How many parabolas can pass through two given points?












2












$begingroup$


I know that in order to uniquely determine a parabola, we require 3 points. So naturally, 2 points will have multiple possible parabolas pass through them.



My question is, how many?






geometry







share|cite|improve this question









$endgroup$








  • 14




    $begingroup$
    Correction: In order to uniquely determine a parabola with an axis in a particular direction, we require 3 points. For why three points aren't sufficient in general, for instance, see, for instance, this answer.
    $endgroup$
    – Blue
    Jan 29 at 7:54








  • 1




    $begingroup$
    In questions like this, the possible answers are, zero, one and infinity. You already said it's more than one, so it must be infinity.
    $endgroup$
    – Oscar Bravo
    Jan 29 at 16:22


















2












$begingroup$


I know that in order to uniquely determine a parabola, we require 3 points. So naturally, 2 points will have multiple possible parabolas pass through them.



My question is, how many?






geometry







share|cite|improve this question









$endgroup$








  • 14




    $begingroup$
    Correction: In order to uniquely determine a parabola with an axis in a particular direction, we require 3 points. For why three points aren't sufficient in general, for instance, see, for instance, this answer.
    $endgroup$
    – Blue
    Jan 29 at 7:54








  • 1




    $begingroup$
    In questions like this, the possible answers are, zero, one and infinity. You already said it's more than one, so it must be infinity.
    $endgroup$
    – Oscar Bravo
    Jan 29 at 16:22
















2












2








2


1



$begingroup$


I know that in order to uniquely determine a parabola, we require 3 points. So naturally, 2 points will have multiple possible parabolas pass through them.



My question is, how many?






geometry







share|cite|improve this question









$endgroup$




I know that in order to uniquely determine a parabola, we require 3 points. So naturally, 2 points will have multiple possible parabolas pass through them.



My question is, how many?






geometry




geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 29 at 7:50









Dhruv GuptaDhruv Gupta

697




697








  • 14




    $begingroup$
    Correction: In order to uniquely determine a parabola with an axis in a particular direction, we require 3 points. For why three points aren't sufficient in general, for instance, see, for instance, this answer.
    $endgroup$
    – Blue
    Jan 29 at 7:54








  • 1




    $begingroup$
    In questions like this, the possible answers are, zero, one and infinity. You already said it's more than one, so it must be infinity.
    $endgroup$
    – Oscar Bravo
    Jan 29 at 16:22
















  • 14




    $begingroup$
    Correction: In order to uniquely determine a parabola with an axis in a particular direction, we require 3 points. For why three points aren't sufficient in general, for instance, see, for instance, this answer.
    $endgroup$
    – Blue
    Jan 29 at 7:54








  • 1




    $begingroup$
    In questions like this, the possible answers are, zero, one and infinity. You already said it's more than one, so it must be infinity.
    $endgroup$
    – Oscar Bravo
    Jan 29 at 16:22










14




14




$begingroup$
Correction: In order to uniquely determine a parabola with an axis in a particular direction, we require 3 points. For why three points aren't sufficient in general, for instance, see, for instance, this answer.
$endgroup$
– Blue
Jan 29 at 7:54






$begingroup$
Correction: In order to uniquely determine a parabola with an axis in a particular direction, we require 3 points. For why three points aren't sufficient in general, for instance, see, for instance, this answer.
$endgroup$
– Blue
Jan 29 at 7:54






1




1




$begingroup$
In questions like this, the possible answers are, zero, one and infinity. You already said it's more than one, so it must be infinity.
$endgroup$
– Oscar Bravo
Jan 29 at 16:22






$begingroup$
In questions like this, the possible answers are, zero, one and infinity. You already said it's more than one, so it must be infinity.
$endgroup$
– Oscar Bravo
Jan 29 at 16:22












3 Answers
3






active

oldest

votes


















6












$begingroup$

Infinitely many.
Draw a line between your two points. Now chose any points on the line as your third point. Since two different points on the line will give you different parabola, you get infinitely many different parabola.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you! When you put it that way, it seems so obvious. Thanks again.
    $endgroup$
    – Dhruv Gupta
    Jan 29 at 7:57






  • 23




    $begingroup$
    You don't get a different parabola for every point. Still infinity, though.
    $endgroup$
    – Spitemaster
    Jan 29 at 12:53










  • $begingroup$
    You can set a third point anywhere not on any parabola you've yet considered.
    $endgroup$
    – Cai
    Jan 29 at 17:06






  • 1




    $begingroup$
    Yeah, naturally, you are right. I rewrote my answer.
    $endgroup$
    – Mundron Schmidt
    Jan 29 at 22:10



















9












$begingroup$

Any size of parabola, with infinitely many orientations for each parabola!



