过抛物线y^2=2px(>0)的对称轴上的定点M（m,0)作直线AB与抛物线相交与A,B两点1试证明A,B两点的纵坐标之积为定值2若点N是定直线l:x=-m上的任一点,试探索三条直线AN,MN,BN的斜率之间的关系,并给出证

Answer 1

1.直线方程设为
y=k(x-m)
带入抛物线方程
k²(x-m)²=2px
整理
k²x²-2(mk²+p)x+k²m²=0
根据韦达定理得到x1x2=m²
y1y2=-2p√x1x2=-2pm
所以A,B两点的纵坐标之积为定值
(2)A(x1,y1) B(x2,y2) M(m,0) N(-m,n)
Kan=(y1-n)/(x1+m)
Kmn=-n/2m
Kbn=(y2-n)/(x2+m)
Kan+Kbn=(y1-n)/(x1+m)+(y2-n)/(x2+m)
=(y1-n)/(y1²/2p+m)+(y2-n)/(y2²/2p+m)
=2p[(y1-n)/(y1²+2pm)+(y2-n)/(y2²+2pm)(将2pm换成-y1y2)
=2p[(y1-n)/(y1²-y1y2)+(y2-n)/(y2²-y1y2)
=2p[(y1-n)y2-(y2-n)y1]/[y1y2(y1-y2)]
=2p[(y1-y2)n]/[y1y2(y1-y2)]
=2pn/(y1y2)(将y1y2换成-2pm)
=-n/m
=2Kmn
上面论证了Kan+Kbn=2Kmn
所以Kan,Kmn,Kbn构成等差数列