Question

抛物线y=ax²(a＞0)与直线y=kx+b(k≠0)有两个公共点，其横坐标分别是x₁、x₂.而直线y=kx+b与x轴交点的横坐标是x₃,则x₁、x₂、x₃之间的关系是

A.x₃=x₁+x₂                                                                                                            

B.x₃=

C.x₁x₃=x₁x₂+x₂x₃                                                                                             D.x₁x₂=x₁x₃+x₂x₃

Answer 1

解法一:(特值法)取a=1,k=1,b=0,

则x₁=0,x₂=1,x₃=0, 可排除A、B.

再取a=1,k=1,b=1,可得x₁+x₂=1,x₁x₂=－1,x₃=－1.检验C、D可知D选项适合.

解法二:(直接法)把y=kx+b代入y=ax²，得

ax²－kx－b=0,x₁+x₂=,x₁x₂=－.

又x₃=－,∴x₁x₂=(x₁+x₂)x₃.

答案:D