过抛物线y2=2px(p>0)上一定点P(x0,y0)(y0>0)作两条直线分别交抛物线于A

过抛物线y2=2px(p>0)上一定点P(x0,y0)(y0>0)作两条直线分别交抛物线于A(x1,y1)、B(x2,y2).

(1)求该抛物线上纵坐标为的点到其焦点F的距离;

(2)当PA与PB的斜率存在且倾斜角互补时,求的值,并证明直线AB的斜率是非零常数.

答案

(1) 点M(,)到F的距离为-(-)=.

(2)证明见解析


解析:

(1)当y=时,x=.

又抛物线y2=2px(p>0)的准线方程为x=-,

则点M(,)到F的距离为-(-)=.

(2)设直线PA的斜率为kPA,直线PB的斜率为kPB.

y12-y02=2p(x1-x0),

则kPA=(x1≠x0).

同理,得kPB=(x2≠x0).

由PA、PB的倾斜角互补知kPA=-kPB,

=-,

即y1+y2=-2y0,故=-2.

设直线AB的斜率为kAB.

y12-y22=2p(x1-x2),

∴kAB=(x1≠x2).

将y1+y2=-2y0(y0>0)代入上式得

kAB=.(P(x0,y0)为一定点,y0>0)

则kAB=-为非零常数.

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