Question

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.

Answer 1

答：一、推导距离表达式设距离是d,则：d²最小的条件既是d最小的条件,下面求d²表达式：d² = (x -m)² + (y - 0)²d² = x² -2mx + m² + y²由x² + 2y² = 2 可得 y² = 1 - x²/2d² = x² -2mx + m² + 1 - x²/2d² = x²/2 - 2mx + m² + 1d² = 1/2(x² -4mx) + m² + 1d² = 1/2(x -2m)² -2m² + m² + 1d² = 1/2(x -2m)² +1 -m²二、讨论距离最小条件第①种情况：x - 2m = 0：d² = 1 - m²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d²要求最小,则要求(x - 2m)²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m0,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)² =(2m - √2)²d² = 1/2(x -2m)² +1 - m² d² = 1/2(2m - √2)² +1 - m²d² = 2m² -2√2m + 1 + 1 -m²d² = m² -2√2m+ 2d² = (m - √2)²综上：① 当 |m| <=√2/2时,最小距离d=√(1 - m²)② 当 |m| >√2/2时,最小距离d=|m - √2|---完---