Question

求解一道数学题，急！！！

已知抛物线y^2=-x与直线y=k(x+1)交于两A,B点(1)求kOA*kOB的值;(2)当S三角形AOB=根号10时,求K的值.答案是-1，+-1/6。要全部过程

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-Yb<>0
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