Question

双曲线x^2/a^2-y^2/b^2=1的离心率为2,坐标原点到直线AB的距离为根号3/2,其中A(0,-b).B(a,

Answer 1

1.AB：bx-ay-ab=0,原点到AB的距离d=ab/√(a^2+b^2)=√3/2,得a^2b^2=3/4*(a^2+b^2)

结合e^2=(a^2+b^2)/a^2=4,得b^2=3a^2,解得a^2=1,b^2=3

方程为：x^2 - y^2/3=1

2.x=1/2为双曲线的右准线,设M(m,n),则M到右准线的距离MM'=m-1/2,设P,Q在右准线的投影分别为P',Q',则PP'=PF/e

QQ'=QF/e,MM'为梯形PP'Q'Q的中位线,故2MM'=PP'+QQ'=(PF+QF)/e=PQ/e=5,MM'=5/2,得m=3

则MN=3-(-2)=5,而NQ垂直于MQ,MQ＝5