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,

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