直線y=kx+1(k∈R)與橢圓x25+y2m=1恒有公共點,則m的取值範圍是() A. [1,5)∪(5,+∞)B.(0,5)C. [1,+∞)D.(1,5)

直線y=kx+1(k∈R)與橢圓x25+y2m=1恒有公共點,則m的取值範圍是() A. [1,5)∪(5,+∞)B.(0,5)C. [1,+∞)D.(1,5)

聯立y=kx+1x25+y2m=1,消去y得到(m+5k2)x2+10kx+5-5m=0,(m>0,m≠5)∵直線y=kx+1(k∈R)與橢圓x25+y2m=1恒有公共點,∴△≥0,即100k2-20(1-m)(m+5k2)≥0,化為m2+5mk2-m≥0,∵m>0,∴m≥-5k2+1,∵…