How many points are needed to uniquely define an ellipse?
Clash Royale CLAN TAG#URR8PPP
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I recently asked a question on this forum regarding why 3 points guaranteed the presence or absence of a unique equation representing a specific circle.
(link here What do "3 different points" have to do with linear dependence in determining a unique circle?)
Shortly after this, I came across a question in my book that provided a picture of 4 red dots (image below) and asked, "How many ellipses do these 4 red points define". Having read the comments on my post with the circle, I thought that this was fairly straight forward.
I chose " 1 ".
This was wrong. The answer was infinite. This caught me as surprising as I didn't think of the equations for a circle and an ellipse as differing by much beyond a scaling factor for each quadratic term.
I know that the general equation for an ellipse is as follows:
$$left(fracx-haright)^2 + left(fracy-kbright)^2 = 1$$
The only thing I can think of is that because of the added scaling factors, there are now technically two additional unknowns (for a total of 4 different unknowns... h, a, k, and b), and therefore I need 4 points to specify an unique ellipse.
However, I thought to myself again, even if the ellipse is not centered at the origin, if all 4 given points happened to coincide with the intersection between the major axis and the ellipse and the minor axis and the ellipse, then certainly that would specify an unique ellipse.
If this is true, then why does the arrangement of the points matter in determining whether or not an unique ellipse is specified?
Visual explanations would be greatly appreciated!
linear-algebra conic-sections
edited Jan 6 at 11:12
polfosol
5,55731945
asked Jan 6 at 8:30
S.CramerS.Cramer
13118
$endgroup$
|
show 1 more comment
$begingroup$
I recently asked a question on this forum regarding why 3 points guaranteed the presence or absence of a unique equation representing a specific circle.
(link here What do "3 different points" have to do with linear dependence in determining a unique circle?)
Shortly after this, I came across a question in my book that provided a picture of 4 red dots (image below) and asked, "How many ellipses do these 4 red points define". Having read the comments on my post with the circle, I thought that this was fairly straight forward.
I chose " 1 ".
This was wrong. The answer was infinite. This caught me as surprising as I didn't think of the equations for a circle and an ellipse as differing by much beyond a scaling factor for each quadratic term.
I know that the general equation for an ellipse is as follows:
$$left(fracx-haright)^2 + left(fracy-kbright)^2 = 1$$
The only thing I can think of is that because of the added scaling factors, there are now technically two additional unknowns (for a total of 4 different unknowns... h, a, k, and b), and therefore I need 4 points to specify an unique ellipse.
However, I thought to myself again, even if the ellipse is not centered at the origin, if all 4 given points happened to coincide with the intersection between the major axis and the ellipse and the minor axis and the ellipse, then certainly that would specify an unique ellipse.
If this is true, then why does the arrangement of the points matter in determining whether or not an unique ellipse is specified?
Visual explanations would be greatly appreciated!
linear-algebra conic-sections
edited Jan 6 at 11:12
polfosol
5,55731945
asked Jan 6 at 8:30
S.CramerS.Cramer
13118
$endgroup$
5
$begingroup$
The general equation of an ellipse whose axes of symmetry are parallel to the co-ordinate axes is$$frac(x-h)^2a^2+frac(y-k)^2b^2=1$$
$endgroup$
– Shubham Johri
Jan 6 at 8:36
1
$begingroup$
see GeoGebra Classic icon
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:37
1
$begingroup$
Search the site for "five points conic".
$endgroup$
– Blue
Jan 6 at 8:57
1
$begingroup$
If you are implicitly excluding all ellipses except those given by your "general equation", then two points are enough: one on the $x$-axis and one on the $y$-axis.
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– TonyK
Jan 6 at 21:55
1
$begingroup$
4 points won't define an ellipse; for the image you provided, imagine one that is longer horizontally than it is tall, then one that't taller than it is horizontally. Both fit with the 4 points!
$endgroup$
– user45266
Jan 7 at 1:00
|
show 1 more comment
$begingroup$
I recently asked a question on this forum regarding why 3 points guaranteed the presence or absence of a unique equation representing a specific circle.
(link here What do "3 different points" have to do with linear dependence in determining a unique circle?)
Shortly after this, I came across a question in my book that provided a picture of 4 red dots (image below) and asked, "How many ellipses do these 4 red points define". Having read the comments on my post with the circle, I thought that this was fairly straight forward.
I chose " 1 ".
This was wrong. The answer was infinite. This caught me as surprising as I didn't think of the equations for a circle and an ellipse as differing by much beyond a scaling factor for each quadratic term.
I know that the general equation for an ellipse is as follows:
$$left(fracx-haright)^2 + left(fracy-kbright)^2 = 1$$
The only thing I can think of is that because of the added scaling factors, there are now technically two additional unknowns (for a total of 4 different unknowns... h, a, k, and b), and therefore I need 4 points to specify an unique ellipse.
However, I thought to myself again, even if the ellipse is not centered at the origin, if all 4 given points happened to coincide with the intersection between the major axis and the ellipse and the minor axis and the ellipse, then certainly that would specify an unique ellipse.
If this is true, then why does the arrangement of the points matter in determining whether or not an unique ellipse is specified?
Visual explanations would be greatly appreciated!
linear-algebra conic-sections
edited Jan 6 at 11:12
polfosol
5,55731945
asked Jan 6 at 8:30
S.CramerS.Cramer
13118
$endgroup$
I recently asked a question on this forum regarding why 3 points guaranteed the presence or absence of a unique equation representing a specific circle.
(link here What do "3 different points" have to do with linear dependence in determining a unique circle?)
Shortly after this, I came across a question in my book that provided a picture of 4 red dots (image below) and asked, "How many ellipses do these 4 red points define". Having read the comments on my post with the circle, I thought that this was fairly straight forward.
I chose " 1 ".
This was wrong. The answer was infinite. This caught me as surprising as I didn't think of the equations for a circle and an ellipse as differing by much beyond a scaling factor for each quadratic term.
I know that the general equation for an ellipse is as follows:
$$left(fracx-haright)^2 + left(fracy-kbright)^2 = 1$$
The only thing I can think of is that because of the added scaling factors, there are now technically two additional unknowns (for a total of 4 different unknowns... h, a, k, and b), and therefore I need 4 points to specify an unique ellipse.
However, I thought to myself again, even if the ellipse is not centered at the origin, if all 4 given points happened to coincide with the intersection between the major axis and the ellipse and the minor axis and the ellipse, then certainly that would specify an unique ellipse.
If this is true, then why does the arrangement of the points matter in determining whether or not an unique ellipse is specified?
Visual explanations would be greatly appreciated!
linear-algebra conic-sections
linear-algebra conic-sections
edited Jan 6 at 11:12
polfosol
5,55731945
asked Jan 6 at 8:30
S.CramerS.Cramer
13118
edited Jan 6 at 11:12
polfosol
5,55731945
asked Jan 6 at 8:30
S.CramerS.Cramer
13118
edited Jan 6 at 11:12
polfosol
5,55731945
edited Jan 6 at 11:12
polfosol
5,55731945
edited Jan 6 at 11:12
polfosol
5,55731945
5,55731945
asked Jan 6 at 8:30
S.CramerS.Cramer
13118
asked Jan 6 at 8:30
S.CramerS.Cramer
13118
asked Jan 6 at 8:30
S.CramerS.Cramer
13118
13118
5
$begingroup$
The general equation of an ellipse whose axes of symmetry are parallel to the co-ordinate axes is$$frac(x-h)^2a^2+frac(y-k)^2b^2=1$$
$endgroup$
– Shubham Johri
Jan 6 at 8:36
1
$begingroup$
see GeoGebra Classic icon
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:37
1
$begingroup$
Search the site for "five points conic".
$endgroup$
– Blue
Jan 6 at 8:57
1
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If you are implicitly excluding all ellipses except those given by your "general equation", then two points are enough: one on the $x$-axis and one on the $y$-axis.
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– TonyK
Jan 6 at 21:55
1
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4 points won't define an ellipse; for the image you provided, imagine one that is longer horizontally than it is tall, then one that't taller than it is horizontally. Both fit with the 4 points!
