Question

点A,B是抛物线y^2=2px（p大于0）原点上以外的两个动点，且OA垂直OB，

设A(x1,y1),B(x2,y2)求y1*y2和x1*x2的值

Answer 1

将A(x1,y1)代入y^2=2px(p>0)中，得x1=y1^2/(2p)
同理，x2=y2^2/(2p)
两式相乘得x1*x2=(y1*y2)^2/(4*p^2)

设OA斜率k1, OB斜率k2
k1=y1/x1,k2=y2/x2
因为OA垂直于OB，所以k1*k2=-1
代入上式得k1*k2=(y1/x1)*(y2/x2)=-1
整理，得x1*x2=-y1*y2

将x1*x2=-y1*y2代入x1*x2=(y1*y2)^2/(4*p^2)中
(y1*y2)^2/(4*p^2)=-y1*y2
化简
y1*y2*(y1*y2+4*p^2)=0
因A、B不在原点,故x1,y1, x2,y2都不为零,
所以y1*y2+4*p^2=0
y1*y2=-4*p^2,x1*x2=4*p^2