How do I convert an expression in terms of the general equation of a conic section to one in the equation of...












1












$begingroup$


In a major assignment I am to determine the semi-major axis of an elliptic orbit for the star S2 around Sagittarius A*. I found some data that I have used to fit the points to an ellipse - however the equation I get is in terms of



$ax^2 + bxy + cy^2 + dx + ey + f = 0$



rather than



$ frac{x^2}{a^2} + frac{y^2}{ b^2}= 1$.



Can any of you smart people teach me how to 'translate' it, or if it's even necessary in order to determine the semi-major axis of the ellipse?






conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    Where is your data relative too, Sagittarius A*?
    $endgroup$
    – Paul
    Dec 12 '18 at 13:00










  • $begingroup$
    Yes. It's measured in arc seconds of declination and right ascension relative to SgrA*.
    $endgroup$
    – William Randrup
    Dec 12 '18 at 13:05










  • $begingroup$
    Well, the first problem is that in the second equation, it's assumed that the ellipse is located at the origin of the coordinate system. Therefore, you have to move it to the right location first ...
    $endgroup$
    – Matti P.
    Dec 12 '18 at 13:05










  • $begingroup$
    I'm not very adept when it comes to ellipses, but I think it is located at the origin of the coordinate system as (0,0) is located at one of the foci.
    $endgroup$
    – William Randrup
    Dec 12 '18 at 13:08










  • $begingroup$
    Another problem is that declination and right ascension are spherical coordinates and are measured in different units, so there's a question of whether "one arc-second" measures the same apparent distance in both directions. Also, it seems we are not looking "straight down" on S2's orbit but rather view it from an oblique angle, which means the major axis of its path on the celestial sphere may be very different from the major axis of the actual orbit. The distance between two points on the orbit depends on the change in distance from us as well as the apparent angle.
    $endgroup$
    – David K
    Dec 12 '18 at 13:33
















1












$begingroup$


In a major assignment I am to determine the semi-major axis of an elliptic orbit for the star S2 around Sagittarius A*. I found some data that I have used to fit the points to an ellipse - however the equation I get is in terms of



$ax^2 + bxy + cy^2 + dx + ey + f = 0$



rather than



$ frac{x^2}{a^2} + frac{y^2}{ b^2}= 1$.



Can any of you smart people teach me how to 'translate' it, or if it's even necessary in order to determine the semi-major axis of the ellipse?






conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    Where is your data relative too, Sagittarius A*?
    $endgroup$
    – Paul
    Dec 12 '18 at 13:00










  • $begingroup$
    Yes. It's measured in arc seconds of declination and right ascension relative to SgrA*.
    $endgroup$
    – William Randrup
    Dec 12 '18 at 13:05










  • $begingroup$
    Well, the first problem is that in the second equation, it's assumed that the ellipse is located at the origin of the coordinate system. Therefore, you have to move it to the right location first ...
    $endgroup$
    – Matti P.
    Dec 12 '18 at 13:05










  • $begingroup$
    I'm not very adept when it comes to ellipses, but I think it is located at the origin of the coordinate system as (0,0) is located at one of the foci.
    $endgroup$
    – William Randrup
    Dec 12 '18 at 13:08










  • $begingroup$
    Another problem is that declination and right ascension are spherical coordinates and are measured in different units, so there's a question of whether "one arc-second" measures the same apparent distance in both directions. Also, it seems we are not looking "straight down" on S2's orbit but rather view it from an oblique angle, which means the major axis of its path on the celestial sphere may be very different from the major axis of the actual orbit. The distance between two points on the orbit depends on the change in distance from us as well as the apparent angle.
    $endgroup$
    – David K
    Dec 12 '18 at 13:33














1












1








1





$begingroup$


In a major assignment I am to determine the semi-major axis of an elliptic orbit for the star S2 around Sagittarius A*. I found some data that I have used to fit the points to an ellipse - however the equation I get is in terms of



$ax^2 + bxy + cy^2 + dx + ey + f = 0$



rather than



$ frac{x^2}{a^2} + frac{y^2}{ b^2}= 1$.



Can any of you smart people teach me how to 'translate' it, or if it's even necessary in order to determine the semi-major axis of the ellipse?






conic-sections







share|cite|improve this question











$endgroup$




In a major assignment I am to determine the semi-major axis of an elliptic orbit for the star S2 around Sagittarius A*. I found some data that I have used to fit the points to an ellipse - however the equation I get is in terms of



$ax^2 + bxy + cy^2 + dx + ey + f = 0$



rather than



$ frac{x^2}{a^2} + frac{y^2}{ b^2}= 1$.



