写成参数方程
x=4cosa
y=2sina
则到直线距离=|4cosa+4sina-√2|/√(1²+2²)
=|4√2sin(a+π/4)-√2|/√5
-1<=sin(a+π/4)<=1
-4√2-√2<=4√2sin(a+π/4)-√2<=4√2-√2
即-5√2<=4√2sin(a+π/4)-√2<=3√2
所以0<=|4√2sin(a+π/4)-√2|<=5√2
所以最大距离=5√2/√5=√10