$endgroup$
– user45266
Jan 7 at 1:00
|
show 1 more comment
5
$begingroup$
The general equation of an ellipse whose axes of symmetry are parallel to the co-ordinate axes is$$frac(x-h)^2a^2+frac(y-k)^2b^2=1$$
$endgroup$
– Shubham Johri
Jan 6 at 8:36
1
$begingroup$
see GeoGebra Classic icon
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:37
1
$begingroup$
Search the site for "five points conic".
$endgroup$
– Blue
Jan 6 at 8:57
1
$begingroup$
If you are implicitly excluding all ellipses except those given by your "general equation", then two points are enough: one on the $x$-axis and one on the $y$-axis.
$endgroup$
– TonyK
Jan 6 at 21:55
1
$begingroup$
4 points won't define an ellipse; for the image you provided, imagine one that is longer horizontally than it is tall, then one that't taller than it is horizontally. Both fit with the 4 points!
$endgroup$
– user45266
Jan 7 at 1:00
5
5
$begingroup$
The general equation of an ellipse whose axes of symmetry are parallel to the co-ordinate axes is$$frac(x-h)^2a^2+frac(y-k)^2b^2=1$$
$endgroup$
– Shubham Johri
Jan 6 at 8:36
$begingroup$
The general equation of an ellipse whose axes of symmetry are parallel to the co-ordinate axes is$$frac(x-h)^2a^2+frac(y-k)^2b^2=1$$
$endgroup$
– Shubham Johri
Jan 6 at 8:36
1
1
$begingroup$
see GeoGebra Classic icon
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:37
$begingroup$
see GeoGebra Classic icon
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:37
1
1
$begingroup$
Search the site for "five points conic".
$endgroup$
– Blue
Jan 6 at 8:57
$begingroup$
Search the site for "five points conic".
$endgroup$
– Blue
Jan 6 at 8:57
1
1
$begingroup$
If you are implicitly excluding all ellipses except those given by your "general equation", then two points are enough: one on the $x$-axis and one on the $y$-axis.
$endgroup$
– TonyK
Jan 6 at 21:55
$begingroup$
If you are implicitly excluding all ellipses except those given by your "general equation", then two points are enough: one on the $x$-axis and one on the $y$-axis.
$endgroup$
– TonyK
Jan 6 at 21:55
1
1
$begingroup$
4 points won't define an ellipse; for the image you provided, imagine one that is longer horizontally than it is tall, then one that't taller than it is horizontally. Both fit with the 4 points!
$endgroup$
– user45266
Jan 7 at 1:00
$begingroup$
4 points won't define an ellipse; for the image you provided, imagine one that is longer horizontally than it is tall, then one that't taller than it is horizontally. Both fit with the 4 points!
$endgroup$
– user45266
Jan 7 at 1:00
|
show 1 more comment
6 Answers
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active
oldest
votes
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The equation of an ellipse is:
$$
ax^2+by^2+cxy+dx+ey+f=0
$$
Hence you need $5$ points to obtain the coefficients: $(a,b,c,d,e,f)$, assuming that the center is unknown.
If, on the the other hand, the center is known then $3$ points are enough, since every point's reflection in respect to the center is also a point of the ellipse and you technically have $6$ known points.
In this case, assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like having the center and $2$ points, which is not enough to uniquely determine an ellipse.
answered Jan 6 at 8:55
JevautJevaut
1,032112
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So it seems to me that the symmetry of these points provided is the issue?
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– S.Cramer
Jan 6 at 9:01
1
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Exactly. Assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like being given the center and $2$ points, which is not enough to uniquely determine an ellipse.
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– Jevaut
Jan 6 at 9:07
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Just out of curiosity then, is it fair to say that 2 of the 4 points are "linearly dependent" on one another?
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– S.Cramer
Jan 6 at 9:08
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There exist $c_1,c_2 neq 0$ s.t.: $$ c_1 cdot (-2,2) + c_2 cdot (2,-2)= 0 $$ So, yes.
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– Jevaut
Jan 6 at 9:24
1
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@ShubhamJohri Right, my bad. Taking your explanation of "linear dependence of points" into account, I shall add that there are two pairs of linearly dependent red points.
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– Jevaut
Jan 6 at 10:54
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show 2 more comments
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The equation $left(fracx-haright)^2 + left( fracy-kbright)^2 = 1$ is the equation for an ellipse with major and minor axes parallel to the coordinate axes. We expect such ellipses to be unchanged under horizontal reflection and under vertical reflection through their axes. In this equation, these reflections are effected by $x mapsto 2h - x$ and $y mapsto 2k -y$.
This means, if all you have is one point on the ellipse and the three reflected images of this point, you do not have $8$ independent coordinates; you have $2$ and uninformative reflections forced by the equation.
We can see this by plotting two ellipses at the same center (same $h$ and $k$), intersecting at $4$ points, with, say, semiaxes of length $1$ and $2$.
These clearly have four points of intersection. But as soon as you know an ellipse is centered at the origin and contains any one of the four points of intersection, by the major and minor axis reflection symmetries, it contains all four. This is still true if you use generic ellipses, which can be rotated.
Remember that the reflections are through the major and minor axes, wherever they are.
Of course, there are other ways for two ellipses to intersect at four points.
So just knowing those four points are on an ellipse cannot possibly tell you which one is intended.
Returning to the first diagram, corresponding to the diagram you fave where the four known points are the vertices of a square... Symmetries force the center of the ellipse to be the center of the square, but that's not a very strong constraint.
answered Jan 6 at 18:44
Eric TowersEric Towers
32.4k22268
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This is exactly the question discussed several months ago here (Chinese). The central problem is the hidden constraints put on the ellipse.
When you claim that ellipses are determined by the equation $frac(x-h)^2a^2+frac(y-k)^2b^2=1$, you are implicitly assuming that there are no 'slant' ellipses. For instance, the equation $x^2+y^2+xy=1$ also characterize an ellipse, but it is not included in your equation.
If by 'determine' you mean that the ellipse is the unique one passing through the points, the answer is 5. Since two ellipses can intersect at four points, these 4 points cannot determine a unique one. On the other hand, you can easily construct the unique quadratic curve passing through any 5 given point.
However, you can, in fact, use only one point to specify an ellipse. Since the set of all ellipses $E$ is equipotent to $mathbb R^5$, as discussed above, and $mathbb R^5$ is equipotent with $mathbb R^2$, there is a one-to-one correspondence between $mathbb R^2$ and $E$. You can see a formal discussion here.
answered Jan 6 at 8:55
TreborTrebor
80013
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12
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your link returns to me the most frightening 404 message I've ever seen i.stack.imgur.com/K8HL1.png though google translate may have had something to do with it.
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– uhoh
Jan 7 at 1:35
4
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@uhoh The link is broken for me too. Yes it's google translate on your end, and it's a pretty accurate translation (although I would render it as "... where no knowledge exists").
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– lastresort
Jan 7 at 2:09
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Is that correspondence establishing the equipotency of R5 with R2 either unique or canonical? If not, then one needs the specific correspondence in addition to the single point.
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– Pieter Geerkens
Jan 7 at 10:47
1
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@uhoh The link is fixed.
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– Trebor
Jan 8 at 6:11
add a comment |
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" I know that the general equation for an ellipse is as follows: $(fracx-ha)^2 + (fracy-kb)^2 = 1$ "
This is not correct. The above equation defines not all the ellipses, but only the ellipses with axis parallel to the (Ox,Oy) axis.
The general equation of ellipses is basically the general equation of quadratic curves (with constraints below) :
$$ax^2+2bxy+cy^2+2dx+2fy+g=0$$
where $a,b,c,d,f,g$ are constants.
To distinguish ellipses from hyperbolas, circles and others degenerate formes also defined by the above general equation, the constrains are :
$$Delta=beginvmatrix
a & b & d \
b & c & f \
d & f & g
endvmatrixneq 0quad;quad
beginvmatrix
a & b \
c & d
endvmatrix>0quad;quad
fracDeltaa+c<0quadtextandquad aneq c.$$
They are 5 independent parameters in the above general equation. Thus five points are necessary to define a unique quadratic curve.
Of course, given five arbitrary points doesn't guaranty that the curve will be an ellipse. One have to check that the above constrains are satisfied.
