How to approach on this - finding minimum distance of point on the ellipse from the centre of it.












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Question



The minimum distance of any point on the ellipse $$x^2+3y^2+4xy=4$$ from its centre is ______.



Attempt



Converted the given expression into $$(x+2y)^2-y^2=4$$. But, this becomes equation of hyperbola of the form $$frac{x^2}{a^2}-frac{y^2}{b^2}=1$$.



Then how it is ellipse?



Any hints or suggestion?






analytic-geometry self-learning conic-sections







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    – Aloizio Macedo♦
    Dec 19 '18 at 2:02
















0












$begingroup$


Question



The minimum distance of any point on the ellipse $$x^2+3y^2+4xy=4$$ from its centre is ______.



Attempt



Converted the given expression into $$(x+2y)^2-y^2=4$$. But, this becomes equation of hyperbola of the form $$frac{x^2}{a^2}-frac{y^2}{b^2}=1$$.



Then how it is ellipse?



Any hints or suggestion?






analytic-geometry self-learning conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 19 '18 at 2:02














0












0








0





$begingroup$


Question



The minimum distance of any point on the ellipse $$x^2+3y^2+4xy=4$$ from its centre is ______.



Attempt



Converted the given expression into $$(x+2y)^2-y^2=4$$. But, this becomes equation of hyperbola of the form $$frac{x^2}{a^2}-frac{y^2}{b^2}=1$$.



Then how it is ellipse?



Any hints or suggestion?






analytic-geometry self-learning conic-sections







share|cite|improve this question











$endgroup$




Question



The minimum distance of any point on the ellipse $$x^2+3y^2+4xy=4$$ from its centre is ______.



Attempt



Converted the given expression into $$(x+2y)^2-y^2=4$$. But, this becomes equation of hyperbola of the form $$frac{x^2}{a^2}-frac{y^2}{b^2}=1$$.



Then how it is ellipse?



Any hints or suggestion?






analytic-geometry self-learning conic-sections




analytic-geometry self-learning conic-sections






share|cite|improve this question















share|cite|improve this question













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edited Dec 18 '18 at 9:07







jayant98

















asked Dec 17 '18 at 18:09









jayant98jayant98

653318




653318












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 19 '18 at 2:02


















  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Dec 19 '18 at 2:02
















$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 19 '18 at 2:02




$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Dec 19 '18 at 2:02










2 Answers
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$begingroup$

The quadratic form is
$$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$
The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic



Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $tan A = 1$ and $ tan B = 3.$ The trig formula for the bisecting angle is just
$$ tan left( frac{A+B}{2} right) = frac{sin A + sin B}{cos A + cos B} $$






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    1












    $begingroup$

    The matrix
    $$
    begin{pmatrix} 1 & 2 \ 2 & 3 end{pmatrix}
    $$
    defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2pmsqrt 5$. The eigenvector associated to the positive eigenvalue is $(sqrt5-1,2)$. Therefore the vertices are obtained by solving
    $$
    left{begin{array}{l}
    x^2+4xy+3y^2=4 \
    (x,y) = t(sqrt5-1,2) \
    end{array}right.
    $$
    The solutions are
    $$
    x_0 = sqrt{frac{14}{sqrt{5
    }}-6}
    qquad
    y_0 = sqrt{frac{6}{sqrt{
    5}}-2}
    $$
    and the opposite point $-(x_0,y_0)$.



    Therefore the distance from the center (which is the origin) is
    $$
    |(x_0,y_0)| = sqrt{x_0^2+y_0^2} = 2 sqrt{sqrt5-2} .
    $$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

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      2












      $begingroup$

      The quadratic form is
      $$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$
      The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic



      Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $tan A = 1$ and $ tan B = 3.$ The trig formula for the bisecting angle is just
      $$ tan left( frac{A+B}{2} right) = frac{sin A + sin B}{cos A + cos B} $$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        The quadratic form is
        $$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$
        The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic



        Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $tan A = 1$ and $ tan B = 3.$ The trig formula for the bisecting angle is just
        $$ tan left( frac{A+B}{2} right) = frac{sin A + sin B}{cos A + cos B} $$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          The quadratic form is
          $$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$
          The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic



          Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $tan A = 1$ and $ tan B = 3.$ The trig formula for the bisecting angle is just
          $$ tan left( frac{A+B}{2} right) = frac{sin A + sin B}{cos A + cos B} $$






          share|cite|improve this answer











          $endgroup$



          The quadratic form is
          $$ x^2 + 4xy + 3 y^2 = (x+y)(x+3y) $$
          The lines that approximate the hyperbola far from the origin are thus $x+y = 0$ and $x+3y = 0.$ These have two angle bisectors, which need a bit of work to find. One angle bisector gives the nearest point to the origin, where it intersects your conic $x^2 + 4xy+3y^2 = 4.$ The other angle bisector does not intersect the conic



