Radius of largest circle in an ellipse












1












$begingroup$


Consider an ellipse whose major and minor axis of length $10$ and $8$ unit respectively. Then the radius of the largest circle which can be inscribed in such an ellipse if the circle's center is one focus of the ellipse.



What I tried:



Assuming that major axis and minor axis of an ellipse along coordinate axis.($x$ and $y$ axis respectively)



Then equation of ellipse is $displaystyle frac{x^2}{a^2}+frac{y^2}{b^2}=1.$ where $a=5,b=4$



Then $displaystyle b^2=a^2(1-e^2)$ . getting $displaystyle e=3/5.$



coordinate of focus is $(pm ae,0)=(pm 3,0)$



Let equation of circle is $(xpm 3)^2+y^2=r^2$



How do I solve it from here?






conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    The nearer vertex is the closest point of the ellipse to the focus, so...
    $endgroup$
    – amd
    Dec 29 '18 at 8:57
















1












$begingroup$


Consider an ellipse whose major and minor axis of length $10$ and $8$ unit respectively. Then the radius of the largest circle which can be inscribed in such an ellipse if the circle's center is one focus of the ellipse.



What I tried:



Assuming that major axis and minor axis of an ellipse along coordinate axis.($x$ and $y$ axis respectively)



Then equation of ellipse is $displaystyle frac{x^2}{a^2}+frac{y^2}{b^2}=1.$ where $a=5,b=4$



Then $displaystyle b^2=a^2(1-e^2)$ . getting $displaystyle e=3/5.$



coordinate of focus is $(pm ae,0)=(pm 3,0)$



Let equation of circle is $(xpm 3)^2+y^2=r^2$



How do I solve it from here?






conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    The nearer vertex is the closest point of the ellipse to the focus, so...
    $endgroup$
    – amd
    Dec 29 '18 at 8:57














1












1








1


2



$begingroup$


Consider an ellipse whose major and minor axis of length $10$ and $8$ unit respectively. Then the radius of the largest circle which can be inscribed in such an ellipse if the circle's center is one focus of the ellipse.



What I tried:



Assuming that major axis and minor axis of an ellipse along coordinate axis.($x$ and $y$ axis respectively)



Then equation of ellipse is $displaystyle frac{x^2}{a^2}+frac{y^2}{b^2}=1.$ where $a=5,b=4$



Then $displaystyle b^2=a^2(1-e^2)$ . getting $displaystyle e=3/5.$



coordinate of focus is $(pm ae,0)=(pm 3,0)$



Let equation of circle is $(xpm 3)^2+y^2=r^2$



How do I solve it from here?






conic-sections







share|cite|improve this question











$endgroup$




Consider an ellipse whose major and minor axis of length $10$ and $8$ unit respectively. Then the radius of the largest circle which can be inscribed in such an ellipse if the circle's center is one focus of the ellipse.



What I tried:



Assuming that major axis and minor axis of an ellipse along coordinate axis.($x$ and $y$ axis respectively)



Then equation of ellipse is $displaystyle frac{x^2}{a^2}+frac{y^2}{b^2}=1.$ where $a=5,b=4$



Then $displaystyle b^2=a^2(1-e^2)$ . getting $displaystyle e=3/5.$



coordinate of focus is $(pm ae,0)=(pm 3,0)$



Let equation of circle is $(xpm 3)^2+y^2=r^2$



How do I solve it from 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 29 '18 at 5:11









David G. Stork

11k41432




11k41432










asked Dec 29 '18 at 4:55









jackyjacky

890613




890613












  • $begingroup$
    The nearer vertex is the closest point of the ellipse to the focus, so...
    $endgroup$
    – amd
    Dec 29 '18 at 8:57


















  • $begingroup$
    The nearer vertex is the closest point of the ellipse to the focus, so...
    $endgroup$
    – amd
    Dec 29 '18 at 8:57
















