Question

已知双曲线x^2-y^2/2=1截直线y=x+m所得弦长4√2,求m的值

Answer 1

设弦与双曲线的交点ＡＢ坐标分别为(x1,y1)(x2,y2) 有：y1=x1+m y2=x2+m 把y=x+m代入双曲线方程化简得：x^2-2mx-m^2-2=0 ＡＢ的横坐标是上方程的两根,由韦达定理有：X1+x2=2m x1X2=-m^2-2 |AB|^2=(x1-x2)^2+(y1-y2)^2 =(x1-x2)^2+(x1-x2)^2 =(x1+x2)^2-4x1x2 =4m^2+4m^2+8 =８m^2+8 |AB|^2=(４√２)^2=32 故：８m^2+8=32 m^2=3 m=√３或－√３