设F为抛物线y²=2px(p>0)的焦点,ABC为该抛物线上三点,当向量FA+FB+FC=0且|FA|+|FB|+|Fc|=3时,求设F为抛物线y²=2px(p>0) 的焦点,A、B、C为该抛物线上三点,当向量FA+FB+FC=0且 |FA|+|FB|+|Fc|=3时,此
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设F为抛物线y²=2px(p>0)的焦点,ABC为该抛物线上三点,当向量FA+FB+FC=0且|FA|+|FB|+|Fc|=3时,求设F为抛物线y²=2px(p>0) 的焦点,A、B、C为该抛物线上三点,当向量FA+FB+FC=0且 |FA|+|FB|+|Fc|=3时,此
设F为抛物线y²=2px(p>0)的焦点,ABC为该抛物线上三点,当向量FA+FB+FC=0且|FA|+|FB|+|Fc|=3时,求
设F为抛物线y²=2px(p>0) 的焦点,A、B、C为该抛物线上三点,当向量FA+FB+FC=0且 |FA|+|FB|+|Fc|=3时,此抛物线的方程为
2010年大连市高三第一次模拟试卷第6题
设F为抛物线y²=2px(p>0)的焦点,ABC为该抛物线上三点,当向量FA+FB+FC=0且|FA|+|FB|+|Fc|=3时,求设F为抛物线y²=2px(p>0) 的焦点,A、B、C为该抛物线上三点,当向量FA+FB+FC=0且 |FA|+|FB|+|Fc|=3时,此
设向量FA FB FC分别为(x1,y1)(x2,y2)(x3,y3) 则x1+x2+x3=0
|FA|+|FB|+|Fc|=3 Xa=x1+ p/2同理Xb=x2+ p/2 Xc=x3+ p/2据抛物线定义
FA=x2+ p/2+p/2=x1+p 所以x1+x2+x3+3p=3 p=1 y^2=2x
焦半径公式r=x+P/2
A(x1,y1)B(x2,y2)
FA=(x1-P/2,y1)
FB=(x2-P/2,y2)
FC=(P-x1-x2,-y1-y2)
C(3P/2-x1-x2,-y1-y2)
|FA|+|FB|+|Fc|=3
=(x1+P/2)+(x2+P/2)+(3P/2-x1-x2+P/2)
=3P
得P=1
(参数法)易知,焦点F(p/2,0).由题意可设点A(2pa²,2pa),B(2pb²,2pb),C(2pc²,2pc).(a,b,c∈R).(1)由抛物线定义可知,|FA|=(p/2)+2pa²,|FB|=(p/2)+2pb²,|FC|=(p/2)+2pc².三式相加可知,(3p/2)+2p(a²+b²+c²...
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(参数法)易知,焦点F(p/2,0).由题意可设点A(2pa²,2pa),B(2pb²,2pb),C(2pc²,2pc).(a,b,c∈R).(1)由抛物线定义可知,|FA|=(p/2)+2pa²,|FB|=(p/2)+2pb²,|FC|=(p/2)+2pc².三式相加可知,(3p/2)+2p(a²+b²+c²)=|FA|+|FB|+|FC|=3.===>a²+b²+c²=(6-3p)/(4p).(2).易知,向量FA=(2pa²-p/2,2pa),FB=(2pb²-p/2,2pb),FC=(2pc²-p/2,2pc).三向量相加可得FA+FB+FC=(2p(a²+b²+c²)-(3p/2),2pa+2pb+2pc)=(0,0).===>2p(a²+b²+c²)-(3p/2)=0.===>a²+b²+c²=3/4.(3)结合(1),(2)可知,(6-3p)/(4p)=a²+b²+c²=3/4.===>(6-3p)/(4p)=3/4.===>p=1.故抛物线方程为y²=2x.
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