$begingroup$
The nearer vertex is the closest point of the ellipse to the focus, so...
$endgroup$
– amd
Dec 29 '18 at 8:57




$begingroup$
The nearer vertex is the closest point of the ellipse to the focus, so...
$endgroup$
– amd
Dec 29 '18 at 8:57










4 Answers
4






active

oldest

votes


















2












$begingroup$

You want to minimize $r^2=(x-3)^2+y^2$ subject to $frac{x^2}{25}+frac{y^2}{16}=1$:
$$L(x,y,lambda)=(x-3)^2+y^2+lambdaleft(1-frac{x^2}{25}-frac{y^2}{16}right)\
begin{cases}L_x=2x-6-frac{2xlambda}{25}=0\ L_y=2y-frac{ylambda}{8}=0\ L_{lambda}=1-frac{x^2}{25}-frac{y^2}{16}=0end{cases} Rightarrow (x,y)=(pm5,0);\
r^2(5,0)=2^2 text{(min)}$$
Notes:



1) WolframAlpha answer.



2) If you minimize $r^2=(x+3)^2+y^2$ s.t. $frac{x^2}{25}+frac{y^2}{16}=1$, you will get $r^2(-5,0)=2^2$.



3) Foci $(pm c,0)$ can be found from $c^2=a^2-b^2=5^2-4^2=3^2 Rightarrow c=pm 3$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Denote $F, F'$ the foci of the ellipse. Assume we want to find the greatest circle inscribed in the ellipse, with center $F.$




    • The circle and the ellipse have a common tangent line and also a common normal in their point (points) of tangency $P,$ eventually also $P'$ if they are two.


    • The normal of an ellipse bisects the angle between the lines to the foci. In our case, the normal at $P$ bisects the angle between $PF$ and $PF'.$


    • The normal of a circle passes through its center.



    From the above properties follows that $P,F,F'$ are collinear. Therefore, $P$ is the vertex of the ellipse lying at the major axis nearest to $F,$ and the equation of the circle is
    $$(x-3)^2+y^2=|PF|^2=4$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Eliminate $y$. Then find the value of $x$ where $r$ is minimised.






      share|cite|improve this answer









      $endgroup$





















        -1












        $begingroup$

        Try to obtain a $r$ such that the circle intersects with the ellipse at exactly 2 points. This is when the circle is tangent to the ellipse.






        share|cite|improve this answer









        $endgroup$









        • 2




          $begingroup$
          There is an infinite number of circles centered at a focus that intersect the ellipse at exactly two points. That condition is insufficient for tangency. In fact, there are exactly 2 tangent circles and their intersections with the ellipse each consist of a single point.
          $endgroup$
          – amd
          Dec 29 '18 at 9:00












        • $begingroup$
          @Trebor, if the center of the circle is in a focus of the ellipse, there are never two points of contact. This is one of properties of ellipse. If there are two common points, the circle cuts the ellipse.
          $endgroup$
          – user376343
          Dec 30 '18 at 18:53











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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

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        active

        oldest

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        active

        oldest

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        2












        $begingroup$

        You want to minimize $r^2=(x-3)^2+y^2$ subject to $frac{x^2}{25}+frac{y^2}{16}=1$:
        $$L(x,y,lambda)=(x-3)^2+y^2+lambdaleft(1-frac{x^2}{25}-frac{y^2}{16}right)\
        begin{cases}L_x=2x-6-frac{2xlambda}{25}=0\ L_y=2y-frac{ylambda}{8}=0\ L_{lambda}=1-frac{x^2}{25}-frac{y^2}{16}=0end{cases} Rightarrow (x,y)=(pm5,0);\
        r^2(5,0)=2^2 text{(min)}$$
        Notes:



        1) WolframAlpha answer.



        2) If you minimize $r^2=(x+3)^2+y^2$ s.t. $frac{x^2}{25}+frac{y^2}{16}=1$, you will get $r^2(-5,0)=2^2$.