NOTE : A more intuitive way to understand why five points are necessary, consider the equation $(fracx-ha)^2 + (fracy-kb)^2 = 1$ and rotate the (Ox,Oy) in order to avoid to forget the "inclined" ellipses. One more parameter is necessary (the angle of rotation). Thus, five parameters in total.
answered Jan 6 at 9:19
JJacquelinJJacquelin
43.1k21752
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add a comment |
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Consider a subset of the set of ellipses passing through the $4$ red points, that contains ellipses that pass through the points and are centred at $(0,0)$ with axes parallel to the coordinate axes. The general equation of an ellipse belonging to this subset is $$fracx^2a^2+fracy^2b^2=1$$ Since the $4$ points lie on the ellipse, plug them into its equation. Notice that plugging any point generates the same equation, $$4/a^2+4/b^2=1$$ You have two unknowns $a,b$ and only one equation. This means there are infinitely many ordered pairs $(a,b)$ that satisfy the condition, hence the subset is infinite, which in turn implies the set of ellipses that pass through the given points is infinite.
answered Jan 6 at 8:50
Shubham JohriShubham Johri
4,760717
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Doesnt this only occur because all the points have the same value when squared (i.e. 2^2 and (-2)^2. What if they didnt have the same value and were scattered around the place.
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– S.Cramer
Jan 6 at 8:59
1
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Yes, it is because they have the same value on squaring. This, and also because you don't seem to be aware of the general equation of an ellipse in the $xy$ plane whose axes of symmetry are not necessarily parallel to the co-ordinate axes, is the reason I chose this specific subset.
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– Shubham Johri
Jan 6 at 9:03
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So, just for clarification, if these 4 points were not symmetrically arranged, 4 would be enough (in the subset where the ellipse is axes are parallel to the coordinate axes and centered at zero).
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– S.Cramer
Jan 6 at 9:06
1
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Yes, a single point can be used to infer three more points for an ellipse centred at $(0,0)$ and with axes parallel to the coordinate axes. If your point is $(a,b)$, then the other three points are $(a,-b),(-a,b)(-a,-b)$.
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– Shubham Johri
Jan 6 at 9:22
2
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There is no such thing as linearly dependent/independent points. What you are trying to say is that the equations you obtained from the four points are all the same, hence linearly dependent. As an example, consider the three equations:$$2x+y=1\x+2y=3\3x+3y=4$$From the looks of it, you have $3$ equations in $2$ unknowns $x,y$. But careful examination reveals that the third equation is really only the sum of the first two, and is thus of 'no use'.$$E_3=E_1+E_2$$Since $E_3$ can be expressed as a linear combination of $E_1,E_2$, we say that the equations are 'linearly' dependent.
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– Shubham Johri
Jan 6 at 9:32
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show 3 more comments
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Let me try:
An ellipse is defined as the locus of a
point $P$ s.t. the sum of distances of P to
2 fixed points $F_1$ and $F_2$ is constant, i.e.
$D_P:= d(P,F_1) +d(P,F_2) = c$.
Hence with $F(x_1,y_1),F(x_2,y_2)$, and $c,$ we have $5$ parameters to determine an ellipse.
Note : $D_P > d(F_1,F_2)$ (Why?)
answered Jan 6 at 10:14
Peter SzilasPeter Szilas
11k2721
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6 Answers
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6 Answers
6
active
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$begingroup$
The equation of an ellipse is:
$$
ax^2+by^2+cxy+dx+ey+f=0
$$
Hence you need $5$ points to obtain the coefficients: $(a,b,c,d,e,f)$, assuming that the center is unknown.
If, on the the other hand, the center is known then $3$ points are enough, since every point's reflection in respect to the center is also a point of the ellipse and you technically have $6$ known points.
In this case, assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like having the center and $2$ points, which is not enough to uniquely determine an ellipse.
answered Jan 6 at 8:55
JevautJevaut
1,032112
$endgroup$
$begingroup$
So it seems to me that the symmetry of these points provided is the issue?
$endgroup$
– S.Cramer
Jan 6 at 9:01
1
$begingroup$
Exactly. Assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like being given the center and $2$ points, which is not enough to uniquely determine an ellipse.
$endgroup$
– Jevaut
Jan 6 at 9:07
$begingroup$
Just out of curiosity then, is it fair to say that 2 of the 4 points are "linearly dependent" on one another?
$endgroup$
– S.Cramer
Jan 6 at 9:08
$begingroup$
There exist $c_1,c_2 neq 0$ s.t.: $$ c_1 cdot (-2,2) + c_2 cdot (2,-2)= 0 $$ So, yes.
$endgroup$
– Jevaut
Jan 6 at 9:24
1
$begingroup$
@ShubhamJohri Right, my bad. Taking your explanation of "linear dependence of points" into account, I shall add that there are two pairs of linearly dependent red points.
$endgroup$
– Jevaut
Jan 6 at 10:54
|
show 2 more comments
$begingroup$
The equation of an ellipse is:
$$
ax^2+by^2+cxy+dx+ey+f=0
$$
Hence you need $5$ points to obtain the coefficients: $(a,b,c,d,e,f)$, assuming that the center is unknown.
If, on the the other hand, the center is known then $3$ points are enough, since every point's reflection in respect to the center is also a point of the ellipse and you technically have $6$ known points.
In this case, assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like having the center and $2$ points, which is not enough to uniquely determine an ellipse.
answered Jan 6 at 8:55
JevautJevaut
1,032112
$endgroup$
$begingroup$
So it seems to me that the symmetry of these points provided is the issue?
$endgroup$
– S.Cramer
Jan 6 at 9:01
1
$begingroup$
Exactly. Assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like being given the center and $2$ points, which is not enough to uniquely determine an ellipse.
$endgroup$
– Jevaut
Jan 6 at 9:07
$begingroup$
Just out of curiosity then, is it fair to say that 2 of the 4 points are "linearly dependent" on one another?
$endgroup$
– S.Cramer
Jan 6 at 9:08
$begingroup$
There exist $c_1,c_2 neq 0$ s.t.: $$ c_1 cdot (-2,2) + c_2 cdot (2,-2)= 0 $$ So, yes.
$endgroup$
– Jevaut
Jan 6 at 9:24
1
$begingroup$
@ShubhamJohri Right, my bad. Taking your explanation of "linear dependence of points" into account, I shall add that there are two pairs of linearly dependent red points.
$endgroup$
– Jevaut
Jan 6 at 10:54
|
show 2 more comments
$begingroup$
The equation of an ellipse is:
$$
ax^2+by^2+cxy+dx+ey+f=0
$$
Hence you need $5$ points to obtain the coefficients: $(a,b,c,d,e,f)$, assuming that the center is unknown.
If, on the the other hand, the center is known then $3$ points are enough, since every point's reflection in respect to the center is also a point of the ellipse and you technically have $6$ known points.
In this case, assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like having the center and $2$ points, which is not enough to uniquely determine an ellipse.
answered Jan 6 at 8:55
JevautJevaut
1,032112
$endgroup$
The equation of an ellipse is:
$$
ax^2+by^2+cxy+dx+ey+f=0
$$
Hence you need $5$ points to obtain the coefficients: $(a,b,c,d,e,f)$, assuming that the center is unknown.
If, on the the other hand, the center is known then $3$ points are enough, since every point's reflection in respect to the center is also a point of the ellipse and you technically have $6$ known points.
In this case, assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like having the center and $2$ points, which is not enough to uniquely determine an ellipse.
answered Jan 6 at 8:55
JevautJevaut
1,032112
edited Jan 7 at 7:30
answered Jan 6 at 8:55
JevautJevaut
1,032112
answered Jan 6 at 8:55
JevautJevaut
1,032112
answered Jan 6 at 8:55
JevautJevaut
1,032112
1,032112
$begingroup$
So it seems to me that the symmetry of these points provided is the issue?
$endgroup$
– S.Cramer
Jan 6 at 9:01
1
$begingroup$
Exactly. Assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like being given the center and $2$ points, which is not enough to uniquely determine an ellipse.
$endgroup$
– Jevaut
Jan 6 at 9:07
$begingroup$
Just out of curiosity then, is it fair to say that 2 of the 4 points are "linearly dependent" on one another?
$endgroup$
– S.Cramer
Jan 6 at 9:08
$begingroup$
There exist $c_1,c_2 neq 0$ s.t.: $$ c_1 cdot (-2,2) + c_2 cdot (2,-2)= 0 $$ So, yes.
