Find the locus of perpendicular drawn from focus upon variable tangent to the parabola...












0












$begingroup$


Find the locus of perpendicular drawn from focus upon variable tangent to the parabola $(2x-y+1)^2=frac{8}{sqrt{5}}(x+2y+3)$.



My approach
I am trying to convert above equation in parabolic form



$frac{(ax+by+c)^2}{a^2+b^2}=(x-alpha)^2+(y-beta)^2$



where $ ax+by+c=0$ is the equation of directrix and ($alpha,beta$) is the focus of the parabola but getting complicated.






conic-sections







share|cite|improve this question









$endgroup$












  • $begingroup$
    $$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
    $endgroup$
    – Jan-Magnus Økland
    Dec 12 '18 at 6:11
















0












$begingroup$


Find the locus of perpendicular drawn from focus upon variable tangent to the parabola $(2x-y+1)^2=frac{8}{sqrt{5}}(x+2y+3)$.



My approach
I am trying to convert above equation in parabolic form



$frac{(ax+by+c)^2}{a^2+b^2}=(x-alpha)^2+(y-beta)^2$



where $ ax+by+c=0$ is the equation of directrix and ($alpha,beta$) is the focus of the parabola but getting complicated.






conic-sections







share|cite|improve this question









$endgroup$












  • $begingroup$
    $$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
    $endgroup$
    – Jan-Magnus Økland
    Dec 12 '18 at 6:11














0












0








0





$begingroup$


Find the locus of perpendicular drawn from focus upon variable tangent to the parabola $(2x-y+1)^2=frac{8}{sqrt{5}}(x+2y+3)$.



My approach
I am trying to convert above equation in parabolic form



$frac{(ax+by+c)^2}{a^2+b^2}=(x-alpha)^2+(y-beta)^2$



where $ ax+by+c=0$ is the equation of directrix and ($alpha,beta$) is the focus of the parabola but getting complicated.






conic-sections







share|cite|improve this question









$endgroup$




Find the locus of perpendicular drawn from focus upon variable tangent to the parabola $(2x-y+1)^2=frac{8}{sqrt{5}}(x+2y+3)$.



My approach
I am trying to convert above equation in parabolic form



$frac{(ax+by+c)^2}{a^2+b^2}=(x-alpha)^2+(y-beta)^2$



where $ ax+by+c=0$ is the equation of directrix and ($alpha,beta$) is the focus of the parabola but getting complicated.






conic-sections




conic-sections






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 4:51









Samar Imam ZaidiSamar Imam Zaidi

1,5501520




1,5501520












  • $begingroup$
    $$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
    $endgroup$
    – Jan-Magnus Økland
    Dec 12 '18 at 6:11


















  • $begingroup$
    $$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
    $endgroup$
    – Jan-Magnus Økland
    Dec 12 '18 at 6:11
















$begingroup$
$$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
$endgroup$
– Jan-Magnus Økland
Dec 12 '18 at 6:11




$begingroup$
$$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
$endgroup$
– Jan-Magnus Økland
Dec 12 '18 at 6:11










3 Answers
3






active

oldest

votes


















2












$begingroup$

Hint :



Let $Y=2x-y+1 (1)$ and $4aX=dfrac{8(x+2y+3)}{sqrt5}$ with $x+2y+3=X
(2), a=?$



whose focus is $(a,0)$



Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$



Find the equation of the tangent at $P$



Find the equation of the normal for the tangent passing through $(a,0)$



Find the intersection of the normal with the tangent.



Eliminate $t$



Replace the values of $X,Y$ with $x,y$ using $(1),(2)$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
      $endgroup$
      – amd
      Dec 12 '18 at 7:54












    • $begingroup$
      I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
      $endgroup$
      – Sri Krishna Sahoo
      Dec 12 '18 at 7:58





















    1












    $begingroup$

    The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.



    That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.






    share|cite|improve this answer











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036241%2ffind-the-locus-of-perpendicular-drawn-from-focus-upon-variable-tangent-to-the-pa%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Hint :



      Let $Y=2x-y+1 (1)$ and $4aX=dfrac{8(x+2y+3)}{sqrt5}$ with $x+2y+3=X
      (2), a=?$



      whose focus is $(a,0)$



      Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$



      Find the equation of the tangent at $P$



      Find the equation of the normal for the tangent passing through $(a,0)$



      Find the intersection of the normal with the tangent.



