过点p(0,2)的直线与y=ax²交于a,b两点,已知∠aob为90°
发布网友
发布时间:2024-10-11 22:58
我来回答
共1个回答
热心网友
时间:2024-10-15 02:18
y-2=kx
y=kx+2
kx+2=ax²
ax²-kx-2=0
x1x2=-2/a
x1+x2=k/a
y1y2
=(kx1+2)(kx2+2)
=k²x1x2+2kx1+2kx2+4
=k²x1x2+2k(x1+x2)+4
=k²(-2/a)+2kk/a+4
=-2k²/a+2k²/a+4
=4