直线x+y-1=0与椭圆x^2/a^2+y^2/b^2=1(a>b>0)交于两点A、B,AB中点M,直线OM斜率k=1/2,则b/a=
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![直线x+y-1=0与椭圆x^2/a^2+y^2/b^2=1(a>b>0)交于两点A、B,AB中点M,直线OM斜率k=1/2,则b/a=](/uploads/image/z/10220445-45-5.jpg?t=%E7%9B%B4%E7%BA%BFx%2By-1%3D0%E4%B8%8E%E6%A4%AD%E5%9C%86x%5E2%2Fa%5E2%2By%5E2%2Fb%5E2%3D1%28a%3Eb%3E0%29%E4%BA%A4%E4%BA%8E%E4%B8%A4%E7%82%B9A%E3%80%81B%2CAB%E4%B8%AD%E7%82%B9M%2C%E7%9B%B4%E7%BA%BFOM%E6%96%9C%E7%8E%87k%3D1%2F2%2C%E5%88%99b%2Fa%3D)
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直线x+y-1=0与椭圆x^2/a^2+y^2/b^2=1(a>b>0)交于两点A、B,AB中点M,直线OM斜率k=1/2,则b/a=
直线x+y-1=0与椭圆x^2/a^2+y^2/b^2=1(a>b>0)交于两点A、B,AB中点M,直线OM斜率k=1/2,则b/a=
直线x+y-1=0与椭圆x^2/a^2+y^2/b^2=1(a>b>0)交于两点A、B,AB中点M,直线OM斜率k=1/2,则b/a=
联立x+y-1=0
x^2/a^2+y^2/b^2=1
解得:(a^2+b^2)x^2-2a^2x+a^2-a^2b^2=0
x1+x2=2a^2/(a^2+b^2) X1X2=(a^2-a^2b^2)/(a2+b^2)
y1+y2=-(x1+x2)+2
所以M点坐标为(X0,y0)=(a^2/(a^2+b^2) ,b^2/(a^2+b^2) )
y0/x0=k=1/2就可求出比来为根2分之一