$endgroup$
– Jevaut
Jan 6 at 9:24
1
$begingroup$
@ShubhamJohri Right, my bad. Taking your explanation of "linear dependence of points" into account, I shall add that there are two pairs of linearly dependent red points.
$endgroup$
– Jevaut
Jan 6 at 10:54
|
show 2 more comments
$begingroup$
So it seems to me that the symmetry of these points provided is the issue?
$endgroup$
– S.Cramer
Jan 6 at 9:01
1
$begingroup$
Exactly. Assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like being given the center and $2$ points, which is not enough to uniquely determine an ellipse.
$endgroup$
– Jevaut
Jan 6 at 9:07
$begingroup$
Just out of curiosity then, is it fair to say that 2 of the 4 points are "linearly dependent" on one another?
$endgroup$
– S.Cramer
Jan 6 at 9:08
$begingroup$
There exist $c_1,c_2 neq 0$ s.t.: $$ c_1 cdot (-2,2) + c_2 cdot (2,-2)= 0 $$ So, yes.
$endgroup$
– Jevaut
Jan 6 at 9:24
1
$begingroup$
@ShubhamJohri Right, my bad. Taking your explanation of "linear dependence of points" into account, I shall add that there are two pairs of linearly dependent red points.
$endgroup$
– Jevaut
Jan 6 at 10:54
$begingroup$
So it seems to me that the symmetry of these points provided is the issue?
$endgroup$
– S.Cramer
Jan 6 at 9:01
$begingroup$
So it seems to me that the symmetry of these points provided is the issue?
$endgroup$
– S.Cramer
Jan 6 at 9:01
1
1
$begingroup$
Exactly. Assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like being given the center and $2$ points, which is not enough to uniquely determine an ellipse.
$endgroup$
– Jevaut
Jan 6 at 9:07
$begingroup$
Exactly. Assuming that the blue-painted point is the center, the points you've been given are symmetric in respect to it and thus it's like being given the center and $2$ points, which is not enough to uniquely determine an ellipse.
$endgroup$
– Jevaut
Jan 6 at 9:07
$begingroup$
Just out of curiosity then, is it fair to say that 2 of the 4 points are "linearly dependent" on one another?
$endgroup$
– S.Cramer
Jan 6 at 9:08
$begingroup$
Just out of curiosity then, is it fair to say that 2 of the 4 points are "linearly dependent" on one another?
$endgroup$
– S.Cramer
Jan 6 at 9:08
$begingroup$
There exist $c_1,c_2 neq 0$ s.t.: $$ c_1 cdot (-2,2) + c_2 cdot (2,-2)= 0 $$ So, yes.
$endgroup$
– Jevaut
Jan 6 at 9:24
$begingroup$
There exist $c_1,c_2 neq 0$ s.t.: $$ c_1 cdot (-2,2) + c_2 cdot (2,-2)= 0 $$ So, yes.
$endgroup$
– Jevaut
Jan 6 at 9:24
1
1
$begingroup$
@ShubhamJohri Right, my bad. Taking your explanation of "linear dependence of points" into account, I shall add that there are two pairs of linearly dependent red points.
$endgroup$
– Jevaut
Jan 6 at 10:54
$begingroup$
@ShubhamJohri Right, my bad. Taking your explanation of "linear dependence of points" into account, I shall add that there are two pairs of linearly dependent red points.
$endgroup$
– Jevaut
Jan 6 at 10:54
|
show 2 more comments
$begingroup$
The equation $left(fracx-haright)^2 + left( fracy-kbright)^2 = 1$ is the equation for an ellipse with major and minor axes parallel to the coordinate axes. We expect such ellipses to be unchanged under horizontal reflection and under vertical reflection through their axes. In this equation, these reflections are effected by $x mapsto 2h - x$ and $y mapsto 2k -y$.
This means, if all you have is one point on the ellipse and the three reflected images of this point, you do not have $8$ independent coordinates; you have $2$ and uninformative reflections forced by the equation.
We can see this by plotting two ellipses at the same center (same $h$ and $k$), intersecting at $4$ points, with, say, semiaxes of length $1$ and $2$.
These clearly have four points of intersection. But as soon as you know an ellipse is centered at the origin and contains any one of the four points of intersection, by the major and minor axis reflection symmetries, it contains all four. This is still true if you use generic ellipses, which can be rotated.
Remember that the reflections are through the major and minor axes, wherever they are.
Of course, there are other ways for two ellipses to intersect at four points.
So just knowing those four points are on an ellipse cannot possibly tell you which one is intended.
Returning to the first diagram, corresponding to the diagram you fave where the four known points are the vertices of a square... Symmetries force the center of the ellipse to be the center of the square, but that's not a very strong constraint.
answered Jan 6 at 18:44
Eric TowersEric Towers
32.4k22268
$endgroup$
add a comment |
$begingroup$
The equation $left(fracx-haright)^2 + left( fracy-kbright)^2 = 1$ is the equation for an ellipse with major and minor axes parallel to the coordinate axes. We expect such ellipses to be unchanged under horizontal reflection and under vertical reflection through their axes. In this equation, these reflections are effected by $x mapsto 2h - x$ and $y mapsto 2k -y$.
This means, if all you have is one point on the ellipse and the three reflected images of this point, you do not have $8$ independent coordinates; you have $2$ and uninformative reflections forced by the equation.
We can see this by plotting two ellipses at the same center (same $h$ and $k$), intersecting at $4$ points, with, say, semiaxes of length $1$ and $2$.
These clearly have four points of intersection. But as soon as you know an ellipse is centered at the origin and contains any one of the four points of intersection, by the major and minor axis reflection symmetries, it contains all four. This is still true if you use generic ellipses, which can be rotated.
Remember that the reflections are through the major and minor axes, wherever they are.
Of course, there are other ways for two ellipses to intersect at four points.
So just knowing those four points are on an ellipse cannot possibly tell you which one is intended.
Returning to the first diagram, corresponding to the diagram you fave where the four known points are the vertices of a square... Symmetries force the center of the ellipse to be the center of the square, but that's not a very strong constraint.
answered Jan 6 at 18:44
Eric TowersEric Towers
32.4k22268
$endgroup$
add a comment |
$begingroup$
The equation $left(fracx-haright)^2 + left( fracy-kbright)^2 = 1$ is the equation for an ellipse with major and minor axes parallel to the coordinate axes. We expect such ellipses to be unchanged under horizontal reflection and under vertical reflection through their axes. In this equation, these reflections are effected by $x mapsto 2h - x$ and $y mapsto 2k -y$.
This means, if all you have is one point on the ellipse and the three reflected images of this point, you do not have $8$ independent coordinates; you have $2$ and uninformative reflections forced by the equation.
We can see this by plotting two ellipses at the same center (same $h$ and $k$), intersecting at $4$ points, with, say, semiaxes of length $1$ and $2$.
These clearly have four points of intersection. But as soon as you know an ellipse is centered at the origin and contains any one of the four points of intersection, by the major and minor axis reflection symmetries, it contains all four. This is still true if you use generic ellipses, which can be rotated.
Remember that the reflections are through the major and minor axes, wherever they are.
Of course, there are other ways for two ellipses to intersect at four points.
So just knowing those four points are on an ellipse cannot possibly tell you which one is intended.
Returning to the first diagram, corresponding to the diagram you fave where the four known points are the vertices of a square... Symmetries force the center of the ellipse to be the center of the square, but that's not a very strong constraint.
answered Jan 6 at 18:44
Eric TowersEric Towers
32.4k22268
$endgroup$
The equation $left(fracx-haright)^2 + left( fracy-kbright)^2 = 1$ is the equation for an ellipse with major and minor axes parallel to the coordinate axes. We expect such ellipses to be unchanged under horizontal reflection and under vertical reflection through their axes. In this equation, these reflections are effected by $x mapsto 2h - x$ and $y mapsto 2k -y$.
This means, if all you have is one point on the ellipse and the three reflected images of this point, you do not have $8$ independent coordinates; you have $2$ and uninformative reflections forced by the equation.
We can see this by plotting two ellipses at the same center (same $h$ and $k$), intersecting at $4$ points, with, say, semiaxes of length $1$ and $2$.
These clearly have four points of intersection. But as soon as you know an ellipse is centered at the origin and contains any one of the four points of intersection, by the major and minor axis reflection symmetries, it contains all four. This is still true if you use generic ellipses, which can be rotated.
