曲线x=根号y^2+1
是双曲线 x²-y²=1 的右支
组成方程组
x²-(kx+1)²=1
(1-k²)x²-2kx-2=0
△=b²-4ac= 4k²+8(1-k²)>0
解得 k²<2
-√2<k<√2有,是一道选择题修改好了不是耶,是-√2<k<-1我开始选的是-√2<k<√2 ,错了= =补充x²-(kx+1)²=1(1-k²)x²-2kx-2=0△=b²-4ac= 4k²+8(1-k²)>0解得 k²<2-√2<k<√2 因为是双曲线 x²-y²=1 的右支,故x>0f(x)=(1-k²)x²-2kx-2根据韦达定理 x1+x2= 2k/ (1-k²)>0,x1*x2= 2/(k²-1)>0解得 k< -1故-√2<k<-1k< -1求不出,卡住了
若直线y=kx+1与曲线x=根号y^2+1有两个不同的交点,则k的取值范围
若直线y=kx+1与曲线x=根号y^2+1有两个不同的交点,则k的取值范围
数学人气:678 ℃时间:2019-09-22 09:49:57
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