设直线为 y=kx+2带入得x²+(kx+2)²=1(1+k²)x²+4kx+3=0判别式△=16k²-12(1+k²)=0得 4k²=12 k=√3 或 k=-√3所以切线方程为y=√3x+2或y=-√3x+2
