在椭圆7x2+4y2=28上求一点,使它到直线l:3x-2y-16=0的距离最短,并求出这一

在椭圆7x2+4y2=28上求一点,使它到直线l:3x-2y-16=0的距离最短,并求出这一最短距离.

答案

:把椭圆方程化为=1的形式,

则可设椭圆上点A坐标为(2cosα,sinα),

A到直线l的距离为(其中β=arcsin).

∴当β-α=时,d有最小值,最小值为.

此时α=β-,∴sinα=-cosβ=-,cosα=sinβ=.

A点坐标为().

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