过点P(-1,1)作直线与椭圆x^2/4+y^2/2=1交于A,B两点若线段AB的中点恰为P点求AB所在直线的方程和线段AB的长度？

设A,B的坐标为：(xa,ya),(xb,yb);
AB的斜率k, AB的中点为P, 过AB的直线方程为：
y = k(x+1) +1;
其中k = (ya-yb)/(xa-xb);
代入椭圆方程：
(1+2k^2)x^2 + 4k(k+1)x + 2(k+1)^2 - 4 = 0;
xa, xb 是其两解；
xa + xb = -4k(k+1)/(1+2k^2);
xa*xb = (2(k+1)^2-4)/(1+2k^2);

|AB|^2 = (xa-xb)^2 + (ya-yb)^2
= (xa-xb)^2(1+k^2)
= [(xa+xb)^2 - 4xa*xb](1+k^2);

利用题设求k值；
xa^2/4 + ya^2/2 = 1,
xb^2/4 + yb^2/2 = 1,

(xa + xb)/2 = -1,
(ya + yb)/2 = 1,

(xa+xb)(xa-xb)/4 + (ya+yb)(ya-yb)/2 = 0;
k = (ya-yb)/(xa-xb) = -((xa+xb)/4)/((ya+yb)/2)
= -((xa+xb)/2)/(ya+yb)
= -(-1)/2 = 1/2;

过AB的直线为：
y = (1/2)(x+1) +1, x - 2y + 3 =0;

|AB|^2 = 20/3, |AB| = 2*sqrt(15)/3。