y^2/2-x^2/4=1 x2/2-y2=1渐近线y=±根号2x/2p(2,-2)在渐近线交叉的下方，所以可设双曲线方程是y^2/a^2-x^2/b^2=1(a>0 b>0) 所以4/a^2-4/b^2=1a/b=根号2/2 b^2=2a^2a^2=2 b^2=4