        3) Foci $(pm c,0)$ can be found from $c^2=a^2-b^2=5^2-4^2=3^2 Rightarrow c=pm 3$.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          You want to minimize $r^2=(x-3)^2+y^2$ subject to $frac{x^2}{25}+frac{y^2}{16}=1$:
          $$L(x,y,lambda)=(x-3)^2+y^2+lambdaleft(1-frac{x^2}{25}-frac{y^2}{16}right)\
          begin{cases}L_x=2x-6-frac{2xlambda}{25}=0\ L_y=2y-frac{ylambda}{8}=0\ L_{lambda}=1-frac{x^2}{25}-frac{y^2}{16}=0end{cases} Rightarrow (x,y)=(pm5,0);\
          r^2(5,0)=2^2 text{(min)}$$
          Notes:



          1) WolframAlpha answer.



          2) If you minimize $r^2=(x+3)^2+y^2$ s.t. $frac{x^2}{25}+frac{y^2}{16}=1$, you will get $r^2(-5,0)=2^2$.



          3) Foci $(pm c,0)$ can be found from $c^2=a^2-b^2=5^2-4^2=3^2 Rightarrow c=pm 3$.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            You want to minimize $r^2=(x-3)^2+y^2$ subject to $frac{x^2}{25}+frac{y^2}{16}=1$:
            $$L(x,y,lambda)=(x-3)^2+y^2+lambdaleft(1-frac{x^2}{25}-frac{y^2}{16}right)\
            begin{cases}L_x=2x-6-frac{2xlambda}{25}=0\ L_y=2y-frac{ylambda}{8}=0\ L_{lambda}=1-frac{x^2}{25}-frac{y^2}{16}=0end{cases} Rightarrow (x,y)=(pm5,0);\
            r^2(5,0)=2^2 text{(min)}$$
            Notes:



            1) WolframAlpha answer.



            2) If you minimize $r^2=(x+3)^2+y^2$ s.t. $frac{x^2}{25}+frac{y^2}{16}=1$, you will get $r^2(-5,0)=2^2$.



            3) Foci $(pm c,0)$ can be found from $c^2=a^2-b^2=5^2-4^2=3^2 Rightarrow c=pm 3$.






            share|cite|improve this answer









            $endgroup$



            You want to minimize $r^2=(x-3)^2+y^2$ subject to $frac{x^2}{25}+frac{y^2}{16}=1$:
            $$L(x,y,lambda)=(x-3)^2+y^2+lambdaleft(1-frac{x^2}{25}-frac{y^2}{16}right)\
            begin{cases}L_x=2x-6-frac{2xlambda}{25}=0\ L_y=2y-frac{ylambda}{8}=0\ L_{lambda}=1-frac{x^2}{25}-frac{y^2}{16}=0end{cases} Rightarrow (x,y)=(pm5,0);\
            r^2(5,0)=2^2 text{(min)}$$
            Notes:



            1) WolframAlpha answer.



            2) If you minimize $r^2=(x+3)^2+y^2$ s.t. $frac{x^2}{25}+frac{y^2}{16}=1$, you will get $r^2(-5,0)=2^2$.



            3) Foci $(pm c,0)$ can be found from $c^2=a^2-b^2=5^2-4^2=3^2 Rightarrow c=pm 3$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 29 '18 at 7:02









            farruhotafarruhota

            20.4k2739




            20.4k2739























                2












                $begingroup$

                Denote $F, F'$ the foci of the ellipse. Assume we want to find the greatest circle inscribed in the ellipse, with center $F.$




                • The circle and the ellipse have a common tangent line and also a common normal in their point (points) of tangency $P,$ eventually also $P'$ if they are two.


                • The normal of an ellipse bisects the angle between the lines to the foci. In our case, the normal at $P$ bisects the angle between $PF$ and $PF'.$


                • The normal of a circle passes through its center.



