ǰλãҳ >> >> 若直线y=kx-1与双曲线x2-y2=4始终有公共点,则k取值范围是

若直线y=kx-1与双曲线x2-y2=4始终有公共点,则k取值范围是

ĵ:

...ֱy=kx-1˫x2-y2=1֧,...

(2015•ͷУĩ)ֱy=kx-1˫x2-y2=,kȡֵΧ( )_𰸽_2015ѧ_һģ/ģ/ģ/_ͼ_ٶȸ߿

ֱy=kx-2˫x2-y2=4ûй,kȡֵΧ( )

ֱy=kx-2˫x2-y2=4ûй,kȡֵΧ( )_𰸽_2012ѧ_һģ/ģ/ģ/_ͼ_ٶȸ߿

...ֱ֪y=kx-1˫x2-y2=1֧ڲͬ,...

(2015•У¿)ֱ֪y=kx-1˫x2-y2=1֧ڲͬ,kȡֵΧ( )_𰸽_2015ѧ_һģ/ģ/ģ/_ͼ_...

ֱy=kx+1˫x2-=1ֻһ,kȡֵΧ___.

ֱy=kx+1˫x2-=1ֻһ,kȡֵΧ___. ȷ𰸼ؽ ȷ {-2,2,,} :,:(4-k2)x2-2kx-5=0 ....

ֱ֪y=kx+1˫x2-2y2=1ҽһ,...

ֱ֪y=kx+1˫x2-2y2=1ҽһ,ʵkֵ( )_𰸽_2016ѧ_һģ/ģ/ģ/_ͼ_ٶȸ߿

(2015•ϽУĩ)ֱy=kx+4+2k...

•ϽУĩ)ֱy=kx+4+2k,kȡֵΧ(...D(-,-1]ȷ𰸼ؽ ȷ B : x2+y2=4,...

ֱy=kxֱx=1,x=2,y=1,y=2Χɵй...

ֱy=kxֱx=1,x=2,y=1,y=2Χɵй,kȡֵΧ___. ȷ𰸼ؽ ȷ k2y=kx...

ֱy=kx+1Բx2+4y2=1ҽһ,kֵΪ___.

ֱy=kx+1Բx2+4y2=1ҽһ,kֵΪ___. ȷ𰸼ؽ ȷ y=kx+1x2+4y2=1(1+4k2)x2+8kx+3=0...

ֱy=kx+1˫C:x2-y2=1ֻ֧һ,kȡֵΪ( )

:ֱ֪y=kx+1˫C:x2-y2=1ڵֻ֧һ,ɵõĺС0. ѷ̢ٴ,÷(1-k2)x2-2kx-2=0ǡһ...

ֱy=kx+2y2=4xһ,ʵk=___.

ѧ ֱԲ׶Ĺϵ ֱy=kx+2y2=4xһ,ʵk=___. ȷ𰸼ؽ ȷ 0, :бk=0 ʱ...
رǩ:
վͼ

ĵϹ nexoncn.com copyright ©right 2010-2020
ĵϹ磬ַϵͷemail:zhit325@126.com