已知直线的参数方程x=-1+根号3t y=2-t,求直线被曲线x^2-y^2=-3+4根号3所截的弦长及弦中点坐标

代入1-2√3t+3t²-4+4t-t²=-3+4√3t²+(2-√3)t-2√3=0t1+t2=√3-2t1t2=-2√3(t1-t2)²=(t1+t2)²-4t1t2=7+4√3=(2+√3)²(x1+x2)/2=(-1+√3t1-1+√3t2)/2=-1+√3(t1+t2)/2=(1-2√3)/2(y1+...