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

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        4 Answers
        4






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        active

        oldest

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        active

        oldest

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        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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                                Are they similar matrix

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                                0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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