已知抛物线y2=4x,点M(1,0)关于y轴对称点为N,直线L过点M交抛物线于AB两点.(1)证明:NA,NB的斜率互为相反数;(2)求△ANB面积最小值(3)第三问不要求过程若M(m,0)时,(1)是否仍成立?△A
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/18 12:55:24
![已知抛物线y2=4x,点M(1,0)关于y轴对称点为N,直线L过点M交抛物线于AB两点.(1)证明:NA,NB的斜率互为相反数;(2)求△ANB面积最小值(3)第三问不要求过程若M(m,0)时,(1)是否仍成立?△A](/uploads/image/z/8688281-41-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFy2%3D4x%2C%E7%82%B9M%EF%BC%881%2C0%EF%BC%89%E5%85%B3%E4%BA%8Ey%E8%BD%B4%E5%AF%B9%E7%A7%B0%E7%82%B9%E4%B8%BAN%2C%E7%9B%B4%E7%BA%BFL%E8%BF%87%E7%82%B9M%E4%BA%A4%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%BA%8EAB%E4%B8%A4%E7%82%B9.%EF%BC%881%EF%BC%89%E8%AF%81%E6%98%8E%EF%BC%9ANA%2CNB%E7%9A%84%E6%96%9C%E7%8E%87%E4%BA%92%E4%B8%BA%E7%9B%B8%E5%8F%8D%E6%95%B0%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E2%96%B3ANB%E9%9D%A2%E7%A7%AF%E6%9C%80%E5%B0%8F%E5%80%BC%EF%BC%883%EF%BC%89%E7%AC%AC%E4%B8%89%E9%97%AE%E4%B8%8D%E8%A6%81%E6%B1%82%E8%BF%87%E7%A8%8B%E8%8B%A5M%EF%BC%88m%2C0%EF%BC%89%E6%97%B6%2C%EF%BC%881%EF%BC%89%E6%98%AF%E5%90%A6%E4%BB%8D%E6%88%90%E7%AB%8B%3F%E2%96%B3A)
已知抛物线y2=4x,点M(1,0)关于y轴对称点为N,直线L过点M交抛物线于AB两点.(1)证明:NA,NB的斜率互为相反数;(2)求△ANB面积最小值(3)第三问不要求过程若M(m,0)时,(1)是否仍成立?△A
已知抛物线y2=4x,点M(1,0)关于y轴对称点为N,直线L过点M交抛物线于AB两点.
(1)证明:NA,NB的斜率互为相反数;
(2)求△ANB面积最小值
(3)第三问不要求过程
若M(m,0)时,(1)是否仍成立?△ANB面积最小值又是多少?
已知抛物线y2=4x,点M(1,0)关于y轴对称点为N,直线L过点M交抛物线于AB两点.(1)证明:NA,NB的斜率互为相反数;(2)求△ANB面积最小值(3)第三问不要求过程若M(m,0)时,(1)是否仍成立?△A
N(-1,0)
直线L:x=ty+1,与抛物线y2=4x联立后得
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)
当t=0,Smin=4
(3)若M(m,0)时,(1)仍成立
直线L:x=ty+m,与抛物线y2=4x联立后得
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,
当t>0,u'>0,当t