x²/4²+y²/2²=1
则设点是(4cosa,2sina)
所以距离d=|4cosa+4sina-√2|/√(1²+2²)
4cosa+4sina-√2
=4√2sin(a+π/4)-√2
则-5√2<=4cosa+4sina-√2<=3√2
0<=d<=5√2/√5
所以最大距离5√2/√5=√10