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

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

Expand Messages
  • Luís Lopes
    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
    Message 1 of 5 , Jun 12, 2009
    • 0 Attachment
      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]
    • Nikolaos Dergiades
      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
      Message 2 of 5 , Jun 12, 2009
      • 0 Attachment
        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
      • 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 3 of 5 , Jun 12, 2009
        • 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 4 of 5 , Jun 12, 2009
          • 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 5 of 5 , Jun 12, 2009
            • 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.