Finding Eccentricity of A Hyperbola












0












$begingroup$


Given an asymptote to an hyperbola and that a line perpendicular to it, intersects it at a single point, we need to find its eccentricity.



Asymptote : $5x-4y+5=0$ and Tangent : $4x+5y-7=0$.



I thought that if we consider asymptote to be limiting tangent at infinity, then the point of intersection $(3/41,55/41)$ should lie on the director circle of the hyperbola as it is the locus of the perpendicular tangents. I am finding too many variables to handle here. And I suspect this question should be solved by an argument for a specific type of hyperbola (Maybe rectangular), but I could use a hint over here






conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    The orthoptic for an ellipse and a hyperbola is its director circle. Please google 'director circle' .
    $endgroup$
    – Sarthak Rout
    Dec 7 '18 at 9:10












  • $begingroup$
    Google “orthoptic” and “director circle” yourself. The orthoptic of an ellipse is its director circle, but the orthoptic of a hyperbola is neither of its director circles.
    $endgroup$
    – amd
    Dec 7 '18 at 9:20












  • $begingroup$
    Look, I googled and I got this Wiki: In geometry, the director circle of an ellipse or hyperbola is a circle consisting of all points where two perpendicular tangent lines to the ellipse or hyperbola cross each other. And Orthoptic of a hyperbola as per wiki" The orthoptic of a hyperbola x^2/a^2 − y^2/b^2 = 1, a > b, is the circle x^2 + y^2 = a^2 − b^2 (in case of a ≤ b there are no orthogonal tangents, see below)". And , that is the director circle of the hyperbola. Please elaborate.
    $endgroup$
    – Sarthak Rout
    Dec 7 '18 at 9:29












  • $begingroup$
    There appear to be conflicting articles and terminology (not the first time). In en.wikipedia.org/wiki/Orthoptic_(geometry) a hyperbola’s orthoptic is pointedly not called its director circle; en.wikipedia.org/wiki/Hyperbola also avoids calling the orthoptic the director circle, but also uses the term “circular directrix” for the latter. Unlike the orthoptic, these circles exist for all hyperbolas, a useful trait if you’re going to use them for construction. A footnote suggests that the term “director circle” is used differently in German and English.
    $endgroup$
    – amd
    Dec 7 '18 at 10:10








  • 1




    $begingroup$
    At any rate, without further information or assumptions, I think that the best you can do here is to state an upper bound for the eccentricity.
    $endgroup$
    – amd
    Dec 7 '18 at 10:11
















0












$begingroup$


Given an asymptote to an hyperbola and that a line perpendicular to it, intersects it at a single point, we need to find its eccentricity.



Asymptote : $5x-4y+5=0$ and Tangent : $4x+5y-7=0$.



I thought that if we consider asymptote to be limiting tangent at infinity, then the point of intersection $(3/41,55/41)$ should lie on the director circle of the hyperbola as it is the locus of the perpendicular tangents. I am finding too many variables to handle here. And I suspect this question should be solved by an argument for a specific type of hyperbola (Maybe rectangular), but I could use a hint over here






conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    The orthoptic for an ellipse and a hyperbola is its director circle. Please google 'director circle' .
    $endgroup$
    – Sarthak Rout
    Dec 7 '18 at 9:10












  • $begingroup$
    Google “orthoptic” and “director circle” yourself. The orthoptic of an ellipse is its director circle, but the orthoptic of a hyperbola is neither of its director circles.
    $endgroup$
    – amd
    Dec 7 '18 at 9:20












  • $begingroup$
    Look, I googled and I got this Wiki: In geometry, the director circle of an ellipse or hyperbola is a circle consisting of all points where two perpendicular tangent lines to the ellipse or hyperbola cross each other. And Orthoptic of a hyperbola as per wiki" The orthoptic of a hyperbola x^2/a^2 − y^2/b^2 = 1, a > b, is the circle x^2 + y^2 = a^2 − b^2 (in case of a ≤ b there are no orthogonal tangents, see below)". And , that is the director circle of the hyperbola. Please elaborate.
    $endgroup$
    – Sarthak Rout
    Dec 7 '18 at 9:29












  • $begingroup$
    There appear to be conflicting articles and terminology (not the first time). In en.wikipedia.org/wiki/Orthoptic_(geometry) a hyperbola’s orthoptic is pointedly not called its director circle; en.wikipedia.org/wiki/Hyperbola also avoids calling the orthoptic the director circle, but also uses the term “circular directrix” for the latter. Unlike the orthoptic, these circles exist for all hyperbolas, a useful trait if you’re going to use them for construction. A footnote suggests that the term “director circle” is used differently in German and English.
    $endgroup$
    – amd
    Dec 7 '18 at 10:10








  • 1




    $begingroup$
    At any rate, without further information or assumptions, I think that the best you can do here is to state an upper bound for the eccentricity.
    $endgroup$
    – amd
    Dec 7 '18 at 10:11














0












0








0


0



$begingroup$


Given an asymptote to an hyperbola and that a line perpendicular to it, intersects it at a single point, we need to find its eccentricity.



