直线y=x+m与双曲线2x2-y2=2交于a,b两点,若以ab为直径的圆经过原点,求m的值?

Question

直线y=x+m与双曲线2x2-y2=2交于a,b两点,若以ab为直径的圆经过原点,求m的值?

Answer 1

把y = x + m和2x2 – y2 = 2联立可得2x2 – (x + m)2 = 2 => x2 – 2mx – m2 – 2 = 0,Δ=4m2 + 4(m2 - 2) = 8(m2 - 1)>0 => m < -1或者m > 1,设A(x1,y1),B(x2,y2),由韦达定理可得x1 + x2 = 2m,x1x2 = -m2 – 2,而y1y2 = (x1 + m)(x2 + m) = x1x2 + m(x1 + x2) + m2 = -m2 – 2 + m·2m + m2 = 2m2 – 2,因为以AB为直径的圆经过原点O,所以OA⊥OB,向量OA⊥向量OB,向量OA·OB = 0 =>（x1,y1）·（x2,y2）= x1x2 + y1y2 = 0 => -m2 – 2 + 2m2 – 2 = 0 => m2 – 4 = 0 => m2 = 4 => m =±2 .