equation of a chord of an ellipse












0












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There is an ellipse that has equation $x^2+4y^2=36$ and one point inside it $A=(2,1)$. The point is a centre of a chord of an ellipse. I don't know how to find equation of a chord using analytic geometry. Thanks for any ideas how about to start.






analytic-geometry







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  • $begingroup$
    The question is arguably bad. There were two close votes, but they aged away. Sorry about forgetting to consider that before starting a bounty.
    $endgroup$
    – Jyrki Lahtonen
    Sep 20 '15 at 19:16
















0












$begingroup$


There is an ellipse that has equation $x^2+4y^2=36$ and one point inside it $A=(2,1)$. The point is a centre of a chord of an ellipse. I don't know how to find equation of a chord using analytic geometry. Thanks for any ideas how about to start.






analytic-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    The question is arguably bad. There were two close votes, but they aged away. Sorry about forgetting to consider that before starting a bounty.
    $endgroup$
    – Jyrki Lahtonen
    Sep 20 '15 at 19:16














0












0








0


0



$begingroup$


There is an ellipse that has equation $x^2+4y^2=36$ and one point inside it $A=(2,1)$. The point is a centre of a chord of an ellipse. I don't know how to find equation of a chord using analytic geometry. Thanks for any ideas how about to start.






analytic-geometry







share|cite|improve this question









$endgroup$




There is an ellipse that has equation $x^2+4y^2=36$ and one point inside it $A=(2,1)$. The point is a centre of a chord of an ellipse. I don't know how to find equation of a chord using analytic geometry. Thanks for any ideas how about to start.






analytic-geometry




analytic-geometry






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asked Sep 13 '15 at 15:27









Tomb22Tomb22

112




112












  • $begingroup$
    The question is arguably bad. There were two close votes, but they aged away. Sorry about forgetting to consider that before starting a bounty.
    $endgroup$
    – Jyrki Lahtonen
    Sep 20 '15 at 19:16


















  • $begingroup$
    The question is arguably bad. There were two close votes, but they aged away. Sorry about forgetting to consider that before starting a bounty.
    $endgroup$
    – Jyrki Lahtonen
    Sep 20 '15 at 19:16
















$begingroup$
The question is arguably bad. There were two close votes, but they aged away. Sorry about forgetting to consider that before starting a bounty.
$endgroup$
– Jyrki Lahtonen
Sep 20 '15 at 19:16




$begingroup$
The question is arguably bad. There were two close votes, but they aged away. Sorry about forgetting to consider that before starting a bounty.
$endgroup$
– Jyrki Lahtonen
Sep 20 '15 at 19:16










4 Answers
4






active

oldest

votes


















7





+100







$begingroup$

I think the trick here is to change coordinates.
If you scale one of the axis, two equal length lines will continue equal after that scaling.
So, if you take $z=2y$, then the ellipse becomes a circle $x^2 + z^2 = 6^2$, and you point becomes $P=(2,2)$.
With that, you can see the chord, in the new coordinates is $z=-x+4$.



So, by returning to the old coordinates it becomes $2y = -x + 4 $.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    that's beautiful
    $endgroup$
    – Mrigank
    Jan 4 '17 at 2:29



















1












$begingroup$

The equation of the line with gradient $m$ passing through $(2,1)$ is $$y-1=m(x-2)$$
Substitute for $y$ into the equation of the ellipse and you get a quadratic equation $$x^2(1+4m^2)+x(8m-16m^2)-16m-32=0$$
The roots of the equation are $x_1$ and $x_2$, and the sum of the roots of a quadratic is $-frac ba$. But we also know that $$frac {x_1+x_2}{2}=2$$ since $(2,1)$ is the midpoint.
Therefore we have $$-frac{8m-16m^2}{1+4m^2}=4$$
$$Rightarrow m=-frac 12$$
Hence you have the equation of the chord.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    suppose you parametrize the ellipse by $x = 6cos t, y = 3 sin t, 0 le t le 2pi.$ suppose take chord is given by points $t_1, t_2$ and whose center is $A= (2,1).$



    then we have
    $$begin{align} cos t_1 + cos t_2 &= 2/3 tag 1\
    sin t_1 + sin t_2 &= 2/3. tag 2end{align}$$



