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[Maths] Maths 2001-LQ-17

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Maths 2001-LQ-17



[ 本帖最後由 wky92hk 於 2010-4-8 11:14 PM 編輯 ]
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it seems like (amaths)locus...

for any point on the circle~
For example, now S=(a,b)
slope of OS = (b-0)/(a-0) =b/a
slope of SP = (b-0)/(a-p)  =b/(a-p)
as OS perpendicular to SP (converse of angle in semi- circle)
angle OSP = 90degree
slope of OS X slope of SP = -1
(b/a)[b/(a-p)]=-1
so (y/x)  [y/(x-p)]=-1

[ 本帖最後由 ericchim 於 2010-4-8 05:37 PM 編輯 ]
http://10ce.fddsa.com/ce.php?sub=IKLMNQSTVWY&.gif

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引用:
原帖由 ericchim 於 2010-4-8 05:27 PM 發表
for any point on the circle~
For example, now S=(a,b)
slope of OS = (b-0)/(a-0) =b/a
slope of SP = (b-0)/(a-p)  =b/(a-p)
as OS perpendicular to SP (converse of angle in semi- circle)
angle O ...
哦..原來係咁..thanks
我而家先知原來圓方程可以咁求

仲有個問題


[ 本帖最後由 wky92hk 於 2010-4-8 06:27 PM 編輯 ]
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alternative answers are enclosed with rectangles
i am not very understood why the 2nd answer is x^2+y^2-px+pa-a^2-b^2=0...
but it just the same as the first one as pa-a^2-b^2 =0
http://10ce.fddsa.com/ce.php?sub=IKLMNQSTVWY&.gif

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引用:
原帖由 ericchim 於 2010-4-8 10:49 PM 發表
alternative answers are enclosed with rectangles
i am not very understood why the 2nd answer is x^2+y^2-px+pa-a^2-b^2=0...
but it just the same as the first one as pa-a^2-b^2 =0
我終於搵到呢舊野點黎..
但我都係唔明點解要加上 pa-a^2-b^2
雖然答案都係一樣

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