Can any of you smart people teach me how to 'translate' it, or if it's even necessary in order to determine the semi-major axis of the ellipse?






conic-sections




conic-sections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 13:02









Emilio Novati

52.1k43474




52.1k43474










asked Dec 12 '18 at 12:58









William RandrupWilliam Randrup

61




61












  • $begingroup$
    Where is your data relative too, Sagittarius A*?
    $endgroup$
    – Paul
    Dec 12 '18 at 13:00










  • $begingroup$
    Yes. It's measured in arc seconds of declination and right ascension relative to SgrA*.
    $endgroup$
    – William Randrup
    Dec 12 '18 at 13:05










  • $begingroup$
    Well, the first problem is that in the second equation, it's assumed that the ellipse is located at the origin of the coordinate system. Therefore, you have to move it to the right location first ...
    $endgroup$
    – Matti P.
    Dec 12 '18 at 13:05










  • $begingroup$
    I'm not very adept when it comes to ellipses, but I think it is located at the origin of the coordinate system as (0,0) is located at one of the foci.
    $endgroup$
    – William Randrup
    Dec 12 '18 at 13:08










  • $begingroup$
    Another problem is that declination and right ascension are spherical coordinates and are measured in different units, so there's a question of whether "one arc-second" measures the same apparent distance in both directions. Also, it seems we are not looking "straight down" on S2's orbit but rather view it from an oblique angle, which means the major axis of its path on the celestial sphere may be very different from the major axis of the actual orbit. The distance between two points on the orbit depends on the change in distance from us as well as the apparent angle.
    $endgroup$
    – David K
    Dec 12 '18 at 13:33


















  • $begingroup$
    Where is your data relative too, Sagittarius A*?
    $endgroup$
    – Paul
    Dec 12 '18 at 13:00










  • $begingroup$
    Yes. It's measured in arc seconds of declination and right ascension relative to SgrA*.
    $endgroup$
    – William Randrup
    Dec 12 '18 at 13:05










  • $begingroup$
    Well, the first problem is that in the second equation, it's assumed that the ellipse is located at the origin of the coordinate system. Therefore, you have to move it to the right location first ...
    $endgroup$
    – Matti P.
    Dec 12 '18 at 13:05










  • $begingroup$
    I'm not very adept when it comes to ellipses, but I think it is located at the origin of the coordinate system as (0,0) is located at one of the foci.
    $endgroup$
    – William Randrup
    Dec 12 '18 at 13:08










  • $begingroup$
    Another problem is that declination and right ascension are spherical coordinates and are measured in different units, so there's a question of whether "one arc-second" measures the same apparent distance in both directions. Also, it seems we are not looking "straight down" on S2's orbit but rather view it from an oblique angle, which means the major axis of its path on the celestial sphere may be very different from the major axis of the actual orbit. The distance between two points on the orbit depends on the change in distance from us as well as the apparent angle.
    $endgroup$
    – David K
    Dec 12 '18 at 13:33
















$begingroup$
Where is your data relative too, Sagittarius A*?
$endgroup$
– Paul
Dec 12 '18 at 13:00




$begingroup$
Where is your data relative too, Sagittarius A*?
$endgroup$
– Paul
Dec 12 '18 at 13:00












$begingroup$
Yes. It's measured in arc seconds of declination and right ascension relative to SgrA*.
$endgroup$
– William Randrup
Dec 12 '18 at 13:05




$begingroup$
Yes. It's measured in arc seconds of declination and right ascension relative to SgrA*.
$endgroup$
– William Randrup
Dec 12 '18 at 13:05












$begingroup$
Well, the first problem is that in the second equation, it's assumed that the ellipse is located at the origin of the coordinate system. Therefore, you have to move it to the right location first ...
$endgroup$
– Matti P.
Dec 12 '18 at 13:05




$begingroup$
Well, the first problem is that in the second equation, it's assumed that the ellipse is located at the origin of the coordinate system. Therefore, you have to move it to the right location first ...
$endgroup$
– Matti P.
Dec 12 '18 at 13:05












$begingroup$
I'm not very adept when it comes to ellipses, but I think it is located at the origin of the coordinate system as (0,0) is located at one of the foci.
$endgroup$
– William Randrup
Dec 12 '18 at 13:08




