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

11.1k41432




11.1k41432










asked Dec 29 '18 at 4:55









jackyjacky

1,235815




1,235815












  • $begingroup$
    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












$begingroup$

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












    $begingroup$

    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









    $endgroup$





















      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

        votes








        4 Answers
        4






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

        votes









        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

            3,9584829




            3,9584829























                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

                    21.3k2841




                    21.3k2841























                        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

                            33.6k12462




                            33.6k12462























                                -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

                                92815




                                92815








                                • 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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                                Grease: Live!

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                                A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
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