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Amath questions
2008-11-24 2:40 am
Consider a fixed point F(-1,0) and a fixed line x+4=0.From a moving point P,draw a prependicular PM to the given line.If PM=2PF ,find the equation of the locus of P
回答 (2)
2008-11-24 3:00 am
✔ 最佳答案
LetP be(x,y)PM=2PF
絕對值x+4/開方1^2+0^2=2[開方(x+1)^2+(y-0)^2]
之後2次左2邊,就會計到,3x^2+4y^2-12=0
以左邊條式你可以搵番書,右邊果條就正常都明.
2008-11-24 3:01 am
Since PM prependicular to x+4, we can set
M(-4,y) and P(x,y)
From PM=2PF
(PM)^2=(2PF)^2
(x+4)^2=4[(x+1)^2+y^2]
x^2+8x+16=4x^2+8x+4+4y^2
3x^2+4y^2-12=0
which is the equation of the locus of P
2008-11-23 19:02:11 補充：
成功在8:00前完成
M(-4,y) and P(x,y)
From PM=2PF
(PM)^2=(2PF)^2
(x+4)^2=4[(x+1)^2+y^2]
x^2+8x+16=4x^2+8x+4+4y^2
3x^2+4y^2-12=0
which is the equation of the locus of P
2008-11-23 19:02:11 補充：
成功在8:00前完成
收錄日期: 2021-04-25 16:58:07
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https://hk.answers.yahoo.com/question/index?qid=20081123000051KK01806