Asymptote : $5x-4y+5=0$ and Tangent : $4x+5y-7=0$.



I thought that if we consider asymptote to be limiting tangent at infinity, then the point of intersection $(3/41,55/41)$ should lie on the director circle of the hyperbola as it is the locus of the perpendicular tangents. I am finding too many variables to handle here. And I suspect this question should be solved by an argument for a specific type of hyperbola (Maybe rectangular), but I could use a hint over here






conic-sections







share|cite|improve this question











$endgroup$




Given an asymptote to an hyperbola and that a line perpendicular to it, intersects it at a single point, we need to find its eccentricity.



Asymptote : $5x-4y+5=0$ and Tangent : $4x+5y-7=0$.



I thought that if we consider asymptote to be limiting tangent at infinity, then the point of intersection $(3/41,55/41)$ should lie on the director circle of the hyperbola as it is the locus of the perpendicular tangents. I am finding too many variables to handle here. And I suspect this question should be solved by an argument for a specific type of hyperbola (Maybe rectangular), but I could use a hint over here






conic-sections




conic-sections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 15:51







Sarthak Rout

















asked Dec 7 '18 at 8:14









Sarthak RoutSarthak Rout

488




488












  • $begingroup$
    The orthoptic for an ellipse and a hyperbola is its director circle. Please google 'director circle' .
    $endgroup$
    – Sarthak Rout
    Dec 7 '18 at 9:10












  • $begingroup$
    Google “orthoptic” and “director circle” yourself. The orthoptic of an ellipse is its director circle, but the orthoptic of a hyperbola is neither of its director circles.
    $endgroup$
    – amd
    Dec 7 '18 at 9:20












  • $begingroup$
    Look, I googled and I got this Wiki: In geometry, the director circle of an ellipse or hyperbola is a circle consisting of all points where two perpendicular tangent lines to the ellipse or hyperbola cross each other. And Orthoptic of a hyperbola as per wiki" The orthoptic of a hyperbola x^2/a^2 − y^2/b^2 = 1, a > b, is the circle x^2 + y^2 = a^2 − b^2 (in case of a ≤ b there are no orthogonal tangents, see below)". And , that is the director circle of the hyperbola. Please elaborate.
    $endgroup$
    – Sarthak Rout
    Dec 7 '18 at 9:29












  • $begingroup$
    There appear to be conflicting articles and terminology (not the first time). In en.wikipedia.org/wiki/Orthoptic_(geometry) a hyperbola’s orthoptic is pointedly not called its director circle; en.wikipedia.org/wiki/Hyperbola also avoids calling the orthoptic the director circle, but also uses the term “circular directrix” for the latter. Unlike the orthoptic, these circles exist for all hyperbolas, a useful trait if you’re going to use them for construction. A footnote suggests that the term “director circle” is used differently in German and English.
    $endgroup$
    – amd
    Dec 7 '18 at 10:10








  • 1




    $begingroup$
    At any rate, without further information or assumptions, I think that the best you can do here is to state an upper bound for the eccentricity.
    $endgroup$
    – amd
    Dec 7 '18 at 10:11


















  • $begingroup$
    The orthoptic for an ellipse and a hyperbola is its director circle. Please google 'director circle' .
    $endgroup$
    – Sarthak Rout
    Dec 7 '18 at 9:10












  • $begingroup$
    Google “orthoptic” and “director circle” yourself. The orthoptic of an ellipse is its director circle, but the orthoptic of a hyperbola is neither of its director circles.
    $endgroup$
    – amd
    Dec 7 '18 at 9:20












