Question

已知定点A(-2,-4),过点A作倾角为45°的直线L交抛物线Y^2=2PX(P>0)与B,C两点,且｜AB｜,｜BC｜

Answer 1

设B（x1,y1）C(x2,y2)

过定点(-2,-4)作倾斜角为45°的直线l

则直线方程为 y=x-2 代入y2=2px

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

x1+x2=2p+4

x1*x2=4

AB BC AC成等比数列

则AB/BC=BC/AC

(x1+2)/(x2-x1)=(x2-x1)/(x2+2)

整理得

x1x2+2(x1+x2)+4=(x1+x2)^2-4x1x2

4+2(2p+4)+4=(2p+4)^2-16

解得p=1

所以抛物线的方程为

y^2=2x