          Let's see, the bisector that does intersect the conic is also the bisector of the lines $y=x$ and $y=3x$ that stays in the first quadrant, meaning positive slope. We have lines with slopes $tan A = 1$ and $ tan B = 3.$ The trig formula for the bisecting angle is just
          $$ tan left( frac{A+B}{2} right) = frac{sin A + sin B}{cos A + cos B} $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 19:48

























          answered Dec 17 '18 at 19:11









          Will JagyWill Jagy

          104k5102201




          104k5102201























              1












              $begingroup$

              The matrix
              $$
              begin{pmatrix} 1 & 2 \ 2 & 3 end{pmatrix}
              $$
              defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2pmsqrt 5$. The eigenvector associated to the positive eigenvalue is $(sqrt5-1,2)$. Therefore the vertices are obtained by solving
              $$
              left{begin{array}{l}
              x^2+4xy+3y^2=4 \
              (x,y) = t(sqrt5-1,2) \
              end{array}right.
              $$
              The solutions are
              $$
              x_0 = sqrt{frac{14}{sqrt{5
              }}-6}
              qquad
              y_0 = sqrt{frac{6}{sqrt{
              5}}-2}
              $$
              and the opposite point $-(x_0,y_0)$.



              Therefore the distance from the center (which is the origin) is
              $$
              |(x_0,y_0)| = sqrt{x_0^2+y_0^2} = 2 sqrt{sqrt5-2} .
              $$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The matrix
                $$
                begin{pmatrix} 1 & 2 \ 2 & 3 end{pmatrix}
                $$
                defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2pmsqrt 5$. The eigenvector associated to the positive eigenvalue is $(sqrt5-1,2)$. Therefore the vertices are obtained by solving
                $$
                left{begin{array}{l}
                x^2+4xy+3y^2=4 \
                (x,y) = t(sqrt5-1,2) \
                end{array}right.
                $$
                The solutions are
                $$
                x_0 = sqrt{frac{14}{sqrt{5
                }}-6}
                qquad
                y_0 = sqrt{frac{6}{sqrt{
                5}}-2}
                $$
                and the opposite point $-(x_0,y_0)$.



                Therefore the distance from the center (which is the origin) is
                $$
                |(x_0,y_0)| = sqrt{x_0^2+y_0^2} = 2 sqrt{sqrt5-2} .
                $$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The matrix
                  $$
                  begin{pmatrix} 1 & 2 \ 2 & 3 end{pmatrix}
                  $$
                  defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2pmsqrt 5$. The eigenvector associated to the positive eigenvalue is $(sqrt5-1,2)$. Therefore the vertices are obtained by solving
                  $$
                  left{begin{array}{l}
                  x^2+4xy+3y^2=4 \
                  (x,y) = t(sqrt5-1,2) \
                  end{array}right.
                  $$
                  The solutions are
                  $$
                  x_0 = sqrt{frac{14}{sqrt{5
                  }}-6}
                  qquad
                  y_0 = sqrt{frac{6}{sqrt{
                  5}}-2}
                  $$
                  and the opposite point $-(x_0,y_0)$.



                  Therefore the distance from the center (which is the origin) is
                  $$
                  |(x_0,y_0)| = sqrt{x_0^2+y_0^2} = 2 sqrt{sqrt5-2} .
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  The matrix
                  $$
                  begin{pmatrix} 1 & 2 \ 2 & 3 end{pmatrix}
                  $$
                  defining the hyperbola $(x,y)B(x,y)^T=4$ has eigenvalues $2pmsqrt 5$. The eigenvector associated to the positive eigenvalue is $(sqrt5-1,2)$. Therefore the vertices are obtained by solving
                  $$
                  left{begin{array}{l}
                  x^2+4xy+3y^2=4 \
                  (x,y) = t(sqrt5-1,2) \
                  end{array}right.
                  $$
                  The solutions are
                  $$
                  x_0 = sqrt{frac{14}{sqrt{5
                  }}-6}
                  qquad
                  y_0 = sqrt{frac{6}{sqrt{
                  5}}-2}
                  $$
                  and the opposite point $-(x_0,y_0)$.



                  Therefore the distance from the center (which is the origin) is
                  $$
                  |(x_0,y_0)| = sqrt{x_0^2+y_0^2} = 2 sqrt{sqrt5-2} .
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 19:25









                  FedericoFederico

                  5,144514




                  5,144514






























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