  • $begingroup$
    Look, I googled and I got this Wiki: In geometry, the director circle of an ellipse or hyperbola is a circle consisting of all points where two perpendicular tangent lines to the ellipse or hyperbola cross each other. And Orthoptic of a hyperbola as per wiki" The orthoptic of a hyperbola x^2/a^2 − y^2/b^2 = 1, a > b, is the circle x^2 + y^2 = a^2 − b^2 (in case of a ≤ b there are no orthogonal tangents, see below)". And , that is the director circle of the hyperbola. Please elaborate.
    $endgroup$
    – Sarthak Rout
    Dec 7 '18 at 9:29












  • $begingroup$
    There appear to be conflicting articles and terminology (not the first time). In en.wikipedia.org/wiki/Orthoptic_(geometry) a hyperbola’s orthoptic is pointedly not called its director circle; en.wikipedia.org/wiki/Hyperbola also avoids calling the orthoptic the director circle, but also uses the term “circular directrix” for the latter. Unlike the orthoptic, these circles exist for all hyperbolas, a useful trait if you’re going to use them for construction. A footnote suggests that the term “director circle” is used differently in German and English.
    $endgroup$
    – amd
    Dec 7 '18 at 10:10








  • 1




    $begingroup$
    At any rate, without further information or assumptions, I think that the best you can do here is to state an upper bound for the eccentricity.
    $endgroup$
    – amd
    Dec 7 '18 at 10:11
















$begingroup$
The orthoptic for an ellipse and a hyperbola is its director circle. Please google 'director circle' .
$endgroup$
– Sarthak Rout
Dec 7 '18 at 9:10






$begingroup$
The orthoptic for an ellipse and a hyperbola is its director circle. Please google 'director circle' .
$endgroup$
– Sarthak Rout
Dec 7 '18 at 9:10














$begingroup$
Google “orthoptic” and “director circle” yourself. The orthoptic of an ellipse is its director circle, but the orthoptic of a hyperbola is neither of its director circles.
$endgroup$
– amd
Dec 7 '18 at 9:20






$begingroup$
Google “orthoptic” and “director circle” yourself. The orthoptic of an ellipse is its director circle, but the orthoptic of a hyperbola is neither of its director circles.
$endgroup$
– amd
Dec 7 '18 at 9:20














$begingroup$
Look, I googled and I got this Wiki: In geometry, the director circle of an ellipse or hyperbola is a circle consisting of all points where two perpendicular tangent lines to the ellipse or hyperbola cross each other. And Orthoptic of a hyperbola as per wiki" The orthoptic of a hyperbola x^2/a^2 − y^2/b^2 = 1, a > b, is the circle x^2 + y^2 = a^2 − b^2 (in case of a ≤ b there are no orthogonal tangents, see below)". And , that is the director circle of the hyperbola. Please elaborate.
$endgroup$
– Sarthak Rout
Dec 7 '18 at 9:29






$begingroup$
Look, I googled and I got this Wiki: In geometry, the director circle of an ellipse or hyperbola is a circle consisting of all points where two perpendicular tangent lines to the ellipse or hyperbola cross each other. And Orthoptic of a hyperbola as per wiki" The orthoptic of a hyperbola x^2/a^2 − y^2/b^2 = 1, a > b, is the circle x^2 + y^2 = a^2 − b^2 (in case of a ≤ b there are no orthogonal tangents, see below)". And , that is the director circle of the hyperbola. Please elaborate.
$endgroup$
– Sarthak Rout
Dec 7 '18 at 9:29














$begingroup$
There appear to be conflicting articles and terminology (not the first time). In en.wikipedia.org/wiki/Orthoptic_(geometry) a hyperbola’s orthoptic is pointedly not called its director circle; en.wikipedia.org/wiki/Hyperbola also avoids calling the orthoptic the director circle, but also uses the term “circular directrix” for the latter. Unlike the orthoptic, these circles exist for all hyperbolas, a useful trait if you’re going to use them for construction. A footnote suggests that the term “director circle” is used differently in German and English.
$endgroup$
– amd
Dec 7 '18 at 10:10






$begingroup$
There appear to be conflicting articles and terminology (not the first time). In en.wikipedia.org/wiki/Orthoptic_(geometry) a hyperbola’s orthoptic is pointedly not called its director circle; en.wikipedia.org/wiki/Hyperbola also avoids calling the orthoptic the director circle, but also uses the term “circular directrix” for the latter. Unlike the orthoptic, these circles exist for all hyperbolas, a useful trait if you’re going to use them for construction. A footnote suggests that the term “director circle” is used differently in German and English.
$endgroup$
– amd
Dec 7 '18 at 10:10






