computes the $phi$ angle of an ellipse
0
$begingroup$
I've an ellipse that is rotated and not centered in the origin. Hence, I've an ellipse equation that is in the form of:
begin{equation}
frac{((x-h)cos(phi) + (y-k)sin(phi))^2}{a^2} + frac{((x-h)sin(phi) - (y-k)cos(phi))^2}{b^2} = 1
end{equation}
Suppose that I know the foci, the center and a and b, how can find $phi$ ?
trigonometry conic-sections curves
asked Dec 11 '18 at 14:00
thinker.92thinker.92
156
$endgroup$
|
show 2 more comments
0
$begingroup$
I've an ellipse that is rotated and not centered in the origin. Hence, I've an ellipse equation that is in the form of:
begin{equation}
frac{((x-h)cos(phi) + (y-k)sin(phi))^2}{a^2} + frac{((x-h)sin(phi) - (y-k)cos(phi))^2}{b^2} = 1
end{equation}
Suppose that I know the foci, the center and a and b, how can find $phi$ ?
trigonometry conic-sections curves
asked Dec 11 '18 at 14:00
thinker.92thinker.92
156
$endgroup$
1
$begingroup$
You mean that you have $h,k,a,b$ and the foci right? If I am right, then you can know easily enough the 4 vertices of the ellipse so you can substitute $x$ and $y$ and get some equations having only $phi$ as an unknown
$endgroup$
– Fareed AF
Dec 11 '18 at 16:00
$begingroup$
First of all thanks for the reply. Yes, I know $h$, $k$, $a$, $b$ and foci. If you are right how can I derive the vertices from these information?
$endgroup$
– thinker.92
Dec 11 '18 at 16:05
1
$begingroup$
If you know the foci, then you’re basically done: they lie on the semimajor axis.
$endgroup$
– amd
Dec 11 '18 at 18:38
1
$begingroup$
If you know the center and the foci then you can get the equation of this line joining the center with the foci (the major axis) and you know that the distance from then center to the a vertex on the ellipse is $a$ so you get the coordinates of 2 vertices by this method
$endgroup$
– Fareed AF
Dec 11 '18 at 19:01
1
$begingroup$
Yes: the slope of that line is precisely $tanphi$.
$endgroup$
– Aretino
Dec 13 '18 at 17:12
|
show 2 more comments
0
0
0
$begingroup$
I've an ellipse that is rotated and not centered in the origin. Hence, I've an ellipse equation that is in the form of:
begin{equation}
frac{((x-h)cos(phi) + (y-k)sin(phi))^2}{a^2} + frac{((x-h)sin(phi) - (y-k)cos(phi))^2}{b^2} = 1
end{equation}
Suppose that I know the foci, the center and a and b, how can find $phi$ ?
trigonometry conic-sections curves
asked Dec 11 '18 at 14:00
thinker.92thinker.92
156
$endgroup$
I've an ellipse that is rotated and not centered in the origin. Hence, I've an ellipse equation that is in the form of:
begin{equation}
frac{((x-h)cos(phi) + (y-k)sin(phi))^2}{a^2} + frac{((x-h)sin(phi) - (y-k)cos(phi))^2}{b^2} = 1
end{equation}
Suppose that I know the foci, the center and a and b, how can find $phi$ ?
trigonometry conic-sections curves
trigonometry conic-sections curves
asked Dec 11 '18 at 14:00
thinker.92thinker.92
156
asked Dec 11 '18 at 14:00
thinker.92thinker.92
156
asked Dec 11 '18 at 14:00
thinker.92thinker.92
156
asked Dec 11 '18 at 14:00
thinker.92thinker.92
156
asked Dec 11 '18 at 14:00
thinker.92thinker.92
156
156
1
$begingroup$
You mean that you have $h,k,a,b$ and the foci right? If I am right, then you can know easily enough the 4 vertices of the ellipse so you can substitute $x$ and $y$ and get some equations having only $phi$ as an unknown
$endgroup$
– Fareed AF
Dec 11 '18 at 16:00
$begingroup$
First of all thanks for the reply. Yes, I know $h$, $k$, $a$, $b$ and foci. If you are right how can I derive the vertices from these information?
