抛物线与直线关系

来源:百度知道 编辑:UC知道 时间:2024/05/11 00:46:22
如图,o为坐标原点,直线l在x轴和y轴上的截距分别是a和b(a>0,b不等于0),且交抛物线y2=2px(p>0)于M(x1,y1)N(x2,y2)两点。证明1/y1+1/y2=1/b;当a=2p时,求∠MON的大小。

l方程: x/a+y/b=1
x=a-ay/b
代人y^2=2px得:
y^2=2pa-2pay/b
y^2+2pay/b-2pa=0
y1+y2=-2pa/b,y1y2=-2pa
所以,
1/y1+1/y2=(y1+y2)/y1y2
=(-2pa/b)/(-2pa)
=1/b

tan∠MON=(Kmo+Kno)/(1-Kmo*Kno)
=(y1/x1+y2/x2)/(1-y1y2/x1x2)
=(2p/y1+2p/y2)/(1-y1y2/x1x2)
=2p(1/y1+1/y2)/(1-y1y2/x1x2)
=2p/b(1-y1y2/x1x2)
=2p/b(1-y1y2/(y1^2*y2^2/4p^2))
=2p/b(1-4p^2/y1y2)
=2p/b(1-4p^2/(-2pa))
=2p/b(1+2p/a)
=2p/b(1+2p/2p)
=2p/2b
=p/b

(1)设直线l的方程截距式为x/a+y/b=1,代入抛物线方程,消去x,得
by²+2pay-2pab=0
y1+y2=-2pa/b
y1y2=-2pa
1/y1+1/y2=(y1+y2)/y1y2=-2pa/b÷(-2pa)=1/b

(2)设l的方程,x-2p=my → (x-my)/2p=1,代入抛物线方程
y²=2px·(x-my)/2p
y²+mxy-x²=0,x≠0,两边同时除以x²,得
(y/x)²+m(y/x)-1=0
y1y2/x1x2=-1
向量OM·向量ON=x1x2+y1y2=0
OM⊥ON,
∠MON=90°