椭圆x^2/16+y^2/4=1上的点到直线x+2y-根号2=0的最大距离是( )?

Question

如题所述

Answer 1

设直线l:x＋2y+k=0 ①
x^2/16+y^2/4=1②
将①代入②得8y2+4ky+k^2-16=0③
③式△=0，即16k^2-4*8(k^2-16)=0 得k=±4√2
-√2-4√2=-5√2 ， -√2+4√2=-3√2
∴直线x＋2y-4√2=0 与椭圆的交点与直线L距离最大，为
（-5√2 ）/√5=√10

Answer 2

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