设椭圆方程x²/a²+y²/b²=1(a>b>0),斜率为1的直线不过原点O,与椭圆交于A,B,M为AB中点直线AB与OM是否垂直,请证明
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![设椭圆方程x²/a²+y²/b²=1(a>b>0),斜率为1的直线不过原点O,与椭圆交于A,B,M为AB中点直线AB与OM是否垂直,请证明](/uploads/image/z/6973434-18-4.jpg?t=%E8%AE%BE%E6%A4%AD%E5%9C%86%E6%96%B9%E7%A8%8Bx%26%23178%3B%2Fa%26%23178%3B%2By%26%23178%3B%2Fb%26%23178%3B%3D1%28a%3Eb%3E0%29%2C%E6%96%9C%E7%8E%87%E4%B8%BA1%E7%9A%84%E7%9B%B4%E7%BA%BF%E4%B8%8D%E8%BF%87%E5%8E%9F%E7%82%B9O%2C%E4%B8%8E%E6%A4%AD%E5%9C%86%E4%BA%A4%E4%BA%8EA%2CB%2CM%E4%B8%BAAB%E4%B8%AD%E7%82%B9%E7%9B%B4%E7%BA%BFAB%E4%B8%8EOM%E6%98%AF%E5%90%A6%E5%9E%82%E7%9B%B4%EF%BC%8C%E8%AF%B7%E8%AF%81%E6%98%8E)
设椭圆方程x²/a²+y²/b²=1(a>b>0),斜率为1的直线不过原点O,与椭圆交于A,B,M为AB中点直线AB与OM是否垂直,请证明
设椭圆方程x²/a²+y²/b²=1(a>b>0),斜率为1的直线不过原点O,与椭圆交于A,B,M为AB中点
直线AB与OM是否垂直,请证明
设椭圆方程x²/a²+y²/b²=1(a>b>0),斜率为1的直线不过原点O,与椭圆交于A,B,M为AB中点直线AB与OM是否垂直,请证明
M(m,n)
xA+xB=2m,yA+yB=2n
[(xA)^2/a^2+(yA)^2/b^2]-[(xB)^2/a^2+(yB)^2/b^2]=1-1=0
(xA+xB)*(xA-xB)/a^2+(yA+yB)*(yA-yB)/b^2=0
b^2/a^2+[(yA+yB)/(xA+xB)]*[(yA-yB)/(xA-xB)]=0
b^2/a^2+[(2m)/(2n)]*k(AB)=0
k(OM)*k(AB)=-b^2/a^2>-1
直线AB与OM不垂直