Suppose the points are distance $d$ apart. Call them $A$ and $B$. Take your chosen parabola and pick two points on it separated by distance $d$. Then place them on $A$ and $B$.



Since you can slide the points along the parabola to wherever you like, the parabola can have any orientation that doesn't put its axis parallel to $AB$. (The sliding only gives you $180°$ worth of orientations, but you can reflect it in $AB$ to get the other $180°$.)



Assuming points on the parabola have real numbers as their coordinates, its size and orientation can be described by two real numbers: a size in the range $(0,infty)$ and an angle in the range $(0,π)cup(-π, 0)$.



Because of the way infinities work, this makes the set of parabolas through $A$ and $B$ uncountably infinite and the same size as the set of real nunbers.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    +1 very pretty geometry. Much nicer than than the algebraic analysis.
    $endgroup$
    – Ethan Bolker
    Jan 29 at 14:24



















1












$begingroup$

General equation of a conic section:



$small {Ax^2+Bxy +Cy^2 +Dx+Ey +F=0}$;



where $A,B,C,D,E,F$ are constants.



Necessary condition for a parabola :



$B^2-4AC =0$, which leaves $5$ constants to be determined.



Cf. Blue's link in his comment above.






share|cite|improve this answer











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    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Infinitely many.
    Draw a line between your two points. Now chose any points on the line as your third point. Since two different points on the line will give you different parabola, you get infinitely many different parabola.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Thank you! When you put it that way, it seems so obvious. Thanks again.
      $endgroup$
      – Dhruv Gupta
      Jan 29 at 7:57






    • 23




      $begingroup$
      You don't get a different parabola for every point. Still infinity, though.
      $endgroup$
      – Spitemaster
      Jan 29 at 12:53










    • $begingroup$
      You can set a third point anywhere not on any parabola you've yet considered.
      $endgroup$
      – Cai
      Jan 29 at 17:06






    • 1




      $begingroup$
      Yeah, naturally, you are right. I rewrote my answer.
      $endgroup$
      – Mundron Schmidt
      Jan 29 at 22:10
















    6












    $begingroup$

    Infinitely many.
    Draw a line between your two points. Now chose any points on the line as your third point. Since two different points on the line will give you different parabola, you get infinitely many different parabola.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Thank you! When you put it that way, it seems so obvious. Thanks again.
      $endgroup$
      – Dhruv Gupta
      Jan 29 at 7:57






    • 23




      $begingroup$
      You don't get a different parabola for every point. Still infinity, though.
      $endgroup$
      – Spitemaster
      Jan 29 at 12:53










    • $begingroup$
      You can set a third point anywhere not on any parabola you've yet considered.
      $endgroup$
      – Cai
      Jan 29 at 17:06






    • 1




      $begingroup$
      Yeah, naturally, you are right. I rewrote my answer.
      $endgroup$
      – Mundron Schmidt
      Jan 29 at 22:10














    6












    6








    6





    $begingroup$

    Infinitely many.
    Draw a line between your two points. Now chose any points on the line as your third point. Since two different points on the line will give you different parabola, you get infinitely many different parabola.






    share|cite|improve this answer











    $endgroup$



    Infinitely many.
    Draw a line between your two points. Now chose any points on the line as your third point. Since two different points on the line will give you different parabola, you get infinitely many different parabola.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 29 at 22:10

























    answered Jan 29 at 7:55









    Mundron SchmidtMundron Schmidt

    7,5042729




    7,5042729








    • 1




      $begingroup$
      Thank you! When you put it that way, it seems so obvious. Thanks again.
      $endgroup$
      – Dhruv Gupta
      Jan 29 at 7:57






    • 23




      $begingroup$
      You don't get a different parabola for every point. Still infinity, though.
      $endgroup$
      – Spitemaster
      Jan 29 at 12:53