                From the above properties follows that $P,F,F'$ are collinear. Therefore, $P$ is the vertex of the ellipse lying at the major axis nearest to $F,$ and the equation of the circle is
                $$(x-3)^2+y^2=|PF|^2=4$$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Denote $F, F'$ the foci of the ellipse. Assume we want to find the greatest circle inscribed in the ellipse, with center $F.$




                  • The circle and the ellipse have a common tangent line and also a common normal in their point (points) of tangency $P,$ eventually also $P'$ if they are two.


                  • The normal of an ellipse bisects the angle between the lines to the foci. In our case, the normal at $P$ bisects the angle between $PF$ and $PF'.$


                  • The normal of a circle passes through its center.



                  From the above properties follows that $P,F,F'$ are collinear. Therefore, $P$ is the vertex of the ellipse lying at the major axis nearest to $F,$ and the equation of the circle is
                  $$(x-3)^2+y^2=|PF|^2=4$$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Denote $F, F'$ the foci of the ellipse. Assume we want to find the greatest circle inscribed in the ellipse, with center $F.$




                    • The circle and the ellipse have a common tangent line and also a common normal in their point (points) of tangency $P,$ eventually also $P'$ if they are two.


                    • The normal of an ellipse bisects the angle between the lines to the foci. In our case, the normal at $P$ bisects the angle between $PF$ and $PF'.$


                    • The normal of a circle passes through its center.



                    From the above properties follows that $P,F,F'$ are collinear. Therefore, $P$ is the vertex of the ellipse lying at the major axis nearest to $F,$ and the equation of the circle is
                    $$(x-3)^2+y^2=|PF|^2=4$$






                    share|cite|improve this answer









                    $endgroup$



                    Denote $F, F'$ the foci of the ellipse. Assume we want to find the greatest circle inscribed in the ellipse, with center $F.$




                    • The circle and the ellipse have a common tangent line and also a common normal in their point (points) of tangency $P,$ eventually also $P'$ if they are two.


                    • The normal of an ellipse bisects the angle between the lines to the foci. In our case, the normal at $P$ bisects the angle between $PF$ and $PF'.$


                    • The normal of a circle passes through its center.



                    From the above properties follows that $P,F,F'$ are collinear. Therefore, $P$ is the vertex of the ellipse lying at the major axis nearest to $F,$ and the equation of the circle is
                    $$(x-3)^2+y^2=|PF|^2=4$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 29 '18 at 22:29









                    user376343user376343

                    3,8133828




                    3,8133828























                        0












                        $begingroup$

                        Eliminate $y$. Then find the value of $x$ where $r$ is minimised.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Eliminate $y$. Then find the value of $x$ where $r$ is minimised.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Eliminate $y$. Then find the value of $x$ where $r$ is minimised.






                            share|cite|improve this answer









                            $endgroup$



                            Eliminate $y$. Then find the value of $x$ where $r$ is minimised.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 29 '18 at 5:47









                            Empy2Empy2

                            33.5k12362




                            33.5k12362























                                -1












                                $begingroup$

                                Try to obtain a $r$ such that the circle intersects with the ellipse at exactly 2 points. This is when the circle is tangent to the ellipse.






                                share|cite|improve this answer









                                $endgroup$









                                • 2




                                  $begingroup$
                                  There is an infinite number of circles centered at a focus that intersect the ellipse at exactly two points. That condition is insufficient for tangency. In fact, there are exactly 2 tangent circles and their intersections with the ellipse each consist of a single point.
                                  $endgroup$
                                  – amd
                                  Dec 29 '18 at 9:00












                                • $begingroup$
                                  @Trebor, if the center of the circle is in a focus of the ellipse, there are never two points of contact. This is one of properties of ellipse. If there are two common points, the circle cuts the ellipse.
                                  $endgroup$
                                  – user376343
                                  Dec 30 '18 at 18:53
















                                -1












                                $begingroup$

                                Try to obtain a $r$ such that the circle intersects with the ellipse at exactly 2 points. This is when the circle is tangent to the ellipse.






