com/upload/e/b6/eb6500f97d133a6f0b478de64947afb9_thumb.jpg)
=1,解得 m =1,
∴
![](https://img.yulucn.com/upload/d/fb/dfb0ca28a24833b361b5d1796abd1163_thumb.jpg)
·
![](https://img.yulucn.com/upload/d/46/d467c48c90330e30cdb600243bb839e7_thumb.jpg)
为定值-2,
当直线 l 的斜率不存在时 P (1,
![](https://img.yulucn.com/upload/6/0c/60c6306b4ebd99307c9f72b25684cfda_thumb.jpg)
), Q (1,-
![](https://img.yulucn.com/upload/6/0c/60c6306b4ebd99307c9f72b25684cfda_thumb.jpg)
),
由 E (1,0)可得
![](https://img.yulucn.com/upload/2/27/2277b58236018b5df38eda2d342c1424_thumb.jpg)
=(0,-
![](https://img.yulucn.com/upload/6/0c/60c6306b4ebd99307c9f72b25684cfda_thumb.jpg)
),
![](https://img.yulucn.com/upload/d/46/d467c48c90330e30cdb600243bb839e7_thumb.jpg)
=(0,
![](https://img.yulucn.com/upload/6/0c/60c6306b4ebd99307c9f72b25684cfda_thumb.jpg)
),
∴
![](https://img.yulucn.com/upload/2/27/2277b58236018b5df38eda2d342c1424_thumb.jpg)
·
1年前
1