1




1




$begingroup$
At any rate, without further information or assumptions, I think that the best you can do here is to state an upper bound for the eccentricity.
$endgroup$
– amd
Dec 7 '18 at 10:11




$begingroup$
At any rate, without further information or assumptions, I think that the best you can do here is to state an upper bound for the eccentricity.
$endgroup$
– amd
Dec 7 '18 at 10:11










1 Answer
1






active

oldest

votes


















0












$begingroup$

The given data are not enough to fix the hyperbola eccentricity. You can better understand that with a simpler example: suppose you are given $y=0$ as an asymptote and $x=a$ as tangent, then take $y=mx$ as the second asymptote (with $m$ any non-vanishing real number). The equation of a hyperbola having those two asymptotes and tangent to line $x=a$ can be easily found to be
$$y(y-mx)=-{a^2m^2over4}.$$
The eccentricity of the above hyperbola is
$$
e={sqrt2over m}sqrt{1+m^2-sqrt{1+m^2}},
$$
hence it can take any value between $1$ and $sqrt{2}$, depending on the value of $m$.



EDIT.



With a rotation and a translation, the above example can be adapted to your data. You may check that the hyperbolas of equation
$$
(5x-4y+5)left((4-5m)y-(5+4m)x-5+{25over4}mright)={9over64}m^2
$$
are all tangent to line $4x+5y-7=0$ and have line $5x-4y+5=0$ as asymptote, for any value of $m$. The eccentricities of these hyperbolas are given by the same formula as before.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have edited the question. Please see, if you can arrive at a result.
    $endgroup$
    – Sarthak Rout
    Dec 9 '18 at 15:52










  • $begingroup$
    I don't see any relevant change in your question: without some other information you cannot find the eccentricity. I used different lines in the above example just to keep calculations easy, but you could do exactly the same reasoning with your data, and find the same result.
    $endgroup$
    – Aretino
    Dec 9 '18 at 16:27











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

The given data are not enough to fix the hyperbola eccentricity. You can better understand that with a simpler example: suppose you are given $y=0$ as an asymptote and $x=a$ as tangent, then take $y=mx$ as the second asymptote (with $m$ any non-vanishing real number). The equation of a hyperbola having those two asymptotes and tangent to line $x=a$ can be easily found to be
$$y(y-mx)=-{a^2m^2over4}.$$
The eccentricity of the above hyperbola is
$$
e={sqrt2over m}sqrt{1+m^2-sqrt{1+m^2}},
$$
hence it can take any value between $1$ and $sqrt{2}$, depending on the value of $m$.



EDIT.



With a rotation and a translation, the above example can be adapted to your data. You may check that the hyperbolas of equation
$$
(5x-4y+5)left((4-5m)y-(5+4m)x-5+{25over4}mright)={9over64}m^2
$$
are all tangent to line $4x+5y-7=0$ and have line $5x-4y+5=0$ as asymptote, for any value of $m$. The eccentricities of these hyperbolas are given by the same formula as before.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have edited the question. Please see, if you can arrive at a result.
    $endgroup$
    – Sarthak Rout
    Dec 9 '18 at 15:52










  • $begingroup$
    I don't see any relevant change in your question: without some other information you cannot find the eccentricity. I used different lines in the above example just to keep calculations easy, but you could do exactly the same reasoning with your data, and find the same result.
    $endgroup$
    – Aretino
    Dec 9 '18 at 16:27
















0












$begingroup$

The given data are not enough to fix the hyperbola eccentricity. You can better understand that with a simpler example: suppose you are given $y=0$ as an asymptote and $x=a$ as tangent, then take $y=mx$ as the second asymptote (with $m$ any non-vanishing real number). The equation of a hyperbola having those two asymptotes and tangent to line $x=a$ can be easily found to be
$$y(y-mx)=-{a^2m^2over4}.$$
The eccentricity of the above hyperbola is
$$
e={sqrt2over m}sqrt{1+m^2-sqrt{1+m^2}},
$$
hence it can take any value between $1$ and $sqrt{2}$, depending on the value of $m$.



EDIT.