    squaring the two equations and subtracting i them gives $$cos (t_1 + t_2) = 0 to t_1 + t_2 = pi/2, 3pi/2, ldots$$
    subbing $t_2 = pi/2 - t_1$ in (1) and $(2)$ gives $sin t_1 + cos t_1 = 2/3$ but subbing $t_2 = 3pi/2 - t_1$ in $(1), (2)$ give $sin t_1 - cos t_1 = 2/3$ and $sin t_1 - cos t_1 = -2/3.$ therefore $$t_1 + t_2=pi/2. $$



    the slope of the chord is $$ frac{3(sin t_1 - sin t_2)}{6(cos t_1 - cos t_2)}=-frac12frac{cos (t_1 + t_2)/2}{sin(t_1+t_2)/2}= -frac12$$ the equation of the required chord is $$frac{y-1}{x-2} = -frac12 to 2y = -x+4.$$






    share|cite|improve this answer









    $endgroup$





















      -1












      $begingroup$

      Equation of chord having midpoint $(h, k)$ is given by
      $T=S^1$



      $T=$ equation of tangent at $(h,k)$, $S =$ putting values of $(h,k)$ in ellipse's equation



      $(h, k)=(2,1)$



      Ellipse is $x^2+4y^2=36$



      $T=x(2)+4y(1)-36$



      $S^1= 2^2+4(1)^2-36$



      Doing $T=S^1$



      we get



      $2x+4y=8$



      $implies x+2y=4$






      share|cite|improve this answer











      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7





        +100







        $begingroup$

        I think the trick here is to change coordinates.
        If you scale one of the axis, two equal length lines will continue equal after that scaling.
        So, if you take $z=2y$, then the ellipse becomes a circle $x^2 + z^2 = 6^2$, and you point becomes $P=(2,2)$.
        With that, you can see the chord, in the new coordinates is $z=-x+4$.



        So, by returning to the old coordinates it becomes $2y = -x + 4 $.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          that's beautiful
          $endgroup$
          – Mrigank
          Jan 4 '17 at 2:29
















        7





        +100







        $begingroup$

        I think the trick here is to change coordinates.
        If you scale one of the axis, two equal length lines will continue equal after that scaling.
        So, if you take $z=2y$, then the ellipse becomes a circle $x^2 + z^2 = 6^2$, and you point becomes $P=(2,2)$.
        With that, you can see the chord, in the new coordinates is $z=-x+4$.



        So, by returning to the old coordinates it becomes $2y = -x + 4 $.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          that's beautiful
          $endgroup$
          – Mrigank
          Jan 4 '17 at 2:29














        7





        +100







        7





        +100



        7




        +100



        $begingroup$

        I think the trick here is to change coordinates.
        If you scale one of the axis, two equal length lines will continue equal after that scaling.
        So, if you take $z=2y$, then the ellipse becomes a circle $x^2 + z^2 = 6^2$, and you point becomes $P=(2,2)$.
        With that, you can see the chord, in the new coordinates is $z=-x+4$.



        So, by returning to the old coordinates it becomes $2y = -x + 4 $.






        share|cite|improve this answer









        $endgroup$



        I think the trick here is to change coordinates.
        If you scale one of the axis, two equal length lines will continue equal after that scaling.
        So, if you take $z=2y$, then the ellipse becomes a circle $x^2 + z^2 = 6^2$, and you point becomes $P=(2,2)$.
        With that, you can see the chord, in the new coordinates is $z=-x+4$.



        So, by returning to the old coordinates it becomes $2y = -x + 4 $.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 13 '15 at 16:02









        DiamantinoDiamantino

        1862




        1862












        • $begingroup$
          that's beautiful
          $endgroup$
          – Mrigank
          Jan 4 '17 at 2:29


















        • $begingroup$
          that's beautiful
          $endgroup$
          – Mrigank
          Jan 4 '17 at 2:29
















        $begingroup$
        that's beautiful
        $endgroup$
        – Mrigank
        Jan 4 '17 at 2:29




        $begingroup$
        that's beautiful
        $endgroup$
        – Mrigank
        Jan 4 '17 at 2:29











