过抛物线y2=2px的焦点的两条相互垂直的弦AB和CD,求证1/AB+1/CD是定值
1个回答
焦点(p/2, 0)
设AB：y=k(x-p/2)
那么CD：y=(-1/k)(x-p/2)
A、B坐标满足方程k^2x^2-(pk^2+2p)x+k^2p^2/4=0
C、D坐标满足方程x^2-(p+2pk^2)x+p^2/4=0
AB=√[(x1-x2)^2+(y1-y2)^2]
=√[(k^2+1)(x1-x2)^2]
=√{(k^2+1)[(x1+x2)^2-4x1x2]}
=√{(k^2+1)[(p+2p/k^2)^2-p^2]}
=|2p(k^2+1)/k^2|
所以1/AB=k^2/[2p(k^2+1)]
CD=√[(x1-x2)^2+(y1-y2)^2]
=√[(1/k^2+1)(x1-x2)^2]
=√{(1/k^2+1)[(x1+x2)^2-4x1x2]}
=√{(1/k^2+1)[(p+2pk^2)^2-p^2]}
=|2p(k^2+1)|
所以1/CD=1/[2p(k^2+1)]
所以1/AB+1/CD=(k^2+1)/[2p(k^2+1)]=1/(2p)
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