21.已知双曲线的中心在原点,右顶点为A(1,0),点P、Q在双曲线的右支

21.已知双曲线的中心在原点,右顶点为A(1,0),点PQ在双曲线的右支上,点Mm,0)到直线AP的距离为1.

(Ⅰ)若直线AP的斜率为k,且|k|∈[

,],求实数m的取值范围;

(Ⅱ)当m =

+1时,△APQ的内心恰好是点M,求此双曲线的方程.

答案

21.本题主要考查直线、双曲线方程和性质等基础知识,考查解析几何的基本思想方法和综合解题能力.

解:(Ⅰ)由条件得直线AP的方程y=kx-1),

kxyk=0.

因为点M到直线AP的距离为1,

=1,

即|m-1|=

=.

∵|k|∈[

,],

≤|m-1|≤2,

解得

+1≤m≤3或-1≤m≤1-.

m的取值范围是[-1,1-

]∪[1+,3].

(Ⅱ)可设双曲线方程为x2

=1(b≠0),

M

+1,0),A(1,0),得|AM|=.

又因为M是△APQ的内心,MAP的距离为1,所以∠MAP=45°,

直线AM是∠PAQ的角平分线,且MAQPQ的距离均为1.

因此,kAP=1,kAQ=-1,(不妨设P在第一象限)

直线PQ方程为x=2+

.直线AP的方程y=x-1,

∴解得P的坐标是(2+

,1+).

P点坐标代入x2

=1得

b2=

.

所以所求双曲线方程为x2

y2=1,

x2-(2

-1)y2=1.

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