      Eliminate $t$



      Replace the values of $X,Y$ with $x,y$ using $(1),(2)$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Hint :



        Let $Y=2x-y+1 (1)$ and $4aX=dfrac{8(x+2y+3)}{sqrt5}$ with $x+2y+3=X
        (2), a=?$



        whose focus is $(a,0)$



        Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$



        Find the equation of the tangent at $P$



        Find the equation of the normal for the tangent passing through $(a,0)$



        Find the intersection of the normal with the tangent.



        Eliminate $t$



        Replace the values of $X,Y$ with $x,y$ using $(1),(2)$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint :



          Let $Y=2x-y+1 (1)$ and $4aX=dfrac{8(x+2y+3)}{sqrt5}$ with $x+2y+3=X
          (2), a=?$



          whose focus is $(a,0)$



          Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$



          Find the equation of the tangent at $P$



          Find the equation of the normal for the tangent passing through $(a,0)$



          Find the intersection of the normal with the tangent.



          Eliminate $t$



          Replace the values of $X,Y$ with $x,y$ using $(1),(2)$






          share|cite|improve this answer









          $endgroup$



          Hint :



          Let $Y=2x-y+1 (1)$ and $4aX=dfrac{8(x+2y+3)}{sqrt5}$ with $x+2y+3=X
          (2), a=?$



          whose focus is $(a,0)$



          Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$



          Find the equation of the tangent at $P$



          Find the equation of the normal for the tangent passing through $(a,0)$



          Find the intersection of the normal with the tangent.



          Eliminate $t$



          Replace the values of $X,Y$ with $x,y$ using $(1),(2)$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 5:21









          lab bhattacharjeelab bhattacharjee

          226k15157275




          226k15157275























              1












              $begingroup$

              There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
                $endgroup$
                – amd
                Dec 12 '18 at 7:54












              • $begingroup$
                I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
                $endgroup$
                – Sri Krishna Sahoo
                Dec 12 '18 at 7:58


















              1












              $begingroup$

              There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
                $endgroup$
                – amd
                Dec 12 '18 at 7:54












              • $begingroup$
                I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
                $endgroup$
                – Sri Krishna Sahoo
                Dec 12 '18 at 7:58
















              1












              1








              1





              $begingroup$

              There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.






              share|cite|improve this answer









              $endgroup$



              There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 12 '18 at 7:31









              Sri Krishna SahooSri Krishna Sahoo

              606217




              606217












              • $begingroup$
                This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
                $endgroup$
                – amd
                Dec 12 '18 at 7:54












              • $begingroup$
                I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
                $endgroup$
                – Sri Krishna Sahoo
                Dec 12 '18 at 7:58




















              • $begingroup$
                This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
                $endgroup$
                – amd
                Dec 12 '18 at 7:54












              • $begingroup$
                I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
                $endgroup$
                – Sri Krishna Sahoo
                Dec 12 '18 at 7:58


















              $begingroup$
              This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
              $endgroup$
              – amd
              Dec 12 '18 at 7:54






              $begingroup$
              This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
              $endgroup$
              – amd
              Dec 12 '18 at 7:54














              $begingroup$
              I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
              $endgroup$
              – Sri Krishna Sahoo
              Dec 12 '18 at 7:58






              $begingroup$
              I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
              $endgroup$
              – Sri Krishna Sahoo
              Dec 12 '18 at 7:58













              1












              $begingroup$

              The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.



              That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.



                That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.



                  That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.






                  share|cite|improve this answer











                  $endgroup$



                  The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.



                  That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 12 '18 at 8:20

























                  answered Dec 12 '18 at 8:00









                  amdamd

                  30.6k21050




                  30.6k21050






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036241%2ffind-the-locus-of-perpendicular-drawn-from-focus-upon-variable-tangent-to-the-pa%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                      Read more

                      Puebla de Zaragoza

                      Image
                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                      Read more

                      Musa

                      Image
                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                      Read more