        1












        $begingroup$

        The equation of the line with gradient $m$ passing through $(2,1)$ is $$y-1=m(x-2)$$
        Substitute for $y$ into the equation of the ellipse and you get a quadratic equation $$x^2(1+4m^2)+x(8m-16m^2)-16m-32=0$$
        The roots of the equation are $x_1$ and $x_2$, and the sum of the roots of a quadratic is $-frac ba$. But we also know that $$frac {x_1+x_2}{2}=2$$ since $(2,1)$ is the midpoint.
        Therefore we have $$-frac{8m-16m^2}{1+4m^2}=4$$
        $$Rightarrow m=-frac 12$$
        Hence you have the equation of the chord.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          The equation of the line with gradient $m$ passing through $(2,1)$ is $$y-1=m(x-2)$$
          Substitute for $y$ into the equation of the ellipse and you get a quadratic equation $$x^2(1+4m^2)+x(8m-16m^2)-16m-32=0$$
          The roots of the equation are $x_1$ and $x_2$, and the sum of the roots of a quadratic is $-frac ba$. But we also know that $$frac {x_1+x_2}{2}=2$$ since $(2,1)$ is the midpoint.
          Therefore we have $$-frac{8m-16m^2}{1+4m^2}=4$$
          $$Rightarrow m=-frac 12$$
          Hence you have the equation of the chord.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            The equation of the line with gradient $m$ passing through $(2,1)$ is $$y-1=m(x-2)$$
            Substitute for $y$ into the equation of the ellipse and you get a quadratic equation $$x^2(1+4m^2)+x(8m-16m^2)-16m-32=0$$
            The roots of the equation are $x_1$ and $x_2$, and the sum of the roots of a quadratic is $-frac ba$. But we also know that $$frac {x_1+x_2}{2}=2$$ since $(2,1)$ is the midpoint.
            Therefore we have $$-frac{8m-16m^2}{1+4m^2}=4$$
            $$Rightarrow m=-frac 12$$
            Hence you have the equation of the chord.






            share|cite|improve this answer









            $endgroup$



            The equation of the line with gradient $m$ passing through $(2,1)$ is $$y-1=m(x-2)$$
            Substitute for $y$ into the equation of the ellipse and you get a quadratic equation $$x^2(1+4m^2)+x(8m-16m^2)-16m-32=0$$
            The roots of the equation are $x_1$ and $x_2$, and the sum of the roots of a quadratic is $-frac ba$. But we also know that $$frac {x_1+x_2}{2}=2$$ since $(2,1)$ is the midpoint.
            Therefore we have $$-frac{8m-16m^2}{1+4m^2}=4$$
            $$Rightarrow m=-frac 12$$
            Hence you have the equation of the chord.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 13 '15 at 16:05









            David QuinnDavid Quinn

            23.9k21141




            23.9k21141























                1












                $begingroup$

                suppose you parametrize the ellipse by $x = 6cos t, y = 3 sin t, 0 le t le 2pi.$ suppose take chord is given by points $t_1, t_2$ and whose center is $A= (2,1).$



                then we have
                $$begin{align} cos t_1 + cos t_2 &= 2/3 tag 1\
                sin t_1 + sin t_2 &= 2/3. tag 2end{align}$$



                squaring the two equations and subtracting i them gives $$cos (t_1 + t_2) = 0 to t_1 + t_2 = pi/2, 3pi/2, ldots$$
                subbing $t_2 = pi/2 - t_1$ in (1) and $(2)$ gives $sin t_1 + cos t_1 = 2/3$ but subbing $t_2 = 3pi/2 - t_1$ in $(1), (2)$ give $sin t_1 - cos t_1 = 2/3$ and $sin t_1 - cos t_1 = -2/3.$ therefore $$t_1 + t_2=pi/2. $$



                the slope of the chord is $$ frac{3(sin t_1 - sin t_2)}{6(cos t_1 - cos t_2)}=-frac12frac{cos (t_1 + t_2)/2}{sin(t_1+t_2)/2}= -frac12$$ the equation of the required chord is $$frac{y-1}{x-2} = -frac12 to 2y = -x+4.$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  suppose you parametrize the ellipse by $x = 6cos t, y = 3 sin t, 0 le t le 2pi.$ suppose take chord is given by points $t_1, t_2$ and whose center is $A= (2,1).$



