(1)设点M(x,y),P(x0,y0),则由题意知P0(x0,0). 由=(x0-x,-y),=(0,-y0),且=,得 (x0-x,-y)= (0,-y0). ∴于是 又+=4,∴x2+y2=4.∴点M的轨迹C的方程为=1. (2)设A(x1,y1),B(x2,y2).联立 得(3+4k2)x2+8mkx+4(m2-3)=0. ∴Δ=(8mk)2-16(3+4k2)(m2-3)>0, 即3+4k2-m2>0.(*)且 依题意,k2=,即k2=. ∴x1x2k2=k2x1x2+km(x1+x2)+m2. ∴km(x1+x2)+m2=0,即km+m2=0. ∵m≠0,∴k+1=0,解得k2=. 将k2=代入(*),得m2<6.∴m的取值范围是(-,0)∪(0,). |