                                share|cite|improve this answer









                                $endgroup$









                                • 2




                                  $begingroup$
                                  There is an infinite number of circles centered at a focus that intersect the ellipse at exactly two points. That condition is insufficient for tangency. In fact, there are exactly 2 tangent circles and their intersections with the ellipse each consist of a single point.
                                  $endgroup$
                                  – amd
                                  Dec 29 '18 at 9:00












                                • $begingroup$
                                  @Trebor, if the center of the circle is in a focus of the ellipse, there are never two points of contact. This is one of properties of ellipse. If there are two common points, the circle cuts the ellipse.
                                  $endgroup$
                                  – user376343
                                  Dec 30 '18 at 18:53














                                -1












                                -1








                                -1





                                $begingroup$

                                Try to obtain a $r$ such that the circle intersects with the ellipse at exactly 2 points. This is when the circle is tangent to the ellipse.






                                share|cite|improve this answer









                                $endgroup$



                                Try to obtain a $r$ such that the circle intersects with the ellipse at exactly 2 points. This is when the circle is tangent to the ellipse.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Dec 29 '18 at 5:16









                                TreborTrebor

                                81013




                                81013








                                • 2




                                  $begingroup$
                                  There is an infinite number of circles centered at a focus that intersect the ellipse at exactly two points. That condition is insufficient for tangency. In fact, there are exactly 2 tangent circles and their intersections with the ellipse each consist of a single point.
                                  $endgroup$
                                  – amd
                                  Dec 29 '18 at 9:00












                                • $begingroup$
                                  @Trebor, if the center of the circle is in a focus of the ellipse, there are never two points of contact. This is one of properties of ellipse. If there are two common points, the circle cuts the ellipse.
                                  $endgroup$
                                  – user376343
                                  Dec 30 '18 at 18:53














                                • 2




                                  $begingroup$
                                  There is an infinite number of circles centered at a focus that intersect the ellipse at exactly two points. That condition is insufficient for tangency. In fact, there are exactly 2 tangent circles and their intersections with the ellipse each consist of a single point.
                                  $endgroup$
                                  – amd
                                  Dec 29 '18 at 9:00












                                • $begingroup$
                                  @Trebor, if the center of the circle is in a focus of the ellipse, there are never two points of contact. This is one of properties of ellipse. If there are two common points, the circle cuts the ellipse.
                                  $endgroup$
                                  – user376343
                                  Dec 30 '18 at 18:53








                                2




                                2




                                $begingroup$
                                There is an infinite number of circles centered at a focus that intersect the ellipse at exactly two points. That condition is insufficient for tangency. In fact, there are exactly 2 tangent circles and their intersections with the ellipse each consist of a single point.
                                $endgroup$
                                – amd
                                Dec 29 '18 at 9:00






                                $begingroup$
                                There is an infinite number of circles centered at a focus that intersect the ellipse at exactly two points. That condition is insufficient for tangency. In fact, there are exactly 2 tangent circles and their intersections with the ellipse each consist of a single point.
                                $endgroup$
                                – amd
                                Dec 29 '18 at 9:00














                                $begingroup$
                                @Trebor, if the center of the circle is in a focus of the ellipse, there are never two points of contact. This is one of properties of ellipse. If there are two common points, the circle cuts the ellipse.
                                $endgroup$
                                – user376343
                                Dec 30 '18 at 18:53




                                $begingroup$
                                @Trebor, if the center of the circle is in a focus of the ellipse, there are never two points of contact. This is one of properties of ellipse. If there are two common points, the circle cuts the ellipse.
                                $endgroup$
                                – user376343
                                Dec 30 '18 at 18:53


















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                                Cabo Verde

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                                  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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                                Gyllenstierna

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