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过抛物线y^2=2px(p>0)焦点的直线交抛物线于A(X1,Y?

2007-12-25 19:12:131***
O为坐标原点,求OA向量*OB向量过抛物线y^2=2px(p>0)焦点的直线交抛物线于A(X1,Y1)B(X2,Y2)O为坐标原点,求OA向量*OB向量:OA向量*OB向量 =x1x1+y1y2?

最佳回答

  • OA向量*OB向量 =x1x1+y1y2 过抛物线y²=2px(p>0)焦点(p/2 ,0)的直线 y=k(x- p/2)和抛物线方程联立 ==> k²(x- p/2)²=2px k²x²-(k²+2p)x+k²p²/4=0 ==>x1x2 =p²/4 ,和抛物线方程联立 ==>y1y2=- p² ==>OA向量*OB向量 =x1x1+y1y2 = -3p²/4
    2007-12-25 19:34:27
  • 过焦点F(p/2,0)的直线方程是y=k(x-p/2) 与y^2=2px消去x得到 ky^2-2py-kp^2=0 所以y1+y2=2p/k,y1y2=-p^2 故x1x2=[y1^2/(2p)]*[y2^2/(2p)] ,,,,,,=(y1y2)^2/(4p^2) ,,,,,,=(-p^2)^2/4p^2 ,,,,,,=p^2/4 因此x1x2+y1y2=p^2/4-p^2=-3p^2/4 此时A(x1,y1),B(x2,y2).向量OA=(x1y1),向量OB=(x2,y2) 所以,向量OA·向量OB=x1x2+y1y2=-3p^2/4.
    2007-12-25 19:51:28
  • 撤消
    2007-12-25 19:36:14
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