Find the locus of perpendicular drawn from focus upon variable tangent to the parabola...












0












$begingroup$


Find the locus of perpendicular drawn from focus upon variable tangent to the parabola $(2x-y+1)^2=frac{8}{sqrt{5}}(x+2y+3)$.



My approach
I am trying to convert above equation in parabolic form



$frac{(ax+by+c)^2}{a^2+b^2}=(x-alpha)^2+(y-beta)^2$



where $ ax+by+c=0$ is the equation of directrix and ($alpha,beta$) is the focus of the parabola but getting complicated.






conic-sections







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  • $begingroup$
    $$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
    $endgroup$
    – Jan-Magnus Økland
    Dec 12 '18 at 6:11
















0












$begingroup$


Find the locus of perpendicular drawn from focus upon variable tangent to the parabola $(2x-y+1)^2=frac{8}{sqrt{5}}(x+2y+3)$.



My approach
I am trying to convert above equation in parabolic form



$frac{(ax+by+c)^2}{a^2+b^2}=(x-alpha)^2+(y-beta)^2$



where $ ax+by+c=0$ is the equation of directrix and ($alpha,beta$) is the focus of the parabola but getting complicated.






conic-sections







share|cite|improve this question









$endgroup$












  • $begingroup$
    $$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
    $endgroup$
    – Jan-Magnus Økland
    Dec 12 '18 at 6:11














0












0








0





$begingroup$


Find the locus of perpendicular drawn from focus upon variable tangent to the parabola $(2x-y+1)^2=frac{8}{sqrt{5}}(x+2y+3)$.



My approach
I am trying to convert above equation in parabolic form



$frac{(ax+by+c)^2}{a^2+b^2}=(x-alpha)^2+(y-beta)^2$



where $ ax+by+c=0$ is the equation of directrix and ($alpha,beta$) is the focus of the parabola but getting complicated.






conic-sections







share|cite|improve this question









$endgroup$




Find the locus of perpendicular drawn from focus upon variable tangent to the parabola $(2x-y+1)^2=frac{8}{sqrt{5}}(x+2y+3)$.



My approach
I am trying to convert above equation in parabolic form



$frac{(ax+by+c)^2}{a^2+b^2}=(x-alpha)^2+(y-beta)^2$



where $ ax+by+c=0$ is the equation of directrix and ($alpha,beta$) is the focus of the parabola but getting complicated.






conic-sections




conic-sections






share|cite|improve this question













share|cite|improve this question











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asked Dec 12 '18 at 4:51









Samar Imam ZaidiSamar Imam Zaidi

1,5501520




1,5501520












  • $begingroup$
    $$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
    $endgroup$
    – Jan-Magnus Økland
    Dec 12 '18 at 6:11


















  • $begingroup$
    $$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
    $endgroup$
    – Jan-Magnus Økland
    Dec 12 '18 at 6:11
















$begingroup$
$$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
$endgroup$
– Jan-Magnus Økland
Dec 12 '18 at 6:11




$begingroup$
$$5,left(-left({{2,y}over{sqrt{5}}}+{{x}over{sqrt{5}}}+{{3, sqrt{5}+2}over{5}}right)^2+left(y-{{4,sqrt{5}-25}over{25}} right)^2+left(x-{{2,sqrt{5}-25}over{25}}right)^2right)$$
$endgroup$
– Jan-Magnus Økland
Dec 12 '18 at 6:11










3 Answers
3






active

oldest

votes


















2












$begingroup$

Hint :



Let $Y=2x-y+1 (1)$ and $4aX=dfrac{8(x+2y+3)}{sqrt5}$ with $x+2y+3=X
(2), a=?$



whose focus is $(a,0)$



Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$



Find the equation of the tangent at $P$



Find the equation of the normal for the tangent passing through $(a,0)$



Find the intersection of the normal with the tangent.



Eliminate $t$



Replace the values of $X,Y$ with $x,y$ using $(1),(2)$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
      $endgroup$
      – amd
      Dec 12 '18 at 7:54












    • $begingroup$
      I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
      $endgroup$
      – Sri Krishna Sahoo
      Dec 12 '18 at 7:58





















    1












    $begingroup$

    The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.



    That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Hint :



      Let $Y=2x-y+1 (1)$ and $4aX=dfrac{8(x+2y+3)}{sqrt5}$ with $x+2y+3=X
      (2), a=?$



      whose focus is $(a,0)$



      Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$



      Find the equation of the tangent at $P$



      Find the equation of the normal for the tangent passing through $(a,0)$



      Find the intersection of the normal with the tangent.



