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问题 解答题

已知a是实数，直线2x-y+5=0与直线x-y+a+4=0的交点不在椭圆x2+2y2=11上，求a的取值范围．

答案

两条直线的交点即方程组

2x-y+5=0
x-y+a+4=0
的解，

此时（x，y）=（a-1，2a-3）．

该点不在椭圆x2+2y2=11上，

当且仅当（a-1）2+2（2a-3）2=11解得a=-2，或a=-

4
9
，

∴a≠-2且a≠-

4
9
．

∴a的取值范围是（-∞，-2）∪（-2，-

4
9
）∪（-
4
9
，+∞）．

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