Loading ...
Sorry, an error occurred while loading the content.

RE: [EMHL] Triangle const. (a,h_a,r) [was: ADMIN.: Links]

Expand Messages
  • Luís Lopes
    Dear Nikos Dergiades, Thank you very much. Garcia Capitan had already given me the same results privately. === and the focii are ( 0 , 2ar^2/(a^2-4r^2) ) ( 0
    Message 1 of 5 , Jun 12 9:10 AM
    • 0 Attachment
      Dear Nikos Dergiades,



      Thank you very much. Garcia Capitan had already

      given me the same results privately.



      ===

      and the focii are
      ( 0 , 2ar^2/(a^2-4r^2) )
      ( 0 ,-2ar^2/(a^2-4r^2) )
      ===

      Hum... didn't you forget the y_0?



      ( 0 , y_0 + 2ar^2/(a^2-4r^2) )
      ( 0 , y_0 -2ar^2/(a^2-4r^2) )
      ===


      The directrices are(?) y = y_0 \pm u/e



      where



      e = eccentricity = \sqrt{1 + (v/u)^2}



      Now I would be able to finish the construction showed in

      the site.



      Best regards,

      Luis



      To: Hyacinthos@yahoogroups.com
      From: ndergiades@...
      Date: Fri, 12 Jun 2009 15:31:12 +0000
      Subject: Re: [EMHL] Triangle const. (a,h_a,r) [was: ADMIN.: Links]








      Dear Luis,

      If I understood what you want, then

      y_0 = a^2r/(a^2-4r^2)

      u = 4r^3/(a^2-4r^2)

      v = 2r^2/sqrt(a^2-4r^2)

      where a^2 > 4r^2 always

      and the focii are
      ( 0 , 2ar^2/(a^2-4r^2) )
      ( 0 ,-2ar^2/(a^2-4r^2) )

      Best regards
      Nikos Dergiades

      > Dear Hyacinthists,
      >
      >
      >
      > In Angel Montesdeoca Delgado's web page below one finds
      >
      > an interesting solution to the problem (a,h_a,r).
      >
      >
      >
      > In it there is the conic (actually a hyperbola)
      >
      >
      >
      > 4r^2x^2 - (a^2 - 4r^2)y^2 + 2a^2 ry - r^2(a^2 + 4r^2) = 0.
      >
      >
      >
      >
      > In order to find its intersection with a line parallel to
      > the
      >
      > directrix I would like to have the hyperbola expressed in
      > the
      >
      > standard form
      >
      >
      >
      > (y - y_0)^2/u^2 - x^2/v^2 = 1. (*)
      >
      >
      >
      > Is it possible to get (*) symbolically?
      >
      >
      >
      > As always, I really appreciate your help.
      >
      > Thank you for your time.
      >
      >
      >
      > Best regards,
      >
      > Luis
      >
      >
      >
      >
      >
      > To: Hyacinthos@yahoogroups.com
      > From: anopolis72@...
      > Date: Thu, 28 May 2009 21:59:12 +0000
      > Subject: [EMHL] ADMIN.: Links
      >
      >
      >
      >
      >
      >
      >
      > I have added to the list's LINKS Page:
      >
      > http://tech.groups.yahoo.com/group/Hyacinthos/links/
      >
      > a link to Angel Montesdeoca Delgado's web page
      >
      > http://webpages.ull.es/users/amontes/otrashtm/varios.htm
      >
      > It contains interesting triangle geometry files
      > in pdf format (in Spanish).
      >
      > Listmembers may add other links (of Geometry interest)
      >
      > Antreas
      >
      >
      >
      >
      >
      >
      >
      >
      >
      > __________________________________________________________
      > Conheça os novos produtos Windows Live! Clique aqui.
      > http://www.windowslive.com.br
      >
      > [Non-text portions of this message have been removed]
      >
      >
      >
      > ------------------------------------
      >
      > Yahoo! Groups Links
      >
      >
      > mailto:Hyacinthos-fullfeatured@yahoogroups.com
      >
      >
      >

      __________________________________________________________
      Χρησιμοποιείτε Yahoo!;
      Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
      διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
      μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr









      _________________________________________________________________
      Conheça os novos produtos Windows Live! Clique aqui.
      http://www.windowslive.com.br

      [Non-text portions of this message have been removed]
    • Nikolaos Dergiades
      Dear Luis, Sorry. You are right. Best regards Nikos Dergiades ... ___________________________________________________________ Χρησιμοποιείτε
      Message 2 of 5 , Jun 12 9:46 AM
      • 0 Attachment
        Dear Luis,
        Sorry. You are right.

