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#1
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ÊÁ¡ÒþÒÃÒâºÅÒ+àÊ鹵ç ¨Ø´µÑ´
¶éÒÁÕÊÁ¡ÒþÒÃÒâºÅÒ $y^2=8x$
ÁÕàÊ鹵ç $2\sqrt{2}x-y-4\sqrt{2}= 0$ ¼èÒ¹¨Ø´â¿¡ÑÊ (2,0) Áըش»ÅÒ·Ñé§ÊͧÍÂÙ躹¾ÒÃÒâºÅÒ àÃÒ¨ÐËҨشµÑ´·Ñé§2ä´éäËÁ¤ÃѺ ¶éÒä´éªèǺ͡ÇÔ¸ÕËÒ´éǤÃѺ>< V V ¢Íº¤Ø³ÁÒ¡¤ÃѺ^^ 18 ¸Ñ¹ÇÒ¤Á 2009 19:50 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ GolFFee |
#2
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ä´é¤ÃѺ µÑ´¡Ñ¹·Õè¨Ø´ $(1,-2\sqrt{2})$ áÅÐ $(4,4\sqrt{2})$
¨Ø´µÑ´¨ÐÊÍ´¤ÅéͧÊͧÊÁ¡ÒþÃéÍÁ¡Ñ¹ ¡çàÍÒ $y=2\sqrt{2}x-4\sqrt{2}$ ä»á·¹ã¹ÊÁ¡Òà $y^2=8x$ áÅéÇá¡éÊÁ¡ÒáÓÅѧÊͧ¤ÃѺ
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site:mathcenter.net ¤Ó¤é¹ 18 ¸Ñ¹ÇÒ¤Á 2009 19:46 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ nooonuii |
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