椭圆x^2+2y^2=2,椭圆上一点P使它到直线2x-y+8=0距离最小,求最小距离设P(m,n)d= |2m-n+8| / √5 令t=2m-n,n=2m-tm²+2n² =2m²+2(2m-t)²=29m²-8tm+2t²-2=0 ①Δ=64t² - 4×9(2t²-2) ≥0解得 -3≤t
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![椭圆x^2+2y^2=2,椭圆上一点P使它到直线2x-y+8=0距离最小,求最小距离设P(m,n)d= |2m-n+8| / √5 令t=2m-n,n=2m-tm²+2n² =2m²+2(2m-t)²=29m²-8tm+2t²-2=0 ①Δ=64t² - 4×9(2t²-2) ≥0解得 -3≤t](/uploads/image/z/5937337-1-7.jpg?t=%E6%A4%AD%E5%9C%86x%5E2%2B2y%5E2%3D2%2C%E6%A4%AD%E5%9C%86%E4%B8%8A%E4%B8%80%E7%82%B9P%E4%BD%BF%E5%AE%83%E5%88%B0%E7%9B%B4%E7%BA%BF2x-y%2B8%3D0%E8%B7%9D%E7%A6%BB%E6%9C%80%E5%B0%8F%2C%E6%B1%82%E6%9C%80%E5%B0%8F%E8%B7%9D%E7%A6%BB%E8%AE%BEP%28m%2Cn%29d%3D+%7C2m-n%2B8%7C+%2F+%E2%88%9A5+%E4%BB%A4t%3D2m-n%2Cn%3D2m-tm%26%23178%3B%2B2n%26%23178%3B+%3D2m%26%23178%3B%2B2%282m-t%29%26%23178%3B%3D29m%26%23178%3B-8tm%2B2t%26%23178%3B-2%3D0+%E2%91%A0%CE%94%3D64t%26%23178%3B+-+4%C3%979%282t%26%23178%3B-2%29+%E2%89%A50%E8%A7%A3%E5%BE%97+-3%E2%89%A4t)
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椭圆x^2+2y^2=2,椭圆上一点P使它到直线2x-y+8=0距离最小,求最小距离设P(m,n)d= |2m-n+8| / √5 令t=2m-n,n=2m-tm²+2n² =2m²+2(2m-t)²=29m²-8tm+2t²-2=0 ①Δ=64t² - 4×9(2t²-2) ≥0解得 -3≤t
椭圆x^2+2y^2=2,椭圆上一点P使它到直线2x-y+8=0距离最小,求最小距离
设P(m,n)d= |2m-n+8| / √5 令t=2m-n,n=2m-tm²+2n² =2m²+2(2m-t)²=29m²-8tm+2t²-2=0 ①Δ=64t² - 4×9(2t²-2) ≥0解得 -3≤t≤3∴2m-n+8∈[5,11]d∈[√5,11/√5]即最小距离√5此时t= -3,代入①解得m= -4/3n=2m-t = 1/3即P(-4/3,1/3)
有查到这个解题过程,但我不理解为什么Δ=64t² - 4×9(2t²-2) ≥0
只要解释这步即可
椭圆x^2+2y^2=2,椭圆上一点P使它到直线2x-y+8=0距离最小,求最小距离设P(m,n)d= |2m-n+8| / √5 令t=2m-n,n=2m-tm²+2n² =2m²+2(2m-t)²=29m²-8tm+2t²-2=0 ①Δ=64t² - 4×9(2t²-2) ≥0解得 -3≤t
因为.(m,n)是p点坐标,他要符合椭圆方程的解,把m,n的关系式,这里设t作为它们的等量关系式子,代入椭圆方程,也同样应有解,二次方程有解,势必戴尔他大于等于零