        Best regards
        Nikos Dergiades

        > Dear Nikos Dergiades,
        >
        >
        >
        > Thank you very much. Garcia Capitan had already
        >
        > given me the same results privately.
        >
        >
        >
        > ===
        >
        > and the focii are
        > ( 0 , 2ar^2/(a^2-4r^2) )
        > ( 0 ,-2ar^2/(a^2-4r^2) )
        > ===
        >
        > Hum... didn't you forget the y_0?
        >
        >
        >
        > ( 0 , y_0 + 2ar^2/(a^2-4r^2) )
        > ( 0 , y_0 -2ar^2/(a^2-4r^2) )
        > ===
        >
        >
        > The directrices are(?) y = y_0 \pm u/e
        >
        >
        >
        > where
        >
        >
        >
        > e = eccentricity = \sqrt{1 + (v/u)^2}
        >
        >
        >
        > Now I would be able to finish the construction showed in
        >
        > the site.
        >
        >
        >
        > Best regards,
        >
        > Luis
        >




        ___________________________________________________________
        Χρησιμοποιείτε Yahoo!;
        Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
        διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
        μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr
      • Luís Lopes
        Dear Hyacinthists, Resending. From: qed_texte@hotmail.com To: hyacinthos@yahoogroups.com Subject: RE: [EMHL] Triangle const. (a,h_a,r) [was: ADMIN.: Links]
        Message 3 of 5 , Jun 12 9:49 AM
        • 0 Attachment
          Dear Hyacinthists,



          Resending.





          From: qed_texte@...
          To: hyacinthos@yahoogroups.com
          Subject: RE: [EMHL] Triangle const. (a,h_a,r) [was: ADMIN.: Links]
          Date: Fri, 12 Jun 2009 16:10:45 +0000



          Dear Nikos Dergiades,

          Thank you very much. Garcia Capitan had already
          given me the same results privately.

          ===
          and the focii are
          ( 0 , 2ar^2/(a^2-4r^2) )
          ( 0 ,-2ar^2/(a^2-4r^2) )
          ===
          Hum... didn't you forget the y_0?

          ( 0 , y_0 + 2ar^2/(a^2-4r^2) )
          ( 0 , y_0 -2ar^2/(a^2-4r^2) )
          ===

          The directrices are(?) y = y_0 \pm u/e

          where

          e = eccentricity = \sqrt{1 + (v/u)^2}

          Now I would be able to finish the construction showed in
          the site.

          Best regards,
          Luis



          To: Hyacinthos@yahoogroups.com
          From: ndergiades@...
          Date: Fri, 12 Jun 2009 15:31:12 +0000
          Subject: Re: [EMHL] Triangle const. (a,h_a,r) [was: ADMIN.: Links]







          Dear Luis,

          If I understood what you want, then

          y_0 = a^2r/(a^2-4r^2)

          u = 4r^3/(a^2-4r^2)

          v = 2r^2/sqrt(a^2-4r^2)

          where a^2 > 4r^2 always

          and the focii are
          ( 0 , 2ar^2/(a^2-4r^2) )
          ( 0 ,-2ar^2/(a^2-4r^2) )

          Best regards
          Nikos Dergiades

          > Dear Hyacinthists,
          >
          >
          >
          > In Angel Montesdeoca Delgado's web page below one finds
          >
          > an interesting solution to the problem (a,h_a,r).
          >
          >
          >
          > In it there is the conic (actually a hyperbola)
          >
          >
          >
          > 4r^2x^2 - (a^2 - 4r^2)y^2 + 2a^2 ry - r^2(a^2 + 4r^2) = 0.
          >
          >
          >
          >
          > In order to find its intersection with a line parallel to
          > the
          >
          > directrix I would like to have the hyperbola expressed in
          > the
          >
          > standard form
          >
          >
          >
          > (y - y_0)^2/u^2 - x^2/v^2 = 1. (*)
          >
          >
          >
          > Is it possible to get (*) symbolically?
          >
          >
          >
          > As always, I really appreciate your help.
          >
          > Thank you for your time.
          >
          >
          >
          > Best regards,
          >
          > Luis
          >
          >
          >
          >
          >
          > To: Hyacinthos@yahoogroups.com
          > From: anopolis72@...
          > Date: Thu, 28 May 2009 21:59:12 +0000
          > Subject: [EMHL] ADMIN.: Links
          >
          >
          >
          >
          >
          >
          >
          > I have added to the list's LINKS Page:
          >
          > http://tech.groups.yahoo.com/group/Hyacinthos/links/
          >
          > a link to Angel Montesdeoca Delgado's web page
          >
          > http://webpages.ull.es/users/amontes/otrashtm/varios.htm
          >
          > It contains interesting triangle geometry files
          > in pdf format (in Spanish).
          >
          > Listmembers may add other links (of Geometry interest)
          >
          > Antreas
          >
          >
          >
          >
          >
          >
          >
          >
          >
          > __________________________________________________________
          > Conheça os novos produtos Windows Live! Clique aqui.
          > http://www.windowslive.com.br
          >
          > [Non-text portions of this message have been removed]
          >
          >
          >
          > ------------------------------------
          >
          > Yahoo! Groups Links
          >
          >
          > mailto:Hyacinthos-fullfeatured@yahoogroups.com
          >
          >
          >

          __________________________________________________________
          Χρησιμοποιείτε Yahoo!;
          Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
          διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
          μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr












          Novo Internet Explorer 8: mais rápido e muito mais seguro. Baixe agora, é grátis!
          _________________________________________________________________
          Deixe suas conversas mais divertidas. Baixe agora mesmo novos emoticons. É grátis!
          http://specials.br.msn.com/ilovemessenger/pacotes.aspx

          [Non-text portions of this message have been removed]
        Your message has been successfully submitted and would be delivered to recipients shortly.