Remember that the reflections are through the major and minor axes, wherever they are.
Of course, there are other ways for two ellipses to intersect at four points.
So just knowing those four points are on an ellipse cannot possibly tell you which one is intended.
Returning to the first diagram, corresponding to the diagram you fave where the four known points are the vertices of a square... Symmetries force the center of the ellipse to be the center of the square, but that's not a very strong constraint.
answered Jan 6 at 18:44
Eric TowersEric Towers
32.4k22268
answered Jan 6 at 18:44
Eric TowersEric Towers
32.4k22268
answered Jan 6 at 18:44
Eric TowersEric Towers
32.4k22268
answered Jan 6 at 18:44
Eric TowersEric Towers
32.4k22268
32.4k22268
add a comment |
add a comment |
$begingroup$
This is exactly the question discussed several months ago here (Chinese). The central problem is the hidden constraints put on the ellipse.
When you claim that ellipses are determined by the equation $frac(x-h)^2a^2+frac(y-k)^2b^2=1$, you are implicitly assuming that there are no 'slant' ellipses. For instance, the equation $x^2+y^2+xy=1$ also characterize an ellipse, but it is not included in your equation.
If by 'determine' you mean that the ellipse is the unique one passing through the points, the answer is 5. Since two ellipses can intersect at four points, these 4 points cannot determine a unique one. On the other hand, you can easily construct the unique quadratic curve passing through any 5 given point.
However, you can, in fact, use only one point to specify an ellipse. Since the set of all ellipses $E$ is equipotent to $mathbb R^5$, as discussed above, and $mathbb R^5$ is equipotent with $mathbb R^2$, there is a one-to-one correspondence between $mathbb R^2$ and $E$. You can see a formal discussion here.
answered Jan 6 at 8:55
TreborTrebor
80013
$endgroup$
12
$begingroup$
your link returns to me the most frightening 404 message I've ever seen i.stack.imgur.com/K8HL1.png though google translate may have had something to do with it.
$endgroup$
– uhoh
Jan 7 at 1:35
4
$begingroup$
@uhoh The link is broken for me too. Yes it's google translate on your end, and it's a pretty accurate translation (although I would render it as "... where no knowledge exists").
$endgroup$
– lastresort
Jan 7 at 2:09
$begingroup$
Is that correspondence establishing the equipotency of R5 with R2 either unique or canonical? If not, then one needs the specific correspondence in addition to the single point.
$endgroup$
– Pieter Geerkens
Jan 7 at 10:47
1
$begingroup$
@uhoh The link is fixed.
$endgroup$
– Trebor
Jan 8 at 6:11
add a comment |
$begingroup$
This is exactly the question discussed several months ago here (Chinese). The central problem is the hidden constraints put on the ellipse.
When you claim that ellipses are determined by the equation $frac(x-h)^2a^2+frac(y-k)^2b^2=1$, you are implicitly assuming that there are no 'slant' ellipses. For instance, the equation $x^2+y^2+xy=1$ also characterize an ellipse, but it is not included in your equation.
If by 'determine' you mean that the ellipse is the unique one passing through the points, the answer is 5. Since two ellipses can intersect at four points, these 4 points cannot determine a unique one. On the other hand, you can easily construct the unique quadratic curve passing through any 5 given point.
However, you can, in fact, use only one point to specify an ellipse. Since the set of all ellipses $E$ is equipotent to $mathbb R^5$, as discussed above, and $mathbb R^5$ is equipotent with $mathbb R^2$, there is a one-to-one correspondence between $mathbb R^2$ and $E$. You can see a formal discussion here.
answered Jan 6 at 8:55
TreborTrebor
80013
$endgroup$
12
$begingroup$
your link returns to me the most frightening 404 message I've ever seen i.stack.imgur.com/K8HL1.png though google translate may have had something to do with it.
$endgroup$
– uhoh
Jan 7 at 1:35
4
$begingroup$
@uhoh The link is broken for me too. Yes it's google translate on your end, and it's a pretty accurate translation (although I would render it as "... where no knowledge exists").
$endgroup$
– lastresort
Jan 7 at 2:09
$begingroup$
Is that correspondence establishing the equipotency of R5 with R2 either unique or canonical? If not, then one needs the specific correspondence in addition to the single point.
$endgroup$
– Pieter Geerkens
Jan 7 at 10:47
1
$begingroup$
@uhoh The link is fixed.
$endgroup$
– Trebor
Jan 8 at 6:11
add a comment |
$begingroup$
This is exactly the question discussed several months ago here (Chinese). The central problem is the hidden constraints put on the ellipse.
When you claim that ellipses are determined by the equation $frac(x-h)^2a^2+frac(y-k)^2b^2=1$, you are implicitly assuming that there are no 'slant' ellipses. For instance, the equation $x^2+y^2+xy=1$ also characterize an ellipse, but it is not included in your equation.
If by 'determine' you mean that the ellipse is the unique one passing through the points, the answer is 5. Since two ellipses can intersect at four points, these 4 points cannot determine a unique one. On the other hand, you can easily construct the unique quadratic curve passing through any 5 given point.
However, you can, in fact, use only one point to specify an ellipse. Since the set of all ellipses $E$ is equipotent to $mathbb R^5$, as discussed above, and $mathbb R^5$ is equipotent with $mathbb R^2$, there is a one-to-one correspondence between $mathbb R^2$ and $E$. You can see a formal discussion here.
answered Jan 6 at 8:55
TreborTrebor
80013
$endgroup$
This is exactly the question discussed several months ago here (Chinese). The central problem is the hidden constraints put on the ellipse.
When you claim that ellipses are determined by the equation $frac(x-h)^2a^2+frac(y-k)^2b^2=1$, you are implicitly assuming that there are no 'slant' ellipses. For instance, the equation $x^2+y^2+xy=1$ also characterize an ellipse, but it is not included in your equation.
If by 'determine' you mean that the ellipse is the unique one passing through the points, the answer is 5. Since two ellipses can intersect at four points, these 4 points cannot determine a unique one. On the other hand, you can easily construct the unique quadratic curve passing through any 5 given point.
However, you can, in fact, use only one point to specify an ellipse. Since the set of all ellipses $E$ is equipotent to $mathbb R^5$, as discussed above, and $mathbb R^5$ is equipotent with $mathbb R^2$, there is a one-to-one correspondence between $mathbb R^2$ and $E$. You can see a formal discussion here.
answered Jan 6 at 8:55
TreborTrebor
80013
edited Jan 8 at 6:09
answered Jan 6 at 8:55
TreborTrebor
80013
answered Jan 6 at 8:55
TreborTrebor
80013
answered Jan 6 at 8:55
TreborTrebor
80013
80013
12
$begingroup$
your link returns to me the most frightening 404 message I've ever seen i.stack.imgur.com/K8HL1.png though google translate may have had something to do with it.
$endgroup$
– uhoh
Jan 7 at 1:35
4
$begingroup$
@uhoh The link is broken for me too. Yes it's google translate on your end, and it's a pretty accurate translation (although I would render it as "... where no knowledge exists").
$endgroup$
– lastresort
Jan 7 at 2:09
$begingroup$
Is that correspondence establishing the equipotency of R5 with R2 either unique or canonical? If not, then one needs the specific correspondence in addition to the single point.
$endgroup$
– Pieter Geerkens
Jan 7 at 10:47
1
$begingroup$
@uhoh The link is fixed.
$endgroup$
– Trebor
Jan 8 at 6:11
add a comment |
12
$begingroup$
your link returns to me the most frightening 404 message I've ever seen i.stack.imgur.com/K8HL1.png though google translate may have had something to do with it.
$endgroup$
– uhoh
Jan 7 at 1:35
4
$begingroup$
@uhoh The link is broken for me too. Yes it's google translate on your end, and it's a pretty accurate translation (although I would render it as "... where no knowledge exists").
$endgroup$
– lastresort
Jan 7 at 2:09
$begingroup$
Is that correspondence establishing the equipotency of R5 with R2 either unique or canonical? If not, then one needs the specific correspondence in addition to the single point.
$endgroup$
– Pieter Geerkens
Jan 7 at 10:47
1
$begingroup$
@uhoh The link is fixed.
$endgroup$
– Trebor
Jan 8 at 6:11
12
12
$begingroup$
your link returns to me the most frightening 404 message I've ever seen i.stack.imgur.com/K8HL1.png though google translate may have had something to do with it.
