設 P(x,y), Q(x,-1)
∵QP*FQ-FP*FQ=0
∴(0,y+1)●(-x,2)-(x,y-1)●(x,-2)=0
∴2(y+1)-(x²-2y+2)=0
∴軌跡為C：x²=4y
設M（t,t²/4),|MD|²=t²+(2-t²/4)²
圓M:(x-t)²+(y+t²/4)²=t²+(2-t²/4)²
令y=0,得 （x-t)²=4,x=t±2
∴A(t-2,0),B(t+2,0)
l1=√(t²-4t+8),l2=√(t²+4t+8)
∴l(xiāng)1/l2+l2/l1=(l1²+l2²)/(l1l2)
=[(t²-4t+8)+ (t²+4t+8)]/ √[(t²+4t+8)(t²-4t+8)]
=(2t²+16)/√[(t²+8)²-16t²]
=(2t²+16)/√(t⁴+64 )
=2√[(t²+8)²/(t⁴+64)]
=2√[(t⁴+64+16t²)/(t⁴+64)]
=2√[1+16t²/(t⁴+64)]
=2√[1+16/(t²+64/t²)]
∵t²+64/t²≥2√64=16∴
∴1+16/(t²+64/t²)≤2
∴2√[1+16/(t²+64/t²)]≤2√2
選項B
在直三棱柱ABC-A1B1C1中,∠BAC=π/2,AB=AC=AA1=1,已知G和E分別為A1B1和CC1的中點,D與F分別為線段AC和AB上的動點(不包括端點),若GD⊥EF,則線段DF的長度的最小值()
A．√5/5 B．1C．2√5/5D．3√5/5
以A為原點,AB,AC,AA1為x,y,z軸j建立坐標系
D(0,y,0)F(x,0,0) ,G(1/2,0,1),E(0,1,1/2)
向量GD=(-1/2,y,-1),向量EF=（x,-1,-1/2)
∵GD⊥EF∴-x/2-y+1/2=0
∴y=1/2-x/2(0
=5/4(x²-2/5x)+1/4=5/4(x-1/5)²+1/5≥1/5
∴|DF|min=√5/5
答 A．√5/5