    • $begingroup$
      You can set a third point anywhere not on any parabola you've yet considered.
      $endgroup$
      – Cai
      Jan 29 at 17:06






    • 1




      $begingroup$
      Yeah, naturally, you are right. I rewrote my answer.
      $endgroup$
      – Mundron Schmidt
      Jan 29 at 22:10














    • 1




      $begingroup$
      Thank you! When you put it that way, it seems so obvious. Thanks again.
      $endgroup$
      – Dhruv Gupta
      Jan 29 at 7:57






    • 23




      $begingroup$
      You don't get a different parabola for every point. Still infinity, though.
      $endgroup$
      – Spitemaster
      Jan 29 at 12:53










    • $begingroup$
      You can set a third point anywhere not on any parabola you've yet considered.
      $endgroup$
      – Cai
      Jan 29 at 17:06






    • 1




      $begingroup$
      Yeah, naturally, you are right. I rewrote my answer.
      $endgroup$
      – Mundron Schmidt
      Jan 29 at 22:10








    1




    1




    $begingroup$
    Thank you! When you put it that way, it seems so obvious. Thanks again.
    $endgroup$
    – Dhruv Gupta
    Jan 29 at 7:57




    $begingroup$
    Thank you! When you put it that way, it seems so obvious. Thanks again.
    $endgroup$
    – Dhruv Gupta
    Jan 29 at 7:57




    23




    23




    $begingroup$
    You don't get a different parabola for every point. Still infinity, though.
    $endgroup$
    – Spitemaster
    Jan 29 at 12:53




    $begingroup$
    You don't get a different parabola for every point. Still infinity, though.
    $endgroup$
    – Spitemaster
    Jan 29 at 12:53












    $begingroup$
    You can set a third point anywhere not on any parabola you've yet considered.
    $endgroup$
    – Cai
    Jan 29 at 17:06




    $begingroup$
    You can set a third point anywhere not on any parabola you've yet considered.
    $endgroup$
    – Cai
    Jan 29 at 17:06




    1




    1




    $begingroup$
    Yeah, naturally, you are right. I rewrote my answer.
    $endgroup$
    – Mundron Schmidt
    Jan 29 at 22:10




    $begingroup$
    Yeah, naturally, you are right. I rewrote my answer.
    $endgroup$
    – Mundron Schmidt
    Jan 29 at 22:10











    9












    $begingroup$

    Any size of parabola, with infinitely many orientations for each parabola!



    Suppose the points are distance $d$ apart. Call them $A$ and $B$. Take your chosen parabola and pick two points on it separated by distance $d$. Then place them on $A$ and $B$.



    Since you can slide the points along the parabola to wherever you like, the parabola can have any orientation that doesn't put its axis parallel to $AB$. (The sliding only gives you $180°$ worth of orientations, but you can reflect it in $AB$ to get the other $180°$.)



    Assuming points on the parabola have real numbers as their coordinates, its size and orientation can be described by two real numbers: a size in the range $(0,infty)$ and an angle in the range $(0,π)cup(-π, 0)$.



    Because of the way infinities work, this makes the set of parabolas through $A$ and $B$ uncountably infinite and the same size as the set of real nunbers.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      +1 very pretty geometry. Much nicer than than the algebraic analysis.
      $endgroup$
      – Ethan Bolker
      Jan 29 at 14:24
















    9












    $begingroup$

    Any size of parabola, with infinitely many orientations for each parabola!



    Suppose the points are distance $d$ apart. Call them $A$ and $B$. Take your chosen parabola and pick two points on it separated by distance $d$. Then place them on $A$ and $B$.



    Since you can slide the points along the parabola to wherever you like, the parabola can have any orientation that doesn't put its axis parallel to $AB$. (The sliding only gives you $180°$ worth of orientations, but you can reflect it in $AB$ to get the other $180°$.)



    Assuming points on the parabola have real numbers as their coordinates, its size and orientation can be described by two real numbers: a size in the range $(0,infty)$ and an angle in the range $(0,π)cup(-π, 0)$.



