也~~~直线y=kx+1与曲线3x2-y2=1相交,交点为A,B.当k为何值时,以AB为直线的圆过坐标原点
y=kx+1 (1)
3x^2-y^2=1 (2)
解方程组： (3-k^2)x^2-2kx-2=0
x=[2k±√(4k^2+24-8k^2)]/2(3-k^2)=[k±√(6-k^2)]/(3-k^2)
x1=[k+√(6-k^2)]/(3-k^2),y1=[3+k√(6-k^2)]/(3-k^2)
x2=[k-√(6-k^2)]/(3-k^2),y2=[3-k√(6-k^2)]/(3-k^2)
|AB|=√[(y1-y2)^2+(x1-x2)^2]=2/(3-k^2)*√[(k^2+1)(6-k^2)]
AB中点即圆心C坐标为(x1/2+x2/2,y...全部
y=kx+1 (1)
3x^2-y^2=1 (2)
解方程组： (3-k^2)x^2-2kx-2=0
x=[2k±√(4k^2+24-8k^2)]/2(3-k^2)=[k±√(6-k^2)]/(3-k^2)
x1=[k+√(6-k^2)]/(3-k^2),y1=[3+k√(6-k^2)]/(3-k^2)
x2=[k-√(6-k^2)]/(3-k^2),y2=[3-k√(6-k^2)]/(3-k^2)
|AB|=√[(y1-y2)^2+(x1-x2)^2]=2/(3-k^2)*√[(k^2+1)(6-k^2)]
AB中点即圆心C坐标为(x1/2+x2/2,y1/2+y2/2)即(k/(3-k^2),3/(3-k^2))
|OA|=1/(3-k^2)*√(k^2+9)
以AB为直径的圆过原点
则OA|=|AB|/2
1/(3-k^2)*√(k^2+9)=2/(3-k^2)*√[(k^2+1)(6-k^2)]/2
√(k^2+9)=√[(k^2+1)(6-k^2)]
k^2+9=(k^2+1)(6-k^2)
k^2+9=-k^4+5k^2+6
k^4-4k^2+3=0
k^2=1或k^2=3
当k^2=3时，y=kx+1与3x2-y2=1之有一个交点，舍
故k^2=1
k=±1
。
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