是以原点为圆心,c(c^2=a^2+b^2为半焦距)为半径的圆与双曲线的交点,满足方程x^2+y^2=a^2+b^2x^2-y^2*(a/b)^2=a^2故y^2*c^2/b^2=b^2y=+-(b^2/c)x=+-a*sqrt(1+b^2/c^2)在此题中,a^2=b^2=6带入计算可得出 x=+-3 ,y=+- sq...