      Eliminate $t$



      Replace the values of $X,Y$ with $x,y$ using $(1),(2)$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Hint :



        Let $Y=2x-y+1 (1)$ and $4aX=dfrac{8(x+2y+3)}{sqrt5}$ with $x+2y+3=X
        (2), a=?$



        whose focus is $(a,0)$



        Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$



        Find the equation of the tangent at $P$



        Find the equation of the normal for the tangent passing through $(a,0)$



        Find the intersection of the normal with the tangent.



        Eliminate $t$



        Replace the values of $X,Y$ with $x,y$ using $(1),(2)$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint :



          Let $Y=2x-y+1 (1)$ and $4aX=dfrac{8(x+2y+3)}{sqrt5}$ with $x+2y+3=X
          (2), a=?$



          whose focus is $(a,0)$



          Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$



          Find the equation of the tangent at $P$



          Find the equation of the normal for the tangent passing through $(a,0)$



          Find the intersection of the normal with the tangent.



          Eliminate $t$



          Replace the values of $X,Y$ with $x,y$ using $(1),(2)$






          share|cite|improve this answer









          $endgroup$



          Hint :



          Let $Y=2x-y+1 (1)$ and $4aX=dfrac{8(x+2y+3)}{sqrt5}$ with $x+2y+3=X
          (2), a=?$



          whose focus is $(a,0)$



          Any point on $Y^2=4aX,$ can be set to $P(at^2,2at)$



          Find the equation of the tangent at $P$



          Find the equation of the normal for the tangent passing through $(a,0)$



          Find the intersection of the normal with the tangent.



          Eliminate $t$



          Replace the values of $X,Y$ with $x,y$ using $(1),(2)$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 5:21









          lab bhattacharjeelab bhattacharjee

          226k15157275




          226k15157275























              1












              $begingroup$

              There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
                $endgroup$
                – amd
                Dec 12 '18 at 7:54












              • $begingroup$
                I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
                $endgroup$
                – Sri Krishna Sahoo
                Dec 12 '18 at 7:58


















              1












              $begingroup$

              There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
                $endgroup$
                – amd
                Dec 12 '18 at 7:54












              • $begingroup$
                I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
                $endgroup$
                – Sri Krishna Sahoo
                Dec 12 '18 at 7:58
















              1












              1








              1





              $begingroup$

              There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.






              share|cite|improve this answer









              $endgroup$



              There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line parallel to the directrix (from the equation) and at equal distance from the focus and directrix. That is the tangent at vertex.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 12 '18 at 7:31









              Sri Krishna SahooSri Krishna Sahoo

              606217




              606217












              • $begingroup$
                This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
                $endgroup$
                – amd
                Dec 12 '18 at 7:54












              • $begingroup$
                I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
                $endgroup$
                – Sri Krishna Sahoo
                Dec 12 '18 at 7:58




















              • $begingroup$
                This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
                $endgroup$
                – amd
                Dec 12 '18 at 7:54












              • $begingroup$
                I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
                $endgroup$
                – Sri Krishna Sahoo
                Dec 12 '18 at 7:58


















              $begingroup$
              This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
              $endgroup$
              – amd
              Dec 12 '18 at 7:54






              $begingroup$
              This begs the question, doesn’t it? It seems to me that this is an exercise in discovering or verifying the property for this particular parabola. Otherwise, the problem can be solved by inspection since the equations of both the axis and tangent at the vertex are explicit in the parabola’s equation.
              $endgroup$
              – amd
              Dec 12 '18 at 7:54














              $begingroup$
              I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
              $endgroup$
              – Sri Krishna Sahoo
              Dec 12 '18 at 7:58






              $begingroup$
              I think that the question doesn't want us to do tedious calculations rather wants to check if the answerer can identify a parabola and know (or think of) this property. I personally don't think they want us to verify the property. If it does, I think the other answer does that.
              $endgroup$
              – Sri Krishna Sahoo
              Dec 12 '18 at 7:58













              1












              $begingroup$

              The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.



              That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.



                That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.



                  That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.






                  share|cite|improve this answer











                  $endgroup$



                  The equation of the parabola’s axis is $2x-y+1=0$ and that of its tangent at the vertex is $x+2y+3=0$. The directrix therefore has the form $x+2y+d=0$ and you can find the focus in various ways, such as using the fact that the vertex is halfway between the focus and directrix.



                  That said, if you’re planning to compute tangents to the parabola and the perpendiculars through the focus, the approach in lab bhattacharjee’s answer is probably a simpler way to go than starting from the form of equation that you propose.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 12 '18 at 8:20

























                  answered Dec 12 '18 at 8:00









                  amdamd

                  30.6k21050




                  30.6k21050






























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