How many parabolas can pass through two given points?
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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.
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– Blue
yesterday
1
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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
yesterday
add a comment |
$begingroup$
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
$endgroup$
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
asked yesterday
Dhruv GuptaDhruv Gupta
696
696
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
yesterday
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
yesterday
add a comment |
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
yesterday
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
yesterday
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
yesterday
$begingroup$
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
yesterday
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
yesterday
$begingroup$
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
yesterday
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
$endgroup$
1
$begingroup$
Thank you! When you put it that way, it seems so obvious. Thanks again.
$endgroup$
– Dhruv Gupta
yesterday
23
$begingroup$
You don't get a different parabola for every point. Still infinity, though.
$endgroup$
– Spitemaster
yesterday
$begingroup$
You can set a third point anywhere not on any parabola you've yet considered.
$endgroup$
– Cai
yesterday
1
$begingroup$
Yeah, naturally, you are right. I rewrote my answer.
$endgroup$
– Mundron Schmidt
yesterday
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
+1 very pretty geometry. Much nicer than than the algebraic analysis.
$endgroup$
– Ethan Bolker
yesterday
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
1
$begingroup$
Thank you! When you put it that way, it seems so obvious. Thanks again.
$endgroup$
– Dhruv Gupta
yesterday
23
$begingroup$
You don't get a different parabola for every point. Still infinity, though.
$endgroup$
– Spitemaster
yesterday
$begingroup$
You can set a third point anywhere not on any parabola you've yet considered.
$endgroup$
– Cai
yesterday
1
$begingroup$
Yeah, naturally, you are right. I rewrote my answer.
$endgroup$
– Mundron Schmidt
yesterday
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
Thank you! When you put it that way, it seems so obvious. Thanks again.
$endgroup$
– Dhruv Gupta
yesterday
23
$begingroup$
You don't get a different parabola for every point. Still infinity, though.
$endgroup$
– Spitemaster
yesterday
$begingroup$
You can set a third point anywhere not on any parabola you've yet considered.
$endgroup$
– Cai
yesterday
1
$begingroup$
Yeah, naturally, you are right. I rewrote my answer.
$endgroup$
– Mundron Schmidt
yesterday
add a comment |
$begingroup$
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.
$endgroup$
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.
edited yesterday
answered yesterday
Mundron SchmidtMundron Schmidt
7,3942729
7,3942729
1
$begingroup$
Thank you! When you put it that way, it seems so obvious. Thanks again.
$endgroup$
– Dhruv Gupta
yesterday
23
$begingroup$
You don't get a different parabola for every point. Still infinity, though.
$endgroup$
– Spitemaster
yesterday
$begingroup$
You can set a third point anywhere not on any parabola you've yet considered.
$endgroup$
– Cai
yesterday
1
$begingroup$
Yeah, naturally, you are right. I rewrote my answer.
$endgroup$
– Mundron Schmidt
yesterday
add a comment |
1
$begingroup$
Thank you! When you put it that way, it seems so obvious. Thanks again.
$endgroup$
– Dhruv Gupta
yesterday
23
$begingroup$
You don't get a different parabola for every point. Still infinity, though.
$endgroup$
– Spitemaster
yesterday
$begingroup$
You can set a third point anywhere not on any parabola you've yet considered.
$endgroup$
– Cai
yesterday
1
$begingroup$
Yeah, naturally, you are right. I rewrote my answer.
$endgroup$
– Mundron Schmidt
yesterday
1
1
$begingroup$
Thank you! When you put it that way, it seems so obvious. Thanks again.
$endgroup$
– Dhruv Gupta
yesterday
$begingroup$
Thank you! When you put it that way, it seems so obvious. Thanks again.
$endgroup$
– Dhruv Gupta
yesterday
23
23
$begingroup$
You don't get a different parabola for every point. Still infinity, though.
$endgroup$
– Spitemaster
yesterday
$begingroup$
You don't get a different parabola for every point. Still infinity, though.
$endgroup$
– Spitemaster
yesterday
$begingroup$
You can set a third point anywhere not on any parabola you've yet considered.
$endgroup$
– Cai
yesterday
$begingroup$
You can set a third point anywhere not on any parabola you've yet considered.
$endgroup$
– Cai
yesterday
1
1
$begingroup$
Yeah, naturally, you are right. I rewrote my answer.
$endgroup$
– Mundron Schmidt
yesterday
$begingroup$
Yeah, naturally, you are right. I rewrote my answer.
$endgroup$
– Mundron Schmidt
yesterday
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
+1 very pretty geometry. Much nicer than than the algebraic analysis.
$endgroup$
– Ethan Bolker
yesterday
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
+1 very pretty geometry. Much nicer than than the algebraic analysis.
$endgroup$
– Ethan Bolker
yesterday
add a comment |
$begingroup$
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.
$endgroup$
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.
edited yesterday
answered yesterday
timtfjtimtfj
1,982419
1,982419
1
$begingroup$
+1 very pretty geometry. Much nicer than than the algebraic analysis.
$endgroup$
– Ethan Bolker
yesterday
add a comment |
1
$begingroup$
+1 very pretty geometry. Much nicer than than the algebraic analysis.
$endgroup$
– Ethan Bolker
yesterday
1
1
$begingroup$
+1 very pretty geometry. Much nicer than than the algebraic analysis.
$endgroup$
– Ethan Bolker
yesterday
$begingroup$
+1 very pretty geometry. Much nicer than than the algebraic analysis.
$endgroup$
– Ethan Bolker
yesterday
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited yesterday
answered yesterday
Peter SzilasPeter Szilas
11.1k2821
11.1k2821
add a comment |
add a comment |
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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
yesterday
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
yesterday