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=4ûй,ʵkȡ...

ֱ֪y=kx-1˫x2-y2=4ûй,ʵkȡֵΧΪ___. ȷ𰸼ؽ ȷ :,ֱy=kx-1˫x2-y2=4,...

ֱ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=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+1˫x2-2y2=1ҽһ,...

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

ֱ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˫x2-y2=1ֻһ,kȡ...

ѡ ѧ ֱԲ׶ߵĹϵ ֱy=kx-1˫x2-y2=1ֻһ,kȡֵ( ) A1 B C-1 D ȷ𰸼ؽ ...

C:x2-y2=1ֱ֧y=kx+1ֻһ,k...

C:x2-y2=1ֱ֧y=kx+1ֻһ,kȡֵΧ ___. ȷ𰸼ؽ ȷ -1<k1,k= :ֱ˫...

ֱy=kx+2y2=4xһ,ʵk=___.

ѧ Բ׶ۺ ֱy=kx+2y2=4xһ,ʵk=___. ȷ𰸼ؽ ȷ бk=0 ʱ,ֱy=kx+2ƽx...
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