易知,直线方程为y=ax+2.与椭圆方程联立可得,(1+3a²)x²+12ax+12-3a²=0.由题设可直,⊿=(12a)²-4(1+3a²)(12-3a²)＞0.(a≠0)===>(a²-1)(3a²+4)＞0.===>a²＞1.===>a＜-1,或...