已知C为y^2=2PX(P大于0)的准线于X轴的交点,点F为焦点.A,B为抛物线上两点若FA+FB+2FC=0求证AB垂直X轴, 向量FA与FB夹角
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![已知C为y^2=2PX(P大于0)的准线于X轴的交点,点F为焦点.A,B为抛物线上两点若FA+FB+2FC=0求证AB垂直X轴, 向量FA与FB夹角](/uploads/image/z/8669997-45-7.jpg?t=%E5%B7%B2%E7%9F%A5C%E4%B8%BAy%5E2%3D2PX%EF%BC%88P%E5%A4%A7%E4%BA%8E0%EF%BC%89%E7%9A%84%E5%87%86%E7%BA%BF%E4%BA%8EX%E8%BD%B4%E7%9A%84%E4%BA%A4%E7%82%B9%2C%E7%82%B9F%E4%B8%BA%E7%84%A6%E7%82%B9.A%2CB%E4%B8%BA%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%B8%8A%E4%B8%A4%E7%82%B9%E8%8B%A5FA%2BFB%2B2FC%3D0%E6%B1%82%E8%AF%81AB%E5%9E%82%E7%9B%B4X%E8%BD%B4%2C+%E5%90%91%E9%87%8FFA%E4%B8%8EFB%E5%A4%B9%E8%A7%92)
已知C为y^2=2PX(P大于0)的准线于X轴的交点,点F为焦点.A,B为抛物线上两点若FA+FB+2FC=0求证AB垂直X轴, 向量FA与FB夹角
已知C为y^2=2PX(P大于0)的准线于X轴的交点,点F为焦点.A,B为抛物线上两点若FA+FB+2FC=0
求证AB垂直X轴, 向量FA与FB夹角
已知C为y^2=2PX(P大于0)的准线于X轴的交点,点F为焦点.A,B为抛物线上两点若FA+FB+2FC=0求证AB垂直X轴, 向量FA与FB夹角
C(-p/2,0) ,F(p/2,0)
A,B为抛物线上两点
设A(x1,y1),B(x2,y2)
∴FA=(x1-p/2,y1),FB=(x2-p/2,y2)
∵FA+FB+2FC=0
∴(x1+x2-p,y1+y2)+2(-p,0)=0向量
∴x1+x2-3p=0,y1+y2=0
∴y²1=2px1=y²2=2px2
∴x1=x2=3p/2
∴AB垂直X轴
∴x1=x2
y²1=y²2=2px1=3p²
不妨设y1>0,y2
设A(x1,y1),B(x2,y2),F(p/2,0),C(-p/2,0)
FA=(x1-p/2,y1),FB=(x2-p/2,y2),FC=(-p,0)
FA+F B+2FC=0
即有(x1+x2-3p,y1+y2)=0
x1+x2-3p=0,y1+y2=0
y1^2=2px1,y2^2=2px2
y1^2+y2^2=2p(x1+x2)=2p*3p...
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设A(x1,y1),B(x2,y2),F(p/2,0),C(-p/2,0)
FA=(x1-p/2,y1),FB=(x2-p/2,y2),FC=(-p,0)
FA+F B+2FC=0
即有(x1+x2-3p,y1+y2)=0
x1+x2-3p=0,y1+y2=0
y1^2=2px1,y2^2=2px2
y1^2+y2^2=2p(x1+x2)=2p*3p=6p^2
(y1+y2)^2=6p^2+2y1y2=0
y1y2=-3p^2
(y1-y2)^2=6p^2-2y1y2=12p^2
y1-y2=2根号3p
故有y1=根号3p,y2=-根号3p
x1=3p/2,x2=3p-x1=3p/2
故有x1=x2,所以有AB垂直于X轴.
FA=(p,根号3p),FB=(p,-根号3p)
cos
故FA.FB的夹角是120度.
收起