|
得(1+4k2)x2+8k(1-k)x+4(1-k2)-16=0
设A(x1,y1),B(x2,y2),则x1+x2=
8k(k-1) |
1+4k2 |
而M(1,1)是AB中点,则
x1+x2 |
2 |
综上,得
8k(k-1) |
1+4k2 |
1 |
4 |
∴直线AB的方程为y-1=-
1 |
4 |
(2)设弦AB的中点为P(x,y)
∵A,B,M,P四点共线,
∴kAB=kMP
即(-
1 |
4 |
x1+x2 |
y1+y2 |
y-1 |
x-1 |
∴(-
1 |
4 |
2x |
2y |
y-1 |
x-1 |
x2 |
16 |
y2 |
4 |
|
8k(k-1) |
1+4k2 |
x1+x2 |
2 |
8k(k-1) |
1+4k2 |
1 |
4 |
1 |
4 |
1 |
4 |
x1+x2 |
y1+y2 |
y-1 |
x-1 |
1 |
4 |
2x |
2y |
y-1 |
x-1 |