$begingroup$
I'm not very adept when it comes to ellipses, but I think it is located at the origin of the coordinate system as (0,0) is located at one of the foci.
$endgroup$
– William Randrup
Dec 12 '18 at 13:08












$begingroup$
Another problem is that declination and right ascension are spherical coordinates and are measured in different units, so there's a question of whether "one arc-second" measures the same apparent distance in both directions. Also, it seems we are not looking "straight down" on S2's orbit but rather view it from an oblique angle, which means the major axis of its path on the celestial sphere may be very different from the major axis of the actual orbit. The distance between two points on the orbit depends on the change in distance from us as well as the apparent angle.
$endgroup$
– David K
Dec 12 '18 at 13:33




$begingroup$
Another problem is that declination and right ascension are spherical coordinates and are measured in different units, so there's a question of whether "one arc-second" measures the same apparent distance in both directions. Also, it seems we are not looking "straight down" on S2's orbit but rather view it from an oblique angle, which means the major axis of its path on the celestial sphere may be very different from the major axis of the actual orbit. The distance between two points on the orbit depends on the change in distance from us as well as the apparent angle.
$endgroup$
– David K
Dec 12 '18 at 13:33










1 Answer
1






active

oldest

votes


















0












$begingroup$

Think of the ellipse as a quadratic form:
$$[x,y,1]begin{bmatrix}a&b/2&d/2\b/2&c&e/2\d/2&e/2&-f end{bmatrix}begin{bmatrix}x\y\1 end{bmatrix}=0 $$. Lets call the matrix in the middle $Q_0$ and then $[x,y,1]Q_0 begin{bmatrix}x\y\1 end{bmatrix}=0 $



Since you are allowed translate and rotate the coordinate system, you can push a 2d "rigid motion" matrix $E$ such that $[x,y,1] E^T Q_1 Ebegin{bmatrix}x\y\1 end{bmatrix}=0 $ and you can look for such and $E$ such that $Q_1=begin{bmatrix} a'&0&0\0&b'&0\0&0&-1 end{bmatrix}$



The general form of $E$ is $begin{bmatrix}cos(theta)&sin(theta)&p_x
\-sin(theta)&cos(theta)&p_y\
0&0&1 end{bmatrix}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In such issues (attraction by a heavy mass), it is preferable to use polar representation $p=p_0/(1+e cos(theta))$ where the center is the center of mass (math.stackexchange.com/q/2389034)
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 20:48










  • $begingroup$
    Why is it preferable?
    $endgroup$
    – user617446
    Dec 13 '18 at 5:36










  • $begingroup$
    if you look for example to the way the 3 Kepler's laws are derived from the (Newton's) inverse square law of gravitation, you will see that it's this form which is used. See for example formula $r(theta)=...$ on page 6 of this document math.utk.edu/~freire/teaching/fall2006/m142f06NewtonKepler.pdf
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 7:25










  • $begingroup$
    In fact, I wanted to make this comment to the OP, not to you...
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 7:26










  • $begingroup$
    I think that for regression fitting purposes, it is in fact more convenient to keep the $x,y$ Cartesian form, but to to each his own.
    $endgroup$
    – user617446
    Dec 13 '18 at 10:11











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Think of the ellipse as a quadratic form:
$$[x,y,1]begin{bmatrix}a&b/2&d/2\b/2&c&e/2\d/2&e/2&-f end{bmatrix}begin{bmatrix}x\y\1 end{bmatrix}=0 $$. Lets call the matrix in the middle $Q_0$ and then $[x,y,1]Q_0 begin{bmatrix}x\y\1 end{bmatrix}=0 $



Since you are allowed translate and rotate the coordinate system, you can push a 2d "rigid motion" matrix $E$ such that $[x,y,1] E^T Q_1 Ebegin{bmatrix}x\y\1 end{bmatrix}=0 $ and you can look for such and $E$ such that $Q_1=begin{bmatrix} a'&0&0\0&b'&0\0&0&-1 end{bmatrix}$



The general form of $E$ is $begin{bmatrix}cos(theta)&sin(theta)&p_x
\-sin(theta)&cos(theta)&p_y\
0&0&1 end{bmatrix}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In such issues (attraction by a heavy mass), it is preferable to use polar representation $p=p_0/(1+e cos(theta))$ where the center is the center of mass (math.stackexchange.com/q/2389034)
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 20:48










  • $begingroup$
    Why is it preferable?
    $endgroup$
    – user617446
    Dec 13 '18 at 5:36










