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过抛物线y^2=4x的准线与轴的交点E作直线交抛物线于A,B两点?

2011-08-18 11:33:262***
过抛物线y^2=4x的准线与轴的交点E作直线交抛物线于A,B两点,F是抛物线焦点,若向量FA*过抛物线y^2=4x的准线与轴的交点E作直线交抛物线于A,B两点,F是抛物线焦点,若向量FA*向量FB=0.求直线AB的方程过抛物线y^2=4x的准线与轴的交点E作直线交抛物线于A,B两点,F是抛物线焦点,若向量FA*过抛物线y^2=4x的准线与轴的交点E作直线交抛物线于A,B两点,?

最佳回答

  • 过抛物线y^2=4x(1) 的准线与轴的交点E(-1,0)作直线:x=my-1(2) 交抛物线于点A(x1,y1),B(x2,y2), 把(2)代入(1),得 y^2-4my+4=0, ∴y1+y2=4m,y1y2=4. 由(2),x1+x2=m(y1+y2)-2=4m^2-2, x1x2=(my1-1)(my2-1)=m^2*y1y2-m(y1+y2)+1 =4m^2-4m^2+1=1, F(1,0), ∴向量FA*FB=(x1-1,y1)*(x2-1,y2) =(x1-1)(x2-1)+y1y2 =x1x2-(x1+x2)+1+y1y2 =1-(4m^2-2)+1+4 =8-4m^2=0, m^2=2,m=土√2. ∴直线AB的方程为x=土√2y-1.
    2011-08-18 12:08:38
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