A：0=-2x+4 x=2 A(2,0)
B：y=-2*0+4=4 B(0,4)
C：-2x+4=2/x
x²-2x+1=0
(x-1)²=0
x=1
y=2/1=2
C(1,2)
设M(m,0)
kCM=(2-0)/(1-m)=2/(1-m)
∵CM⊥CN
∴kCN=(m-1)/2
CN方程：y-2=(m-1)/2*(x-1)
y=2+(m-1)/2*(x-1)
N：x=0
y=2+(m-1)/2*(0-1)=2+(1-m)/2=(5-m)/2
N(0,(5-m)/2)
BN²+AM²=(√((0-0)²+(4-(5-m)/2)²))²+(√((2-m)²+(0-0)²))²
=(9+6m+m²)/4+4-4m+m²
=(5m²-10m+25)/4
MN²=(√((m-0)²+(0-(5-m)/2)²))²
=m²+(25-10m+m²)/4
=(5m²-10m+25)/4
MN²/(BN²+AM²)=((5m²-10m+25)/4)/((5m²-10m+25)/4)=1
可见,无论M在何处,MN²/(BN²+AM²)均等于定值1