平面解析几何随m,n 变化 .坐标原点到直线(2m+n)x+(3m-n)y-m+2n=0 距离的最大值是

平面解析几何随m,n 变化 .坐标原点到直线(2m+n)x+(3m-n)y-m+2n=0 距离的最大值是

d=|m-2n|/(13m^2+2n^2-2mn)^(1/2)

=|1-2n/m|/[13+2(n/m)^2-2n/m]^(1/2)

令n/m=x,x可以取任意实数

d=|1-2x|/(13+2x^2-2x)^(1/2)

=[(4x^2-4x+1)/(2x^2-2x+13)]^(1/2)

={2-25/[2(x-1/2)^2+25/2]}^(1/2)

显然2(x-1/2)^2+25/2越大,d就越大,x取无穷时,d的极限值为根号2

(2m+n)x+(3m-n)y-m+2n=0 即

(2m/n+1)x+(3m/n-1)y-m/n+2=0

m/n=0

所以方程为x-y+2=0

最大值为根号2