  • $begingroup$
    if you look for example to the way the 3 Kepler's laws are derived from the (Newton's) inverse square law of gravitation, you will see that it's this form which is used. See for example formula $r(theta)=...$ on page 6 of this document math.utk.edu/~freire/teaching/fall2006/m142f06NewtonKepler.pdf
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 7:25










  • $begingroup$
    In fact, I wanted to make this comment to the OP, not to you...
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 7:26










  • $begingroup$
    I think that for regression fitting purposes, it is in fact more convenient to keep the $x,y$ Cartesian form, but to to each his own.
    $endgroup$
    – user617446
    Dec 13 '18 at 10:11
















0












$begingroup$

Think of the ellipse as a quadratic form:
$$[x,y,1]begin{bmatrix}a&b/2&d/2\b/2&c&e/2\d/2&e/2&-f end{bmatrix}begin{bmatrix}x\y\1 end{bmatrix}=0 $$. Lets call the matrix in the middle $Q_0$ and then $[x,y,1]Q_0 begin{bmatrix}x\y\1 end{bmatrix}=0 $



Since you are allowed translate and rotate the coordinate system, you can push a 2d "rigid motion" matrix $E$ such that $[x,y,1] E^T Q_1 Ebegin{bmatrix}x\y\1 end{bmatrix}=0 $ and you can look for such and $E$ such that $Q_1=begin{bmatrix} a'&0&0\0&b'&0\0&0&-1 end{bmatrix}$



The general form of $E$ is $begin{bmatrix}cos(theta)&sin(theta)&p_x
\-sin(theta)&cos(theta)&p_y\
0&0&1 end{bmatrix}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In such issues (attraction by a heavy mass), it is preferable to use polar representation $p=p_0/(1+e cos(theta))$ where the center is the center of mass (math.stackexchange.com/q/2389034)
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 20:48










  • $begingroup$
    Why is it preferable?
    $endgroup$
    – user617446
    Dec 13 '18 at 5:36










  • $begingroup$
    if you look for example to the way the 3 Kepler's laws are derived from the (Newton's) inverse square law of gravitation, you will see that it's this form which is used. See for example formula $r(theta)=...$ on page 6 of this document math.utk.edu/~freire/teaching/fall2006/m142f06NewtonKepler.pdf
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 7:25










  • $begingroup$
    In fact, I wanted to make this comment to the OP, not to you...
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 7:26










  • $begingroup$
    I think that for regression fitting purposes, it is in fact more convenient to keep the $x,y$ Cartesian form, but to to each his own.
    $endgroup$
    – user617446
    Dec 13 '18 at 10:11














0












0








0





$begingroup$

Think of the ellipse as a quadratic form:
$$[x,y,1]begin{bmatrix}a&b/2&d/2\b/2&c&e/2\d/2&e/2&-f end{bmatrix}begin{bmatrix}x\y\1 end{bmatrix}=0 $$. Lets call the matrix in the middle $Q_0$ and then $[x,y,1]Q_0 begin{bmatrix}x\y\1 end{bmatrix}=0 $



Since you are allowed translate and rotate the coordinate system, you can push a 2d "rigid motion" matrix $E$ such that $[x,y,1] E^T Q_1 Ebegin{bmatrix}x\y\1 end{bmatrix}=0 $ and you can look for such and $E$ such that $Q_1=begin{bmatrix} a'&0&0\0&b'&0\0&0&-1 end{bmatrix}$



The general form of $E$ is $begin{bmatrix}cos(theta)&sin(theta)&p_x
\-sin(theta)&cos(theta)&p_y\
0&0&1 end{bmatrix}$






share|cite|improve this answer









$endgroup$



Think of the ellipse as a quadratic form:
$$[x,y,1]begin{bmatrix}a&b/2&d/2\b/2&c&e/2\d/2&e/2&-f end{bmatrix}begin{bmatrix}x\y\1 end{bmatrix}=0 $$. Lets call the matrix in the middle $Q_0$ and then $[x,y,1]Q_0 begin{bmatrix}x\y\1 end{bmatrix}=0 $



Since you are allowed translate and rotate the coordinate system, you can push a 2d "rigid motion" matrix $E$ such that $[x,y,1] E^T Q_1 Ebegin{bmatrix}x\y\1 end{bmatrix}=0 $ and you can look for such and $E$ such that $Q_1=begin{bmatrix} a'&0&0\0&b'&0\0&0&-1 end{bmatrix}$