$endgroup$
– uhoh
Jan 7 at 1:35
$begingroup$
your link returns to me the most frightening 404 message I've ever seen i.stack.imgur.com/K8HL1.png though google translate may have had something to do with it.
$endgroup$
– uhoh
Jan 7 at 1:35
4
4
$begingroup$
@uhoh The link is broken for me too. Yes it's google translate on your end, and it's a pretty accurate translation (although I would render it as "... where no knowledge exists").
$endgroup$
– lastresort
Jan 7 at 2:09
$begingroup$
@uhoh The link is broken for me too. Yes it's google translate on your end, and it's a pretty accurate translation (although I would render it as "... where no knowledge exists").
$endgroup$
– lastresort
Jan 7 at 2:09
$begingroup$
Is that correspondence establishing the equipotency of R5 with R2 either unique or canonical? If not, then one needs the specific correspondence in addition to the single point.
$endgroup$
– Pieter Geerkens
Jan 7 at 10:47
$begingroup$
Is that correspondence establishing the equipotency of R5 with R2 either unique or canonical? If not, then one needs the specific correspondence in addition to the single point.
$endgroup$
– Pieter Geerkens
Jan 7 at 10:47
1
1
$begingroup$
@uhoh The link is fixed.
$endgroup$
– Trebor
Jan 8 at 6:11
$begingroup$
@uhoh The link is fixed.
$endgroup$
– Trebor
Jan 8 at 6:11
add a comment |
$begingroup$
" I know that the general equation for an ellipse is as follows: $(fracx-ha)^2 + (fracy-kb)^2 = 1$ "
This is not correct. The above equation defines not all the ellipses, but only the ellipses with axis parallel to the (Ox,Oy) axis.
The general equation of ellipses is basically the general equation of quadratic curves (with constraints below) :
$$ax^2+2bxy+cy^2+2dx+2fy+g=0$$
where $a,b,c,d,f,g$ are constants.
To distinguish ellipses from hyperbolas, circles and others degenerate formes also defined by the above general equation, the constrains are :
$$Delta=beginvmatrix
a & b & d \
b & c & f \
d & f & g
endvmatrixneq 0quad;quad
beginvmatrix
a & b \
c & d
endvmatrix>0quad;quad
fracDeltaa+c<0quadtextandquad aneq c.$$
They are 5 independent parameters in the above general equation. Thus five points are necessary to define a unique quadratic curve.
Of course, given five arbitrary points doesn't guaranty that the curve will be an ellipse. One have to check that the above constrains are satisfied.
NOTE : A more intuitive way to understand why five points are necessary, consider the equation $(fracx-ha)^2 + (fracy-kb)^2 = 1$ and rotate the (Ox,Oy) in order to avoid to forget the "inclined" ellipses. One more parameter is necessary (the angle of rotation). Thus, five parameters in total.
answered Jan 6 at 9:19
JJacquelinJJacquelin
43.1k21752
$endgroup$
add a comment |
$begingroup$
" I know that the general equation for an ellipse is as follows: $(fracx-ha)^2 + (fracy-kb)^2 = 1$ "
This is not correct. The above equation defines not all the ellipses, but only the ellipses with axis parallel to the (Ox,Oy) axis.
The general equation of ellipses is basically the general equation of quadratic curves (with constraints below) :
$$ax^2+2bxy+cy^2+2dx+2fy+g=0$$
where $a,b,c,d,f,g$ are constants.
To distinguish ellipses from hyperbolas, circles and others degenerate formes also defined by the above general equation, the constrains are :
$$Delta=beginvmatrix
a & b & d \
b & c & f \
d & f & g
endvmatrixneq 0quad;quad
beginvmatrix
a & b \
c & d
endvmatrix>0quad;quad
fracDeltaa+c<0quadtextandquad aneq c.$$
They are 5 independent parameters in the above general equation. Thus five points are necessary to define a unique quadratic curve.
Of course, given five arbitrary points doesn't guaranty that the curve will be an ellipse. One have to check that the above constrains are satisfied.
NOTE : A more intuitive way to understand why five points are necessary, consider the equation $(fracx-ha)^2 + (fracy-kb)^2 = 1$ and rotate the (Ox,Oy) in order to avoid to forget the "inclined" ellipses. One more parameter is necessary (the angle of rotation). Thus, five parameters in total.
answered Jan 6 at 9:19
JJacquelinJJacquelin
43.1k21752
$endgroup$
add a comment |
$begingroup$
" I know that the general equation for an ellipse is as follows: $(fracx-ha)^2 + (fracy-kb)^2 = 1$ "
This is not correct. The above equation defines not all the ellipses, but only the ellipses with axis parallel to the (Ox,Oy) axis.
The general equation of ellipses is basically the general equation of quadratic curves (with constraints below) :
$$ax^2+2bxy+cy^2+2dx+2fy+g=0$$
where $a,b,c,d,f,g$ are constants.
To distinguish ellipses from hyperbolas, circles and others degenerate formes also defined by the above general equation, the constrains are :
$$Delta=beginvmatrix
a & b & d \
b & c & f \
d & f & g
endvmatrixneq 0quad;quad
beginvmatrix
a & b \
c & d
endvmatrix>0quad;quad
fracDeltaa+c<0quadtextandquad aneq c.$$
They are 5 independent parameters in the above general equation. Thus five points are necessary to define a unique quadratic curve.
Of course, given five arbitrary points doesn't guaranty that the curve will be an ellipse. One have to check that the above constrains are satisfied.
NOTE : A more intuitive way to understand why five points are necessary, consider the equation $(fracx-ha)^2 + (fracy-kb)^2 = 1$ and rotate the (Ox,Oy) in order to avoid to forget the "inclined" ellipses. One more parameter is necessary (the angle of rotation). Thus, five parameters in total.
answered Jan 6 at 9:19
JJacquelinJJacquelin
43.1k21752
$endgroup$
" I know that the general equation for an ellipse is as follows: $(fracx-ha)^2 + (fracy-kb)^2 = 1$ "
This is not correct. The above equation defines not all the ellipses, but only the ellipses with axis parallel to the (Ox,Oy) axis.
The general equation of ellipses is basically the general equation of quadratic curves (with constraints below) :
$$ax^2+2bxy+cy^2+2dx+2fy+g=0$$
where $a,b,c,d,f,g$ are constants.
To distinguish ellipses from hyperbolas, circles and others degenerate formes also defined by the above general equation, the constrains are :
$$Delta=beginvmatrix
a & b & d \
b & c & f \
d & f & g
endvmatrixneq 0quad;quad
beginvmatrix
a & b \
c & d
endvmatrix>0quad;quad
fracDeltaa+c<0quadtextandquad aneq c.$$
They are 5 independent parameters in the above general equation. Thus five points are necessary to define a unique quadratic curve.
Of course, given five arbitrary points doesn't guaranty that the curve will be an ellipse. One have to check that the above constrains are satisfied.
NOTE : A more intuitive way to understand why five points are necessary, consider the equation $(fracx-ha)^2 + (fracy-kb)^2 = 1$ and rotate the (Ox,Oy) in order to avoid to forget the "inclined" ellipses. One more parameter is necessary (the angle of rotation). Thus, five parameters in total.
answered Jan 6 at 9:19
JJacquelinJJacquelin
43.1k21752
edited Jan 7 at 11:51
answered Jan 6 at 9:19
JJacquelinJJacquelin
43.1k21752
answered Jan 6 at 9:19
JJacquelinJJacquelin
43.1k21752
answered Jan 6 at 9:19
JJacquelinJJacquelin
43.1k21752
43.1k21752
add a comment |
add a comment |
$begingroup$
Consider a subset of the set of ellipses passing through the $4$ red points, that contains ellipses that pass through the points and are centred at $(0,0)$ with axes parallel to the coordinate axes. The general equation of an ellipse belonging to this subset is $$fracx^2a^2+fracy^2b^2=1$$ Since the $4$ points lie on the ellipse, plug them into its equation. Notice that plugging any point generates the same equation, $$4/a^2+4/b^2=1$$ You have two unknowns $a,b$ and only one equation. This means there are infinitely many ordered pairs $(a,b)$ that satisfy the condition, hence the subset is infinite, which in turn implies the set of ellipses that pass through the given points is infinite.
answered Jan 6 at 8:50
Shubham JohriShubham Johri
4,760717
$endgroup$
$begingroup$
Doesnt this only occur because all the points have the same value when squared (i.e. 2^2 and (-2)^2. What if they didnt have the same value and were scattered around the place.
