Question

有关于抛物线设抛物线y^2=4px(p＞0)的准线与x轴的交点为M,过M点作直线L交抛物线于A,B两点,求线段AB中点的

Answer 1

设直线方程为y=k(x+p)

与y^2=4px 联立

得k^2(x^2+2px+p^2)=4px

k^2x^2-(4p-2k^2p)x+k^2p^2=0

则x1+x2=(4p-2k^2p)/k^2 (y1+y2)=k*(x1+x2)+2kp=(4p+2k^2p)/k+2kp=4p/k

故x0=(x1+x2)/2=2p/k^2+p y0=(x1+x2)/2=2p/k

既k=2p/y0

故x0=2p/(2p/y0)^2-p=y0^2/2p-p

故y0^2=2p(x0+p)

y^2=2p(x+p)

又方程k^2x^2-(4p-2k^2p)x+k^2p^2=0 有解

故(4p-2k^2p)^2-4k^4p^2>0

16p^2-16k^2p^2+4k^4p^2-4k^4p^2>0

故16k^24p^2

2p(x+p)>4p^2

x+p>2p x>p