求点M（m,0)(m≠±√2)与椭圆x^2+2y^2=2上的点的距离的最小值.

求点M（m,0)(m≠±√2)与椭圆x^2+2y^2=2上的点的距离的最小值.

答：一、推导距离表达式设距离是d,则：d&sup2;最小的条件既是d最小的条件,下面求d&sup2;表达式：d&sup2; = (x -m)&sup2; + (y - 0)&sup2;d&sup2; = x&sup2; -2mx + m&sup2; + y&sup2;由x&sup2; + 2y&sup2; = 2 可得 y&sup2; = 1 - x&sup2;/2d&sup2; = x&sup2; -2mx + m&sup2; + 1 - x&sup2;/2d&sup2; = x&sup2;/2 - 2mx + m&sup2; + 1d&sup2; = 1/2(x&sup2; -4mx) + m&sup2; + 1d&sup2; = 1/2(x -2m)&sup2; -2m&sup2; + m&sup2; + 1d&sup2; = 1/2(x -2m)&sup2; +1 -m&sup2;二、讨论距离最小条件第①种情况：x - 2m = 0：d&sup2; = 1 - m&sup2;由于 -√2 <= x <= √2,所以：-√2/2 <= m <= √2/2即:|m| <= √2/2第②种情况： |m| > √2/2:这时,x - 2m ≠ 0,如d&sup2;要求最小,则要求(x - 2m)&sup2;最小,所以要求|x-2m|最小.当m>+√2/2时,2m-x >0,所以|x-2m|=2m-x,这个值最小则要求x最大,x=√2,即：2m -√2当m<-√2/2时,-2m + x>0,所以|x-2m|=-2m+x,这个值最小则要求x最小,x=-√2,即：-2m +√2总之两种情况：(x - 2m)&sup2; = (2m - √2)&sup2;d&sup2; = 1/2(x -2m)&sup2; +1 - m&sup2; d&sup2; = 1/2(2m - √2)&sup2; +1 - m&sup2;d&sup2; = 2m&sup2; -2√2m + 1 + 1 -m&sup2;d&sup2; = m&sup2; -2√2m + 2d&sup2; = (m - √2)&sup2;综上：① 当 |m| <=√2/2时,最小距离d=√(1 - m&sup2;)② 当 |m| >√2/2时,最小距离d=|m - √2|---完---