Foot of perpendicular on a chord of a conic












1












$begingroup$


For a standard ellipse, a chord subtends an angle of $90^{circ}$ with the centre $(0,0)$ . To find the locus of the foot of perpendicular to this chord from the centre of the ellipse, I wrote the equation of chord with parameters $ A , B$:



$$frac xaspacecos{A+Bover 2} + frac ybspacesin{A+Bover 2} = cos{A-Bover 2}$$
Since an angle of $90^{circ}$ is subtended at the centre, I get the condition that $tan Atan B = -dfrac {a^2}{ b^2}$



By imposing the condition that FoP is perpendicular to the chord, I get $dfrac {ax}btan{A+Bover 2} = y$.



Eliminiating A and B is a hassle here. Is there a more general way to solving such kinds of problems? This is an ellipse here, it can be a hyperbola, or parabola next.






analytic-geometry conic-sections







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  • $begingroup$
    please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
    $endgroup$
    – Sat D
    Nov 14 '15 at 7:06
















1












$begingroup$


For a standard ellipse, a chord subtends an angle of $90^{circ}$ with the centre $(0,0)$ . To find the locus of the foot of perpendicular to this chord from the centre of the ellipse, I wrote the equation of chord with parameters $ A , B$:



$$frac xaspacecos{A+Bover 2} + frac ybspacesin{A+Bover 2} = cos{A-Bover 2}$$
Since an angle of $90^{circ}$ is subtended at the centre, I get the condition that $tan Atan B = -dfrac {a^2}{ b^2}$



By imposing the condition that FoP is perpendicular to the chord, I get $dfrac {ax}btan{A+Bover 2} = y$.



Eliminiating A and B is a hassle here. Is there a more general way to solving such kinds of problems? This is an ellipse here, it can be a hyperbola, or parabola next.






analytic-geometry conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
    $endgroup$
    – Sat D
    Nov 14 '15 at 7:06














1












1








1





$begingroup$


For a standard ellipse, a chord subtends an angle of $90^{circ}$ with the centre $(0,0)$ . To find the locus of the foot of perpendicular to this chord from the centre of the ellipse, I wrote the equation of chord with parameters $ A , B$:



$$frac xaspacecos{A+Bover 2} + frac ybspacesin{A+Bover 2} = cos{A-Bover 2}$$
Since an angle of $90^{circ}$ is subtended at the centre, I get the condition that $tan Atan B = -dfrac {a^2}{ b^2}$



By imposing the condition that FoP is perpendicular to the chord, I get $dfrac {ax}btan{A+Bover 2} = y$.



Eliminiating A and B is a hassle here. Is there a more general way to solving such kinds of problems? This is an ellipse here, it can be a hyperbola, or parabola next.






analytic-geometry conic-sections







share|cite|improve this question











$endgroup$




For a standard ellipse, a chord subtends an angle of $90^{circ}$ with the centre $(0,0)$ . To find the locus of the foot of perpendicular to this chord from the centre of the ellipse, I wrote the equation of chord with parameters $ A , B$:



$$frac xaspacecos{A+Bover 2} + frac ybspacesin{A+Bover 2} = cos{A-Bover 2}$$
Since an angle of $90^{circ}$ is subtended at the centre, I get the condition that $tan Atan B = -dfrac {a^2}{ b^2}$



By imposing the condition that FoP is perpendicular to the chord, I get $dfrac {ax}btan{A+Bover 2} = y$.



Eliminiating A and B is a hassle here. Is there a more general way to solving such kinds of problems? This is an ellipse here, it can be a hyperbola, or parabola next.






analytic-geometry conic-sections




analytic-geometry conic-sections






share|cite|improve this question















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edited Nov 14 '15 at 7:17









G-man

4,49331344




4,49331344










asked Nov 14 '15 at 7:03









Sat DSat D

423212




423212












  • $begingroup$
    please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
    $endgroup$
    – Sat D
    Nov 14 '15 at 7:06


















  • $begingroup$
    please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
    $endgroup$
    – Sat D
    Nov 14 '15 at 7:06
















$begingroup$
please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
$endgroup$
– Sat D
Nov 14 '15 at 7:06




$begingroup$
please excuse the fact that I havent used Latex. I'm not well-versed with it as of yet
$endgroup$
– Sat D
Nov 14 '15 at 7:06










2 Answers
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$begingroup$

Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.






share|cite|improve this answer











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    0












    $begingroup$

    Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at
    $$
    A=left({aboversqrt{m^2a^2+b^2}},{maboversqrt{m^2a^2+b^2}}right).
    $$
    A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at
    $$
    B=left({maboversqrt{a^2+m^2b^2}},{-aboversqrt{a^2+m^2b^2}}right).
    $$
    You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.






    share|cite|improve this answer









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      2 Answers
      2






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      2 Answers
      2






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      active

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      active

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      1












      $begingroup$

      Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.






          share|cite|improve this answer











          $endgroup$



          Okay to solve questions like these, let the chord be $y = mx + c$. Homogenize this chord with the ellipse to get a POSL. Since angle between lines is π/2, apply condition $coeff(x^2) + coeff(y^2) = 0$, to get $c = φ(m)$. Then put the condition that slope of FoP will be $-1/m$ and solve it with $y = mx + c$, and eliminate m to get locus.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 26 '16 at 5:19

























          answered Nov 17 '15 at 7:50









          Sat DSat D

          423212




          423212























              0












              $begingroup$

              Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at
              $$
              A=left({aboversqrt{m^2a^2+b^2}},{maboversqrt{m^2a^2+b^2}}right).
              $$
              A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at
              $$
              B=left({maboversqrt{a^2+m^2b^2}},{-aboversqrt{a^2+m^2b^2}}right).
              $$
              You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at
                $$
                A=left({aboversqrt{m^2a^2+b^2}},{maboversqrt{m^2a^2+b^2}}right).
                $$
                A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at
                $$
                B=left({maboversqrt{a^2+m^2b^2}},{-aboversqrt{a^2+m^2b^2}}right).
                $$
                You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at
                  $$
                  A=left({aboversqrt{m^2a^2+b^2}},{maboversqrt{m^2a^2+b^2}}right).
                  $$
                  A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at
                  $$
                  B=left({maboversqrt{a^2+m^2b^2}},{-aboversqrt{a^2+m^2b^2}}right).
                  $$
                  You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.






                  share|cite|improve this answer









                  $endgroup$



                  Let $r$ be a line of equation $y=mx$, passing through the origin and intersecting the ellipse at
                  $$
                  A=left({aboversqrt{m^2a^2+b^2}},{maboversqrt{m^2a^2+b^2}}right).
                  $$
                  A line perpendicular to $r$ has equation $y=(-1/m)x$ and intersects the ellipse at
                  $$
                  B=left({maboversqrt{a^2+m^2b^2}},{-aboversqrt{a^2+m^2b^2}}right).
                  $$
                  You can now compute the distance of the origin from line $AB$, to find that it doesn't depend on $m$: the desired locus is then a circle, centred at the origin and having that distance as radius.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 18:59









                  AretinoAretino

                  23.1k21443




                  23.1k21443






























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