已知抛物线y2=-2x,直线l经过点(-1,0),且与抛物线交于A、B,若|AB|=2根号6,则直线l的倾斜角为----

设直线的方程为y=kx+k代入y²=-2xk²(x+1)²=-2xk²x²+(2k²+2)x+k²=0设A B两点坐标分别为（x1,y1）（x2,y2）x1+x2=-(2k²+2)/k²x1x2=1(x1-x2)²=(x1+x2)²-4x1x2=4(k²+1)²/(k²k²)-4|AB|²=(x1-x2)²+（y1-y2)²=(1+k²)(x1-x2)²=(1+k²)[4(k²+1)²/(k²k²)-4]=24解得k=1或-1所以直线l的倾斜角为45°或者135°