Question

已知抛物线y^2=-x与直线y=k(x+1)相交于AB两点，求Koa×Kob的值

Answer 1

设A点坐标为（Xa,Ya),B点坐标为(Xb,Yb)
因为它们在抛物线y^2=-x上,则
A:(-Ya^2,Ya),B(-Yb^2,Yb)
又因为它们在直线y=k(x+1)上,则
Ya=k(Xa+1)
Yb=k(Xb+1)
两都相除得
Ya/Yb=(Xa+1)/(Xb+1)
Ya*(Xb+1)=Yb(Xa+1)
Ya*(-Yb^2+1)=Yb(-Ya^2+1)
-YaYb^2+Ya=-Ya^2Yb+Yb
YaYb(Ya-Yb)+(Ya-Yb)=0
(Ya-Yb)(YaYb+1)=0
因为AB不同点,所以Ya-Yb0
则YaYb=-1
KOA=(Ya-0)/(-Ya^2-0)=-1/Ya
KOB=(Yb-0)/(-Yb^2-0)=-1/Yb
KOA*KOB=-1/YA*(-1/Yb)=1/(YaYb)=-1
请采纳回答

Answer 2

联立
y
²=
-x
与
y=k（x+1）,得到k²
x²
+（2k
²
+1）x
+
k²
=0,然后利用韦达定理,
求出
x1
+
x2,x1
x2,y1+y2,y1
y2,
Koa
x
Kob
=
（y1
y2）/（x1
x2）,求出答案