直线l在双曲线x^2\3-Y^2\2=1上截得弦长为4,其斜率为2,求直线l在y轴上的截距m

设斜率为2的直线方程是y=2x+m代入双曲线方程得x^2/3-(2x+m)^2/2=1 两边乘62x^2-3(4x^2+4mx+m^2)=62x^2-12x^2-12mx-3m^2=610x^2+12mx+3m^2+6=0x1+x2=-12m/10=-6m/5x1x2=(3m^2+6)/10(x1-x2)^2=(x1+x2)^2-4x1x2=(-6m/5)...