已知抛物线y^2=-x与直线y=k(x+1)相交于A.B两点 1.求证OA垂直OB 2.当三角形OAB面积为根号10时,

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
所以OA垂直OB
2:
OA=√(Ya^2+Ya^4) OB=√(Yb^2+Yb^4)
S三角形OAB=1/2*OA*OB=1/2*√(Ya^2+Ya^4) * √(Yb^2+Yb^4)=√10
√(Ya^2+Ya^4) * √(Yb^2+Yb^4)=2√10 两边平方得
(Ya^2+Ya^4) * (Yb^2+Yb^4)=40
Ya^2Yb^2+Ya^2Yb^4+Yb^2Ya^4+Ya^4Yb^4=40
(YaYb)^2+(YaYb)^2Yb^2+(YaYb)^2Ya^2+(YaYb)^4=40
(-1)^2+(-1)^2Yb^2+(-1)^2Ya^2+(-1)^4=40
1+Yb^2+Ya^2+1=40
Ya^2-2+Yb^2=36
Ya^2+2*(-1)+Yb^2=36
Ya^2+2YaYb+Yb^2=36
(Ya+Yb)^2=36
Ya+Yb=6 或 Ya+Yb=-6
因为A,B在直线y=k(x+1)上,则
k=(Ya-Yb)/(Xa-Xb)
=(Ya-Yb)/(-Ya^2+Yb^2)
=(Ya-Yb)/(Ya+Yb)(Yb-Ya)
=-1/(Ya+Yb)
所以 k=1/6 或 k=-1/6