题目内容

双曲线- =1(a>0,b>0)的离心率为AF分别是双曲线的左顶点、右焦点,过点F的直线l交双曲线的右支于PQ两点,交y轴于R点,APAQ分别交右准线于MN两点.

(1)若=5,求直线l的斜率;

(2)证明MN两点的纵坐标之积为-a2.

(1)解:设P(x1,y1),Q(x2,y2),因为双曲线的离心率为,?

所以c=a,b=a,双曲线方程为2x2-y2=2a2.                                                     ?

因为=5,所以x2=c.                                                                               ?

因为直线l:y=k(x-c),?

所以y2=-.                                                                                                   ?

Q是双曲线上一点,所以2()2-(-)2=2a2,                                                    ?

整理,得e2-e2k2=2,解得k.                                                           ?

(2)证明:设P(x1,y1),Q(x2,y2),?

由已知AP:y=(x+a),AQ:y=(x+a),?

所以ym=(+a),yn==(+a).                                                   ?

所以ymyn=·(+a)2=(+a)2.?

得(2-k2)x2+2k2cx-k2c2-2a2=0,?

所以x1+x2=,x1x2=,?

y1y2=k2(x1-c)(x2-c)=k2x1x2-c(x1+x2)+c2]=k2,                                       ?

x1x2+a(x1+x2)+a2=k2.                                                                                  ?

所以ymyn=·=-a2.      

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