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• 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 6:05 AM
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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.

Best regards,

Luis

To: Hyacinthos@yahoogroups.com
From: anopolis72@...
Date: Thu, 28 May 2009 21:59:12 +0000

http://webpages.ull.es/users/amontes/otrashtm/varios.htm

It contains interesting triangle geometry files
in pdf format (in Spanish).

Antreas

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

[Non-text portions of this message have been removed]
• 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 8:31 AM
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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

> 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
>
>
>
>
>
>
>
>
>
>
> http://webpages.ull.es/users/amontes/otrashtm/varios.htm
>
> It contains interesting triangle geometry files
> in pdf format (in Spanish).
>
>
> Antreas
>
>
>
>
>
>
>
>
>
> _________________________________________________________________
> Conheça os novos produtos Windows Live! Clique aqui.
> http://www.windowslive.com.br
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
>
>
>     mailto:Hyacinthos-fullfeatured@yahoogroups.com
>
>
>

___________________________________________________________
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• 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 9:10 AM
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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
Date: Fri, 12 Jun 2009 15:31:12 +0000

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

> 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
>
>
>
>
>
>
>
>
>
>
> http://webpages.ull.es/users/amontes/otrashtm/varios.htm
>
> It contains interesting triangle geometry files
> in pdf format (in Spanish).
>
>
> Antreas
>
>
>
>
>
>
>
>
>
> __________________________________________________________
> Conheça os novos produtos Windows Live! Clique aqui.
> http://www.windowslive.com.br
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
>
>
> mailto:Hyacinthos-fullfeatured@yahoogroups.com
>
>
>

__________________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
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_________________________________________________________________
Conheça os novos produtos Windows Live! Clique aqui.
http://www.windowslive.com.br

[Non-text portions of this message have been removed]
• Dear Luis, Sorry. You are right. Best regards Nikos Dergiades ... ___________________________________________________________ Χρησιμοποιείτε
Message 4 of 5 , Jun 12 9:46 AM
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Dear Luis,
Sorry. You are right.

Best regards

>
>
>
>
> 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
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• 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 9:49 AM
• 0 Attachment
Dear Hyacinthists,

Resending.

From: qed_texte@...
To: hyacinthos@yahoogroups.com
Date: Fri, 12 Jun 2009 16:10:45 +0000

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
Date: Fri, 12 Jun 2009 15:31:12 +0000

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

> 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
>
>
>
>
>
>
>
>
>
>
> http://webpages.ull.es/users/amontes/otrashtm/varios.htm
>
> It contains interesting triangle geometry files
> in pdf format (in Spanish).
>
>
> Antreas
>
>
>
>
>
>
>
>
>
> __________________________________________________________
> Conheça os novos produtos Windows Live! Clique aqui.
> http://www.windowslive.com.br
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
>
>
> mailto:Hyacinthos-fullfeatured@yahoogroups.com
>
>
>

__________________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών

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