With a rotation and a translation, the above example can be adapted to your data. You may check that the hyperbolas of equation
$$
(5x-4y+5)left((4-5m)y-(5+4m)x-5+{25over4}mright)={9over64}m^2
$$
are all tangent to line $4x+5y-7=0$ and have line $5x-4y+5=0$ as asymptote, for any value of $m$. The eccentricities of these hyperbolas are given by the same formula as before.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have edited the question. Please see, if you can arrive at a result.
    $endgroup$
    – Sarthak Rout
    Dec 9 '18 at 15:52










  • $begingroup$
    I don't see any relevant change in your question: without some other information you cannot find the eccentricity. I used different lines in the above example just to keep calculations easy, but you could do exactly the same reasoning with your data, and find the same result.
    $endgroup$
    – Aretino
    Dec 9 '18 at 16:27














0












0








0





$begingroup$

The given data are not enough to fix the hyperbola eccentricity. You can better understand that with a simpler example: suppose you are given $y=0$ as an asymptote and $x=a$ as tangent, then take $y=mx$ as the second asymptote (with $m$ any non-vanishing real number). The equation of a hyperbola having those two asymptotes and tangent to line $x=a$ can be easily found to be
$$y(y-mx)=-{a^2m^2over4}.$$
The eccentricity of the above hyperbola is
$$
e={sqrt2over m}sqrt{1+m^2-sqrt{1+m^2}},
$$
hence it can take any value between $1$ and $sqrt{2}$, depending on the value of $m$.



EDIT.



With a rotation and a translation, the above example can be adapted to your data. You may check that the hyperbolas of equation
$$
(5x-4y+5)left((4-5m)y-(5+4m)x-5+{25over4}mright)={9over64}m^2
$$
are all tangent to line $4x+5y-7=0$ and have line $5x-4y+5=0$ as asymptote, for any value of $m$. The eccentricities of these hyperbolas are given by the same formula as before.






share|cite|improve this answer











$endgroup$



The given data are not enough to fix the hyperbola eccentricity. You can better understand that with a simpler example: suppose you are given $y=0$ as an asymptote and $x=a$ as tangent, then take $y=mx$ as the second asymptote (with $m$ any non-vanishing real number). The equation of a hyperbola having those two asymptotes and tangent to line $x=a$ can be easily found to be
$$y(y-mx)=-{a^2m^2over4}.$$
The eccentricity of the above hyperbola is
$$
e={sqrt2over m}sqrt{1+m^2-sqrt{1+m^2}},
$$
hence it can take any value between $1$ and $sqrt{2}$, depending on the value of $m$.



EDIT.



With a rotation and a translation, the above example can be adapted to your data. You may check that the hyperbolas of equation
$$
(5x-4y+5)left((4-5m)y-(5+4m)x-5+{25over4}mright)={9over64}m^2
$$
are all tangent to line $4x+5y-7=0$ and have line $5x-4y+5=0$ as asymptote, for any value of $m$. The eccentricities of these hyperbolas are given by the same formula as before.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 13:41

























answered Dec 8 '18 at 17:07









AretinoAretino

23.5k21443




23.5k21443












  • $begingroup$
    I have edited the question. Please see, if you can arrive at a result.
    $endgroup$
    – Sarthak Rout
    Dec 9 '18 at 15:52










  • $begingroup$
    I don't see any relevant change in your question: without some other information you cannot find the eccentricity. I used different lines in the above example just to keep calculations easy, but you could do exactly the same reasoning with your data, and find the same result.
    $endgroup$
    – Aretino
    Dec 9 '18 at 16:27


















  • $begingroup$
    I have edited the question. Please see, if you can arrive at a result.
    $endgroup$
    – Sarthak Rout
    Dec 9 '18 at 15:52










  • $begingroup$
    I don't see any relevant change in your question: without some other information you cannot find the eccentricity. I used different lines in the above example just to keep calculations easy, but you could do exactly the same reasoning with your data, and find the same result.
    $endgroup$
    – Aretino
    Dec 9 '18 at 16:27
















$begingroup$
I have edited the question. Please see, if you can arrive at a result.
$endgroup$
– Sarthak Rout
Dec 9 '18 at 15:52




$begingroup$
I have edited the question. Please see, if you can arrive at a result.
$endgroup$
– Sarthak Rout
Dec 9 '18 at 15:52












$begingroup$
I don't see any relevant change in your question: without some other information you cannot find the eccentricity. I used different lines in the above example just to keep calculations easy, but you could do exactly the same reasoning with your data, and find the same result.
$endgroup$
– Aretino
Dec 9 '18 at 16:27




$begingroup$
I don't see any relevant change in your question: without some other information you cannot find the eccentricity. I used different lines in the above example just to keep calculations easy, but you could do exactly the same reasoning with your data, and find the same result.
$endgroup$
– Aretino
Dec 9 '18 at 16:27


















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Bundesstraße 106

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Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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Verónica Boquete

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Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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