    Because of the way infinities work, this makes the set of parabolas through $A$ and $B$ uncountably infinite and the same size as the set of real nunbers.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      +1 very pretty geometry. Much nicer than than the algebraic analysis.
      $endgroup$
      – Ethan Bolker
      Jan 29 at 14:24














    9












    9








    9





    $begingroup$

    Any size of parabola, with infinitely many orientations for each parabola!



    Suppose the points are distance $d$ apart. Call them $A$ and $B$. Take your chosen parabola and pick two points on it separated by distance $d$. Then place them on $A$ and $B$.



    Since you can slide the points along the parabola to wherever you like, the parabola can have any orientation that doesn't put its axis parallel to $AB$. (The sliding only gives you $180°$ worth of orientations, but you can reflect it in $AB$ to get the other $180°$.)



    Assuming points on the parabola have real numbers as their coordinates, its size and orientation can be described by two real numbers: a size in the range $(0,infty)$ and an angle in the range $(0,π)cup(-π, 0)$.



    Because of the way infinities work, this makes the set of parabolas through $A$ and $B$ uncountably infinite and the same size as the set of real nunbers.






    share|cite|improve this answer











    $endgroup$



    Any size of parabola, with infinitely many orientations for each parabola!



    Suppose the points are distance $d$ apart. Call them $A$ and $B$. Take your chosen parabola and pick two points on it separated by distance $d$. Then place them on $A$ and $B$.



    Since you can slide the points along the parabola to wherever you like, the parabola can have any orientation that doesn't put its axis parallel to $AB$. (The sliding only gives you $180°$ worth of orientations, but you can reflect it in $AB$ to get the other $180°$.)



    Assuming points on the parabola have real numbers as their coordinates, its size and orientation can be described by two real numbers: a size in the range $(0,infty)$ and an angle in the range $(0,π)cup(-π, 0)$.



    Because of the way infinities work, this makes the set of parabolas through $A$ and $B$ uncountably infinite and the same size as the set of real nunbers.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 29 at 9:40

























    answered Jan 29 at 9:31









    timtfjtimtfj

    2,458420




    2,458420








    • 1




      $begingroup$
      +1 very pretty geometry. Much nicer than than the algebraic analysis.
      $endgroup$
      – Ethan Bolker
      Jan 29 at 14:24














    • 1




      $begingroup$
      +1 very pretty geometry. Much nicer than than the algebraic analysis.
      $endgroup$
      – Ethan Bolker
      Jan 29 at 14:24








    1




    1




    $begingroup$
    +1 very pretty geometry. Much nicer than than the algebraic analysis.
    $endgroup$
    – Ethan Bolker
    Jan 29 at 14:24




    $begingroup$
    +1 very pretty geometry. Much nicer than than the algebraic analysis.
    $endgroup$
    – Ethan Bolker
    Jan 29 at 14:24











    1












    $begingroup$

    General equation of a conic section:



    $small {Ax^2+Bxy +Cy^2 +Dx+Ey +F=0}$;



    where $A,B,C,D,E,F$ are constants.



    Necessary condition for a parabola :



    $B^2-4AC =0$, which leaves $5$ constants to be determined.



    Cf. Blue's link in his comment above.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      General equation of a conic section:



      $small {Ax^2+Bxy +Cy^2 +Dx+Ey +F=0}$;



      where $A,B,C,D,E,F$ are constants.



      Necessary condition for a parabola :



      $B^2-4AC =0$, which leaves $5$ constants to be determined.



      Cf. Blue's link in his comment above.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        General equation of a conic section:



        $small {Ax^2+Bxy +Cy^2 +Dx+Ey +F=0}$;



        where $A,B,C,D,E,F$ are constants.



        Necessary condition for a parabola :



        $B^2-4AC =0$, which leaves $5$ constants to be determined.



        Cf. Blue's link in his comment above.






        share|cite|improve this answer











        $endgroup$



        General equation of a conic section:



        $small {Ax^2+Bxy +Cy^2 +Dx+Ey +F=0}$;



        where $A,B,C,D,E,F$ are constants.



        Necessary condition for a parabola :



        $B^2-4AC =0$, which leaves $5$ constants to be determined.



        Cf. Blue's link in his comment above.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 29 at 10:09

























        answered Jan 29 at 10:02









        Peter SzilasPeter Szilas

        11.5k2822




        11.5k2822






























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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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