$endgroup$
– S.Cramer
Jan 6 at 8:59
1
$begingroup$
Yes, it is because they have the same value on squaring. This, and also because you don't seem to be aware of the general equation of an ellipse in the $xy$ plane whose axes of symmetry are not necessarily parallel to the co-ordinate axes, is the reason I chose this specific subset.
$endgroup$
– Shubham Johri
Jan 6 at 9:03
$begingroup$
So, just for clarification, if these 4 points were not symmetrically arranged, 4 would be enough (in the subset where the ellipse is axes are parallel to the coordinate axes and centered at zero).
$endgroup$
– S.Cramer
Jan 6 at 9:06
1
$begingroup$
Yes, a single point can be used to infer three more points for an ellipse centred at $(0,0)$ and with axes parallel to the coordinate axes. If your point is $(a,b)$, then the other three points are $(a,-b),(-a,b)(-a,-b)$.
$endgroup$
– Shubham Johri
Jan 6 at 9:22
2
$begingroup$
There is no such thing as linearly dependent/independent points. What you are trying to say is that the equations you obtained from the four points are all the same, hence linearly dependent. As an example, consider the three equations:$$2x+y=1\x+2y=3\3x+3y=4$$From the looks of it, you have $3$ equations in $2$ unknowns $x,y$. But careful examination reveals that the third equation is really only the sum of the first two, and is thus of 'no use'.$$E_3=E_1+E_2$$Since $E_3$ can be expressed as a linear combination of $E_1,E_2$, we say that the equations are 'linearly' dependent.
$endgroup$
– Shubham Johri
Jan 6 at 9:32
|
show 3 more comments
$begingroup$
Consider a subset of the set of ellipses passing through the $4$ red points, that contains ellipses that pass through the points and are centred at $(0,0)$ with axes parallel to the coordinate axes. The general equation of an ellipse belonging to this subset is $$fracx^2a^2+fracy^2b^2=1$$ Since the $4$ points lie on the ellipse, plug them into its equation. Notice that plugging any point generates the same equation, $$4/a^2+4/b^2=1$$ You have two unknowns $a,b$ and only one equation. This means there are infinitely many ordered pairs $(a,b)$ that satisfy the condition, hence the subset is infinite, which in turn implies the set of ellipses that pass through the given points is infinite.
answered Jan 6 at 8:50
Shubham JohriShubham Johri
4,760717
$endgroup$
$begingroup$
Doesnt this only occur because all the points have the same value when squared (i.e. 2^2 and (-2)^2. What if they didnt have the same value and were scattered around the place.
$endgroup$
– S.Cramer
Jan 6 at 8:59
1
$begingroup$
Yes, it is because they have the same value on squaring. This, and also because you don't seem to be aware of the general equation of an ellipse in the $xy$ plane whose axes of symmetry are not necessarily parallel to the co-ordinate axes, is the reason I chose this specific subset.
$endgroup$
– Shubham Johri
Jan 6 at 9:03
$begingroup$
So, just for clarification, if these 4 points were not symmetrically arranged, 4 would be enough (in the subset where the ellipse is axes are parallel to the coordinate axes and centered at zero).
$endgroup$
– S.Cramer
Jan 6 at 9:06
1
$begingroup$
Yes, a single point can be used to infer three more points for an ellipse centred at $(0,0)$ and with axes parallel to the coordinate axes. If your point is $(a,b)$, then the other three points are $(a,-b),(-a,b)(-a,-b)$.
$endgroup$
– Shubham Johri
Jan 6 at 9:22
2
$begingroup$
There is no such thing as linearly dependent/independent points. What you are trying to say is that the equations you obtained from the four points are all the same, hence linearly dependent. As an example, consider the three equations:$$2x+y=1\x+2y=3\3x+3y=4$$From the looks of it, you have $3$ equations in $2$ unknowns $x,y$. But careful examination reveals that the third equation is really only the sum of the first two, and is thus of 'no use'.$$E_3=E_1+E_2$$Since $E_3$ can be expressed as a linear combination of $E_1,E_2$, we say that the equations are 'linearly' dependent.
$endgroup$
– Shubham Johri
Jan 6 at 9:32
|
show 3 more comments
$begingroup$
Consider a subset of the set of ellipses passing through the $4$ red points, that contains ellipses that pass through the points and are centred at $(0,0)$ with axes parallel to the coordinate axes. The general equation of an ellipse belonging to this subset is $$fracx^2a^2+fracy^2b^2=1$$ Since the $4$ points lie on the ellipse, plug them into its equation. Notice that plugging any point generates the same equation, $$4/a^2+4/b^2=1$$ You have two unknowns $a,b$ and only one equation. This means there are infinitely many ordered pairs $(a,b)$ that satisfy the condition, hence the subset is infinite, which in turn implies the set of ellipses that pass through the given points is infinite.
answered Jan 6 at 8:50
Shubham JohriShubham Johri
4,760717
$endgroup$
Consider a subset of the set of ellipses passing through the $4$ red points, that contains ellipses that pass through the points and are centred at $(0,0)$ with axes parallel to the coordinate axes. The general equation of an ellipse belonging to this subset is $$fracx^2a^2+fracy^2b^2=1$$ Since the $4$ points lie on the ellipse, plug them into its equation. Notice that plugging any point generates the same equation, $$4/a^2+4/b^2=1$$ You have two unknowns $a,b$ and only one equation. This means there are infinitely many ordered pairs $(a,b)$ that satisfy the condition, hence the subset is infinite, which in turn implies the set of ellipses that pass through the given points is infinite.
answered Jan 6 at 8:50
Shubham JohriShubham Johri
4,760717
answered Jan 6 at 8:50
Shubham JohriShubham Johri
4,760717
answered Jan 6 at 8:50
Shubham JohriShubham Johri
4,760717
answered Jan 6 at 8:50
Shubham JohriShubham Johri
4,760717
4,760717
$begingroup$
Doesnt this only occur because all the points have the same value when squared (i.e. 2^2 and (-2)^2. What if they didnt have the same value and were scattered around the place.
$endgroup$
– S.Cramer
Jan 6 at 8:59
1
$begingroup$
Yes, it is because they have the same value on squaring. This, and also because you don't seem to be aware of the general equation of an ellipse in the $xy$ plane whose axes of symmetry are not necessarily parallel to the co-ordinate axes, is the reason I chose this specific subset.
$endgroup$
– Shubham Johri
Jan 6 at 9:03
$begingroup$
So, just for clarification, if these 4 points were not symmetrically arranged, 4 would be enough (in the subset where the ellipse is axes are parallel to the coordinate axes and centered at zero).
$endgroup$
– S.Cramer
Jan 6 at 9:06
1
$begingroup$
Yes, a single point can be used to infer three more points for an ellipse centred at $(0,0)$ and with axes parallel to the coordinate axes. If your point is $(a,b)$, then the other three points are $(a,-b),(-a,b)(-a,-b)$.
$endgroup$
– Shubham Johri
Jan 6 at 9:22
2
$begingroup$
There is no such thing as linearly dependent/independent points. What you are trying to say is that the equations you obtained from the four points are all the same, hence linearly dependent. As an example, consider the three equations:$$2x+y=1\x+2y=3\3x+3y=4$$From the looks of it, you have $3$ equations in $2$ unknowns $x,y$. But careful examination reveals that the third equation is really only the sum of the first two, and is thus of 'no use'.$$E_3=E_1+E_2$$Since $E_3$ can be expressed as a linear combination of $E_1,E_2$, we say that the equations are 'linearly' dependent.
$endgroup$
– Shubham Johri
Jan 6 at 9:32
|
show 3 more comments
$begingroup$
Doesnt this only occur because all the points have the same value when squared (i.e. 2^2 and (-2)^2. What if they didnt have the same value and were scattered around the place.
$endgroup$
– S.Cramer
Jan 6 at 8:59
1
$begingroup$
Yes, it is because they have the same value on squaring. This, and also because you don't seem to be aware of the general equation of an ellipse in the $xy$ plane whose axes of symmetry are not necessarily parallel to the co-ordinate axes, is the reason I chose this specific subset.
