已知椭圆x²/a²+y²/b²=1(a>b>0)与直线x+y-1=0交于A、B两点,|AB|=2√2,AB的中点M与椭圆中心连线的斜率是√2/2,求椭圆方程
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![已知椭圆x²/a²+y²/b²=1(a>b>0)与直线x+y-1=0交于A、B两点,|AB|=2√2,AB的中点M与椭圆中心连线的斜率是√2/2,求椭圆方程](/uploads/image/z/14759032-40-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%A4%AD%E5%9C%86x%26%23178%3B%2Fa%26%23178%3B%2By%26%23178%3B%2Fb%26%23178%3B%3D1%EF%BC%88a%3Eb%3E0%EF%BC%89%E4%B8%8E%E7%9B%B4%E7%BA%BFx%2By-1%3D0%E4%BA%A4%E4%BA%8EA%E3%80%81B%E4%B8%A4%E7%82%B9%2C%7CAB%7C%3D2%E2%88%9A2%2CAB%E7%9A%84%E4%B8%AD%E7%82%B9M%E4%B8%8E%E6%A4%AD%E5%9C%86%E4%B8%AD%E5%BF%83%E8%BF%9E%E7%BA%BF%E7%9A%84%E6%96%9C%E7%8E%87%E6%98%AF%E2%88%9A2%2F2%2C%E6%B1%82%E6%A4%AD%E5%9C%86%E6%96%B9%E7%A8%8B)
已知椭圆x²/a²+y²/b²=1(a>b>0)与直线x+y-1=0交于A、B两点,|AB|=2√2,AB的中点M与椭圆中心连线的斜率是√2/2,求椭圆方程
已知椭圆x²/a²+y²/b²=1(a>b>0)与直线x+y-1=0交于A、B两点,|AB|=2√2,AB的中点M与椭圆中心连线的斜率是√2/2,求椭圆方程
已知椭圆x²/a²+y²/b²=1(a>b>0)与直线x+y-1=0交于A、B两点,|AB|=2√2,AB的中点M与椭圆中心连线的斜率是√2/2,求椭圆方程
设A(x1,1-x1),B(x2,1-x2)
将直线x+y-1=0代入椭圆x²/a²+y²/b²=1(a>b>0)消去y并整理得:(1/a²+1/b²)x²-2x/b²+1/b²-1=0 由韦达定理:x1+x2=2a²/(a²+b²)①,x1x2=(a²-a²b²)/(a²+b²)②
|AB|=√[(x2-x1)²+(1-x2-1+x1)²]=√|x2-x1|=2√2,所以|x2-x1|=1
又|x2-x1|²=(x2+x1)²-4x1x2=1 ③
AB的中点M与椭圆中心连线的斜率是(2-x1-x2)/(x1+x2)=2/(x1+x2)-1=√2/2 ④
将①,②代入③.④可以解得a²=2,b²=√2
所以椭圆方程x²/2+y²/√2=1