Question

设点M(2,3)N(1,0),直线x=ty+1与抛物线y^2=4x交于AB两点,若t∈[0,1],则△AMN与△BMN的

Answer 1

x=ty+1,①

y^2=4x,②

把①代入②,y^-4ty-4=0,

y1y2=-4,y1=2t-2√(t^+1),

MN:3x-y-3=0,

点(x1,y1)到MN的距离d1=|3(ty1+1)-y1-3|/√10

=|（3t-1)y1|/√10,

点(x2,y2)到MN的距离d2=|（3t-1)y2|/√10,

∴△AMN与△BMN的面积之比=d1/d2=|y1/y2|=|y1^/(y1y2)|

=[√(t^+1)-t]^,↓,

t=1时它取最小值（√2-1）^=3-2√2.