N(-1,0)
直線L:x=ty+1,與拋物線y2=4x聯(lián)立后得
y^2-4ty-4=0,
y1+y2=4t,y1y2=-4
(1)kNA+kNB=y1/(y1^2/4 + 1) +y2/(y2^2/4 + 1)
=[1/4y1y2^2+1/4y1^2y2+y1+y2]/(y1^2/4 + 1)(y2^2/4 + 1)
=(y1y2/4 +1)(y1+y2)/(y1^2/4 + 1)(y2^2/4 + 1)
=(-1+1)(y1+y2)/(y1^2/4 + 1)(y2^2/4 + 1) =0
(2)S=1/2*|AB|*d
d=|-2|/√(1+t^2)=2/√(1+t^2)
|AB|=√(1+t^2)|y1-y2|=√(1+t^2)*√[(y1+y2)^2-4y1y2]
=√(1+t^2)*√16(1+t^2)
=4(1+t^2)
S=1/2*|AB|*d
=1/2*4(1+t^2)*2/√(1+t^2)
=4√(1+t^2)
當(dāng)t=0,Smin=4
(3)若M（m,0）時(shí),（1）仍成立
直線L:x=ty+m,與拋物線y2=4x聯(lián)立后得
y^2-4ty-4m=0,
y1+y2=4t,y1y2=-4m
(1)kNA+kNB=y1/(y1^2/4 + m) +y2/(y2^2/4 + m)
=[1/4y1y2^2+1/4y1^2y2+my1+my2]/(y1^2/4 + m)(y2^2/4 + m)
=(y1y2/4 +m)(y1+y2)/(y1^2/4 + 1)(y2^2/4 + 1)
=(-m+m)(y1+y2)/(y1^2/4 + m)(y2^2/4 + m) =0
(2)S=1/2*|AB|*d
d=|-2m|/√(1+t^2)=|2m|/√(1+t^2)
|AB|=√(1+t^2)|y1-y2|=√(1+t^2)*√[(y1+y2)^2-4y1y2]
=√(1+t^2)*√16(m+t^2)
S=1/2*|AB|*d
=1/2*√(1+t^2)*√16(m+t^2)*|2m|/√(1+t^2)
=|m|*√16(m+t^2)
=4√m^2(m+t^2)
令u=m^2(m+t^2),u'=2m^2*t=0,
當(dāng)t>0,u'>0,當(dāng)t