直线y=kx(k>0)与双曲线y=4/x交与a(x1,y1),(x2,y2)两点,求2x1y2-7x2y1
问题描述:
直线y=kx(k>0)与双曲线y=4/x交与a(x1,y1),(x2,y2)两点,求2x1y2-7x2y1
答
解:2x1y2-7x2y1=2kx1x2-7kx1x2=-5kx1x2 联立y=kx y=4/x kx^2-4=0 x1x2=-4/k
-5kx1x2=20
答
kx=4/x
x²=4/k
x1=2√k/k,x2=-2√k/k
y1=2√k,y2=-2√k
2x1y2-7x2y1=-8+28=20