The general form of $E$ is $begin{bmatrix}cos(theta)&sin(theta)&p_x
\-sin(theta)&cos(theta)&p_y\
0&0&1 end{bmatrix}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 13:26









user617446user617446

4443




4443












  • $begingroup$
    In such issues (attraction by a heavy mass), it is preferable to use polar representation $p=p_0/(1+e cos(theta))$ where the center is the center of mass (math.stackexchange.com/q/2389034)
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 20:48










  • $begingroup$
    Why is it preferable?
    $endgroup$
    – user617446
    Dec 13 '18 at 5:36










  • $begingroup$
    if you look for example to the way the 3 Kepler's laws are derived from the (Newton's) inverse square law of gravitation, you will see that it's this form which is used. See for example formula $r(theta)=...$ on page 6 of this document math.utk.edu/~freire/teaching/fall2006/m142f06NewtonKepler.pdf
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 7:25










  • $begingroup$
    In fact, I wanted to make this comment to the OP, not to you...
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 7:26










  • $begingroup$
    I think that for regression fitting purposes, it is in fact more convenient to keep the $x,y$ Cartesian form, but to to each his own.
    $endgroup$
    – user617446
    Dec 13 '18 at 10:11


















  • $begingroup$
    In such issues (attraction by a heavy mass), it is preferable to use polar representation $p=p_0/(1+e cos(theta))$ where the center is the center of mass (math.stackexchange.com/q/2389034)
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 20:48










  • $begingroup$
    Why is it preferable?
    $endgroup$
    – user617446
    Dec 13 '18 at 5:36










  • $begingroup$
    if you look for example to the way the 3 Kepler's laws are derived from the (Newton's) inverse square law of gravitation, you will see that it's this form which is used. See for example formula $r(theta)=...$ on page 6 of this document math.utk.edu/~freire/teaching/fall2006/m142f06NewtonKepler.pdf
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 7:25










  • $begingroup$
    In fact, I wanted to make this comment to the OP, not to you...
    $endgroup$
    – Jean Marie
    Dec 13 '18 at 7:26










  • $begingroup$
    I think that for regression fitting purposes, it is in fact more convenient to keep the $x,y$ Cartesian form, but to to each his own.
    $endgroup$
    – user617446
    Dec 13 '18 at 10:11
















$begingroup$
In such issues (attraction by a heavy mass), it is preferable to use polar representation $p=p_0/(1+e cos(theta))$ where the center is the center of mass (math.stackexchange.com/q/2389034)
$endgroup$
– Jean Marie
Dec 12 '18 at 20:48




$begingroup$
In such issues (attraction by a heavy mass), it is preferable to use polar representation $p=p_0/(1+e cos(theta))$ where the center is the center of mass (math.stackexchange.com/q/2389034)
$endgroup$
– Jean Marie
Dec 12 '18 at 20:48












$begingroup$
Why is it preferable?
$endgroup$
– user617446
Dec 13 '18 at 5:36




$begingroup$
Why is it preferable?
$endgroup$
– user617446
Dec 13 '18 at 5:36












$begingroup$
if you look for example to the way the 3 Kepler's laws are derived from the (Newton's) inverse square law of gravitation, you will see that it's this form which is used. See for example formula $r(theta)=...$ on page 6 of this document math.utk.edu/~freire/teaching/fall2006/m142f06NewtonKepler.pdf
$endgroup$
– Jean Marie
Dec 13 '18 at 7:25




$begingroup$
if you look for example to the way the 3 Kepler's laws are derived from the (Newton's) inverse square law of gravitation, you will see that it's this form which is used. See for example formula $r(theta)=...$ on page 6 of this document math.utk.edu/~freire/teaching/fall2006/m142f06NewtonKepler.pdf
$endgroup$
– Jean Marie
Dec 13 '18 at 7:25












$begingroup$
In fact, I wanted to make this comment to the OP, not to you...
$endgroup$
– Jean Marie
Dec 13 '18 at 7:26




$begingroup$
In fact, I wanted to make this comment to the OP, not to you...
$endgroup$
– Jean Marie
Dec 13 '18 at 7:26












$begingroup$
I think that for regression fitting purposes, it is in fact more convenient to keep the $x,y$ Cartesian form, but to to each his own.
$endgroup$
– user617446
Dec 13 '18 at 10:11




$begingroup$
I think that for regression fitting purposes, it is in fact more convenient to keep the $x,y$ Cartesian form, but to to each his own.
$endgroup$
– user617446
Dec 13 '18 at 10:11


















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