                  then we have
                  $$begin{align} cos t_1 + cos t_2 &= 2/3 tag 1\
                  sin t_1 + sin t_2 &= 2/3. tag 2end{align}$$



                  squaring the two equations and subtracting i them gives $$cos (t_1 + t_2) = 0 to t_1 + t_2 = pi/2, 3pi/2, ldots$$
                  subbing $t_2 = pi/2 - t_1$ in (1) and $(2)$ gives $sin t_1 + cos t_1 = 2/3$ but subbing $t_2 = 3pi/2 - t_1$ in $(1), (2)$ give $sin t_1 - cos t_1 = 2/3$ and $sin t_1 - cos t_1 = -2/3.$ therefore $$t_1 + t_2=pi/2. $$



                  the slope of the chord is $$ frac{3(sin t_1 - sin t_2)}{6(cos t_1 - cos t_2)}=-frac12frac{cos (t_1 + t_2)/2}{sin(t_1+t_2)/2}= -frac12$$ the equation of the required chord is $$frac{y-1}{x-2} = -frac12 to 2y = -x+4.$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    suppose you parametrize the ellipse by $x = 6cos t, y = 3 sin t, 0 le t le 2pi.$ suppose take chord is given by points $t_1, t_2$ and whose center is $A= (2,1).$



                    then we have
                    $$begin{align} cos t_1 + cos t_2 &= 2/3 tag 1\
                    sin t_1 + sin t_2 &= 2/3. tag 2end{align}$$



                    squaring the two equations and subtracting i them gives $$cos (t_1 + t_2) = 0 to t_1 + t_2 = pi/2, 3pi/2, ldots$$
                    subbing $t_2 = pi/2 - t_1$ in (1) and $(2)$ gives $sin t_1 + cos t_1 = 2/3$ but subbing $t_2 = 3pi/2 - t_1$ in $(1), (2)$ give $sin t_1 - cos t_1 = 2/3$ and $sin t_1 - cos t_1 = -2/3.$ therefore $$t_1 + t_2=pi/2. $$



                    the slope of the chord is $$ frac{3(sin t_1 - sin t_2)}{6(cos t_1 - cos t_2)}=-frac12frac{cos (t_1 + t_2)/2}{sin(t_1+t_2)/2}= -frac12$$ the equation of the required chord is $$frac{y-1}{x-2} = -frac12 to 2y = -x+4.$$






                    share|cite|improve this answer









                    $endgroup$



                    suppose you parametrize the ellipse by $x = 6cos t, y = 3 sin t, 0 le t le 2pi.$ suppose take chord is given by points $t_1, t_2$ and whose center is $A= (2,1).$



                    then we have
                    $$begin{align} cos t_1 + cos t_2 &= 2/3 tag 1\
                    sin t_1 + sin t_2 &= 2/3. tag 2end{align}$$



                    squaring the two equations and subtracting i them gives $$cos (t_1 + t_2) = 0 to t_1 + t_2 = pi/2, 3pi/2, ldots$$
                    subbing $t_2 = pi/2 - t_1$ in (1) and $(2)$ gives $sin t_1 + cos t_1 = 2/3$ but subbing $t_2 = 3pi/2 - t_1$ in $(1), (2)$ give $sin t_1 - cos t_1 = 2/3$ and $sin t_1 - cos t_1 = -2/3.$ therefore $$t_1 + t_2=pi/2. $$



                    the slope of the chord is $$ frac{3(sin t_1 - sin t_2)}{6(cos t_1 - cos t_2)}=-frac12frac{cos (t_1 + t_2)/2}{sin(t_1+t_2)/2}= -frac12$$ the equation of the required chord is $$frac{y-1}{x-2} = -frac12 to 2y = -x+4.$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 20 '15 at 18:09









                    abelabel

                    26.5k11948




                    26.5k11948























                        -1












                        $begingroup$

                        Equation of chord having midpoint $(h, k)$ is given by
                        $T=S^1$



                        $T=$ equation of tangent at $(h,k)$, $S =$ putting values of $(h,k)$ in ellipse's equation