$endgroup$
– Shubham Johri
Jan 6 at 9:03
$begingroup$
So, just for clarification, if these 4 points were not symmetrically arranged, 4 would be enough (in the subset where the ellipse is axes are parallel to the coordinate axes and centered at zero).
$endgroup$
– S.Cramer
Jan 6 at 9:06
1
$begingroup$
Yes, a single point can be used to infer three more points for an ellipse centred at $(0,0)$ and with axes parallel to the coordinate axes. If your point is $(a,b)$, then the other three points are $(a,-b),(-a,b)(-a,-b)$.
$endgroup$
– Shubham Johri
Jan 6 at 9:22
2
$begingroup$
There is no such thing as linearly dependent/independent points. What you are trying to say is that the equations you obtained from the four points are all the same, hence linearly dependent. As an example, consider the three equations:$$2x+y=1\x+2y=3\3x+3y=4$$From the looks of it, you have $3$ equations in $2$ unknowns $x,y$. But careful examination reveals that the third equation is really only the sum of the first two, and is thus of 'no use'.$$E_3=E_1+E_2$$Since $E_3$ can be expressed as a linear combination of $E_1,E_2$, we say that the equations are 'linearly' dependent.
$endgroup$
– Shubham Johri
Jan 6 at 9:32
$begingroup$
Doesnt this only occur because all the points have the same value when squared (i.e. 2^2 and (-2)^2. What if they didnt have the same value and were scattered around the place.
$endgroup$
– S.Cramer
Jan 6 at 8:59
$begingroup$
Doesnt this only occur because all the points have the same value when squared (i.e. 2^2 and (-2)^2. What if they didnt have the same value and were scattered around the place.
$endgroup$
– S.Cramer
Jan 6 at 8:59
1
1
$begingroup$
Yes, it is because they have the same value on squaring. This, and also because you don't seem to be aware of the general equation of an ellipse in the $xy$ plane whose axes of symmetry are not necessarily parallel to the co-ordinate axes, is the reason I chose this specific subset.
$endgroup$
– Shubham Johri
Jan 6 at 9:03
$begingroup$
Yes, it is because they have the same value on squaring. This, and also because you don't seem to be aware of the general equation of an ellipse in the $xy$ plane whose axes of symmetry are not necessarily parallel to the co-ordinate axes, is the reason I chose this specific subset.
$endgroup$
– Shubham Johri
Jan 6 at 9:03
$begingroup$
So, just for clarification, if these 4 points were not symmetrically arranged, 4 would be enough (in the subset where the ellipse is axes are parallel to the coordinate axes and centered at zero).
$endgroup$
– S.Cramer
Jan 6 at 9:06
$begingroup$
So, just for clarification, if these 4 points were not symmetrically arranged, 4 would be enough (in the subset where the ellipse is axes are parallel to the coordinate axes and centered at zero).
$endgroup$
– S.Cramer
Jan 6 at 9:06
1
1
$begingroup$
Yes, a single point can be used to infer three more points for an ellipse centred at $(0,0)$ and with axes parallel to the coordinate axes. If your point is $(a,b)$, then the other three points are $(a,-b),(-a,b)(-a,-b)$.
$endgroup$
– Shubham Johri
Jan 6 at 9:22
$begingroup$
Yes, a single point can be used to infer three more points for an ellipse centred at $(0,0)$ and with axes parallel to the coordinate axes. If your point is $(a,b)$, then the other three points are $(a,-b),(-a,b)(-a,-b)$.
$endgroup$
– Shubham Johri
Jan 6 at 9:22
2
2
$begingroup$
There is no such thing as linearly dependent/independent points. What you are trying to say is that the equations you obtained from the four points are all the same, hence linearly dependent. As an example, consider the three equations:$$2x+y=1\x+2y=3\3x+3y=4$$From the looks of it, you have $3$ equations in $2$ unknowns $x,y$. But careful examination reveals that the third equation is really only the sum of the first two, and is thus of 'no use'.$$E_3=E_1+E_2$$Since $E_3$ can be expressed as a linear combination of $E_1,E_2$, we say that the equations are 'linearly' dependent.
$endgroup$
– Shubham Johri
Jan 6 at 9:32
$begingroup$
There is no such thing as linearly dependent/independent points. What you are trying to say is that the equations you obtained from the four points are all the same, hence linearly dependent. As an example, consider the three equations:$$2x+y=1\x+2y=3\3x+3y=4$$From the looks of it, you have $3$ equations in $2$ unknowns $x,y$. But careful examination reveals that the third equation is really only the sum of the first two, and is thus of 'no use'.$$E_3=E_1+E_2$$Since $E_3$ can be expressed as a linear combination of $E_1,E_2$, we say that the equations are 'linearly' dependent.
$endgroup$
– Shubham Johri
Jan 6 at 9:32
|
show 3 more comments
$begingroup$
Let me try:
An ellipse is defined as the locus of a
point $P$ s.t. the sum of distances of P to
2 fixed points $F_1$ and $F_2$ is constant, i.e.
$D_P:= d(P,F_1) +d(P,F_2) = c$.
Hence with $F(x_1,y_1),F(x_2,y_2)$, and $c,$ we have $5$ parameters to determine an ellipse.
Note : $D_P > d(F_1,F_2)$ (Why?)
answered Jan 6 at 10:14
Peter SzilasPeter Szilas
11k2721
$endgroup$
add a comment |
$begingroup$
Let me try:
An ellipse is defined as the locus of a
point $P$ s.t. the sum of distances of P to
2 fixed points $F_1$ and $F_2$ is constant, i.e.
$D_P:= d(P,F_1) +d(P,F_2) = c$.
Hence with $F(x_1,y_1),F(x_2,y_2)$, and $c,$ we have $5$ parameters to determine an ellipse.
Note : $D_P > d(F_1,F_2)$ (Why?)
answered Jan 6 at 10:14
Peter SzilasPeter Szilas
11k2721
$endgroup$
add a comment |
$begingroup$
Let me try:
An ellipse is defined as the locus of a
point $P$ s.t. the sum of distances of P to
2 fixed points $F_1$ and $F_2$ is constant, i.e.
$D_P:= d(P,F_1) +d(P,F_2) = c$.
Hence with $F(x_1,y_1),F(x_2,y_2)$, and $c,$ we have $5$ parameters to determine an ellipse.
Note : $D_P > d(F_1,F_2)$ (Why?)
answered Jan 6 at 10:14
Peter SzilasPeter Szilas
11k2721
$endgroup$
Let me try:
An ellipse is defined as the locus of a
point $P$ s.t. the sum of distances of P to
2 fixed points $F_1$ and $F_2$ is constant, i.e.
$D_P:= d(P,F_1) +d(P,F_2) = c$.
Hence with $F(x_1,y_1),F(x_2,y_2)$, and $c,$ we have $5$ parameters to determine an ellipse.
Note : $D_P > d(F_1,F_2)$ (Why?)
answered Jan 6 at 10:14
Peter SzilasPeter Szilas
11k2721
edited Jan 6 at 10:19
answered Jan 6 at 10:14
Peter SzilasPeter Szilas
11k2721
answered Jan 6 at 10:14
Peter SzilasPeter Szilas
11k2721
answered Jan 6 at 10:14
Peter SzilasPeter Szilas
11k2721
11k2721
add a comment |
add a comment |
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$begingroup$
The general equation of an ellipse whose axes of symmetry are parallel to the co-ordinate axes is$$frac(x-h)^2a^2+frac(y-k)^2b^2=1$$
$endgroup$
– Shubham Johri
Jan 6 at 8:36
1
$begingroup$
see GeoGebra Classic icon
$endgroup$
– Aleksas Domarkas
Jan 6 at 8:37
1
$begingroup$
Search the site for "five points conic".
$endgroup$
– Blue
Jan 6 at 8:57
1
$begingroup$
If you are implicitly excluding all ellipses except those given by your "general equation", then two points are enough: one on the $x$-axis and one on the $y$-axis.
$endgroup$
– TonyK
Jan 6 at 21:55
1
$begingroup$
4 points won't define an ellipse; for the image you provided, imagine one that is longer horizontally than it is tall, then one that't taller than it is horizontally. Both fit with the 4 points!
$endgroup$
– user45266
Jan 7 at 1:00