求文档:已知：抛物线y2=2px(p›0),过焦点弦AB,A（x1,y1已知：抛物线y2=2px(p›0),过焦点弦AB,A（x1,y1）,B（x2,y2）θ是弦与X轴的倾斜角.求证：y1.y2=-p2X1.x2=-p2/4AB=2p/sin2θ

Question

求文档:已知：抛物线y2=2px(p›0),过焦点弦AB,A（x1,y1
已知：抛物线y2=2px(p›0),过焦点弦AB,A（x1,y1）,B（x2,y2）
θ是弦与X轴的倾斜角.
求证：y1.y2=-p2
X1.x2=-p2/4
AB=2p/sin2θ

Answer 1

焦点坐标为(p/2,0)
1.当θ＝90度时,x1=x2=p/2,y1=-y2=p,故x1*x2=p^2/4,y1*y2=-p^2.你的题应该抄错了.x1*x2不会是负的.
2.当θ不等于90度时,则设直线方程为y=k(x-p/2)
代入到抛物线方程得,
k^2(x-p/2)^2=2px
即k^2*x^2-(k^2*p+2p)*x+k^2*p^2/4=0
x1+x2=(k^2*p+2p)/k^2=p+2p/k^2
x1*x2=p^2/4,
y1*y2=k(x1-p/2)*k(x2-p/2)=k^2*[ x1*x2 - (x1+x2)*p/2 + p^2/4]
=k^2 (p^2/4-p^2/2-p^2/k^2+p^2/4)
=k^2 (p^2/2-p^2/2-p^2/k^2)
=k^2* -p^2/k^2=-p^2
得证.
AB＝√ [ (1+k^2)(x1+x2)^2-4x1*x2) ]
= √ [ (1+k^2)( p^2_4p^2/k^4 +4p^2/k^2 -p^2 ) ]
=√[ 4p^2*(1+k^2)^2 /k^4 ]
=2p*(1+k^2)/k^2
又k=tanθ
所以
AB＝2p(1+tanθ^2)/tanθ^2
=2p/sinθ^2
你的题中的sin2θ应该是“(sinθ)的平方”吧?