设x=2cost,y=sint
m=(y-3)/(x-4)=(sint-3)/(2cost-4)
2mcost-4m=sint-3
2mcost-sint= 4m-3
√（4m^2+1)*[2m/√（4m^2+1) *cost-1/√（4m^2+1)*sint]=4m-3
令2m/√（4m^2+1)=sinu,1/√（4m^2+1)=cosu
所以√（4m^2+1)*（sinucost-cosusint)=4m-3
√（4m^2+1)*sin(u-t)=4m-3
sin(u-t)(4m-3)/√（4m^2+1)
|sin(u-t)|<=1
所以 |(4m-3)/√(4m^2+1)|<=1
解得： 1-(√3)/3<= m<=1+(√3)/3赞同1| 评论
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