                        $(h, k)=(2,1)$



                        Ellipse is $x^2+4y^2=36$



                        $T=x(2)+4y(1)-36$



                        $S^1= 2^2+4(1)^2-36$



                        Doing $T=S^1$



                        we get



                        $2x+4y=8$



                        $implies x+2y=4$






                        share|cite|improve this answer











                        $endgroup$


















                          -1












                          $begingroup$

                          Equation of chord having midpoint $(h, k)$ is given by
                          $T=S^1$



                          $T=$ equation of tangent at $(h,k)$, $S =$ putting values of $(h,k)$ in ellipse's equation



                          $(h, k)=(2,1)$



                          Ellipse is $x^2+4y^2=36$



                          $T=x(2)+4y(1)-36$



                          $S^1= 2^2+4(1)^2-36$



                          Doing $T=S^1$



                          we get



                          $2x+4y=8$



                          $implies x+2y=4$






                          share|cite|improve this answer











                          $endgroup$
















                            -1












                            -1








                            -1





                            $begingroup$

                            Equation of chord having midpoint $(h, k)$ is given by
                            $T=S^1$



                            $T=$ equation of tangent at $(h,k)$, $S =$ putting values of $(h,k)$ in ellipse's equation



                            $(h, k)=(2,1)$



                            Ellipse is $x^2+4y^2=36$



                            $T=x(2)+4y(1)-36$



                            $S^1= 2^2+4(1)^2-36$



                            Doing $T=S^1$



                            we get



                            $2x+4y=8$



                            $implies x+2y=4$






                            share|cite|improve this answer











                            $endgroup$



                            Equation of chord having midpoint $(h, k)$ is given by
                            $T=S^1$



                            $T=$ equation of tangent at $(h,k)$, $S =$ putting values of $(h,k)$ in ellipse's equation



                            $(h, k)=(2,1)$



                            Ellipse is $x^2+4y^2=36$



                            $T=x(2)+4y(1)-36$



                            $S^1= 2^2+4(1)^2-36$



                            Doing $T=S^1$



                            we get



                            $2x+4y=8$



                            $implies x+2y=4$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 6 '18 at 14:02









                            Brahadeesh

                            6,22242361




                            6,22242361










                            answered Dec 6 '18 at 12:48









                            sunny52525sunny52525

                            1




                            1






























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                                Province de Neuquén

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                                Province de Neuquén Provincia del Neuquén .mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Héraldique Drapeau Localisation de la province de Neuquén Administration Pays   Argentine Capitale Neuquén Gouverneur Omar Gutiérrez ISO 3166-2 AR-Q Démographie Gentilé Neuquino/a Population 550 344  hab. (2010) Densité 5,8  hab./km 2 Géographie Superficie 94 078  km 2 Liens Site web http://www.neuquen.gov.ar modifier   La province de Neuquén est une province d'Argentine, qui se trouve à l'extrémité nord-ouest de la Patagonie argentine. Avec une superficie de 94 378  km 2 , elle est la plus petite province des régions australes du pays. Sa géographie s'abîme sur l'imposante Cordillère des Andes. Sommaire 1 Histoire 1.1 Avant la découverte par les Espagnols 1.2 ...
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                                Journaliste

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                                Pour les articles homonymes, voir Journaliste (homonymie). Journaliste Journalistes Appellation Journaliste Secteur d'activité Média - Communication Professions voisines Journaliste sportif - Reporter-photographe - Éditorialiste Codes CITP 2451, 2642 IDEO (France) 10197 ROME (France) E1106 modifier - modifier le code - modifier Wikidata Un journaliste est une personne qui recueille, écrit ou distribue des informations. L'activité d'un journaliste est le journalisme. Un journaliste peut travailler sur des sujets généraux ou se spécialiser dans certaines questions. La plupart des journalistes ont tendance à se spécialiser et, en collaborant avec d'autres journalistes, rédigent un article ou mettent en forme un reportage qui couvre l'actualité. Par exemple, un journaliste sportif couvre les nouvelles dans le monde du sport, mais ce journaliste peut aussi faire partie d'un journal qui couvre d'autres su...
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