过(0,-2)的直线与抛物线y^2=8x交于AB两点,若线段AB中点的纵坐标为2,求|AB|

显示全部楼层 · 2012-8-5 00:05:38

设直线斜率为k直线方程：y=kx-2设A(x1,y1)
B(x2,y2)则y1+y2=4
y^2=8xy=kx-2
x=y/k+2/ky^2=8y/k+16/ky^2-8y/k-16/k=0y1+y2=8/k=4
k=2 直线方程 y=2x-2
y^2=4x^2-8x+4y^2=8x4x^2-8x+4=8xx^2-4x+1=0x1+x2=4
x1x2=1|AB|=√(1+k^2)*|x1-x2|=√(1+k^2)*√[(x1+x2)^2-4x1x2]=√5*√12=2√15...

千问 · 2012-8-5 00:05:38

那个，你确定这直线与抛物线有交点？？？把X=0带进去会发现不成立哦...