求直线x=2+t,y=根号3*t（t为参数）被双曲线x^2-y^2=1上截得的弦长?

将直线的参数方程代入双曲线方程,整理后,得2t^2-4t-3=0,根据韦达定理可得 (t1-t2)^2=(t1+t2)^2-4t1t2=10 设截得的弦长为d,那么 d^2=(y1-y2)^2+(x1-x2)^2,将参数方程代入,可得d^2=3(t1-t2)^2+(t1-t2)^2=4(t1-t2)^2=40...