Question

已知抛物线y^2=4x,M为其上面的动点，O(0,0),F为焦点，则MO/MF的最大值是？

Answer 1

设M(x,y)
MF=x+1
MO=√(x^2+y^2)=√(x^2+4x)
MO/MF=√(x^2+4x)/(x+1)
=√[(x^2+4x)/(x+1)^2]
=√[(x+1)^2+2(x+1)-3]/(x+1)^2
=√[-3/(x+1)^2+2/(x+1)+1]
=√-3[(1/(x+1)-1/3)^2+4/3]
所以x=2 最大值=2√3/3

Answer 2

设M(x,y) MO=√(X²+Y²)=√(X²+4X) MF=M到准线的距离=x+1
t=MO/MF=√(X²+4X)/(X+1)
t²=(x²+4x)/(x+1)²整理得
（1-t²)x²+(4-2t²)x-t²=0
△=(4-2t²)²+4t²(1-t²)
=16-12t²≥0
t²≤4/3
t≤2√3/3
所以MO/MF的最大值是2√3/3

追问

thanks!