$endgroup$
– thinker.92
Dec 11 '18 at 16:05
1
$begingroup$
If you know the foci, then you’re basically done: they lie on the semimajor axis.
$endgroup$
– amd
Dec 11 '18 at 18:38
1
$begingroup$
If you know the center and the foci then you can get the equation of this line joining the center with the foci (the major axis) and you know that the distance from then center to the a vertex on the ellipse is $a$ so you get the coordinates of 2 vertices by this method
$endgroup$
– Fareed AF
Dec 11 '18 at 19:01
1
$begingroup$
Yes: the slope of that line is precisely $tanphi$.
$endgroup$
– Aretino
Dec 13 '18 at 17:12
|
show 2 more comments
1
$begingroup$
You mean that you have $h,k,a,b$ and the foci right? If I am right, then you can know easily enough the 4 vertices of the ellipse so you can substitute $x$ and $y$ and get some equations having only $phi$ as an unknown
$endgroup$
– Fareed AF
Dec 11 '18 at 16:00
$begingroup$
First of all thanks for the reply. Yes, I know $h$, $k$, $a$, $b$ and foci. If you are right how can I derive the vertices from these information?
$endgroup$
– thinker.92
Dec 11 '18 at 16:05
1
$begingroup$
If you know the foci, then you’re basically done: they lie on the semimajor axis.
$endgroup$
– amd
Dec 11 '18 at 18:38
1
$begingroup$
If you know the center and the foci then you can get the equation of this line joining the center with the foci (the major axis) and you know that the distance from then center to the a vertex on the ellipse is $a$ so you get the coordinates of 2 vertices by this method
$endgroup$
– Fareed AF
Dec 11 '18 at 19:01
1
$begingroup$
Yes: the slope of that line is precisely $tanphi$.
$endgroup$
– Aretino
Dec 13 '18 at 17:12
1
1
$begingroup$
You mean that you have $h,k,a,b$ and the foci right? If I am right, then you can know easily enough the 4 vertices of the ellipse so you can substitute $x$ and $y$ and get some equations having only $phi$ as an unknown
$endgroup$
– Fareed AF
Dec 11 '18 at 16:00
$begingroup$
You mean that you have $h,k,a,b$ and the foci right? If I am right, then you can know easily enough the 4 vertices of the ellipse so you can substitute $x$ and $y$ and get some equations having only $phi$ as an unknown
$endgroup$
– Fareed AF
Dec 11 '18 at 16:00
$begingroup$
First of all thanks for the reply. Yes, I know $h$, $k$, $a$, $b$ and foci. If you are right how can I derive the vertices from these information?
$endgroup$
– thinker.92
Dec 11 '18 at 16:05
$begingroup$
First of all thanks for the reply. Yes, I know $h$, $k$, $a$, $b$ and foci. If you are right how can I derive the vertices from these information?
$endgroup$
– thinker.92
Dec 11 '18 at 16:05
1
1
$begingroup$
If you know the foci, then you’re basically done: they lie on the semimajor axis.
$endgroup$
– amd
Dec 11 '18 at 18:38
$begingroup$
If you know the foci, then you’re basically done: they lie on the semimajor axis.
$endgroup$
– amd
Dec 11 '18 at 18:38
1
1
$begingroup$
If you know the center and the foci then you can get the equation of this line joining the center with the foci (the major axis) and you know that the distance from then center to the a vertex on the ellipse is $a$ so you get the coordinates of 2 vertices by this method
$endgroup$
– Fareed AF
Dec 11 '18 at 19:01
$begingroup$
If you know the center and the foci then you can get the equation of this line joining the center with the foci (the major axis) and you know that the distance from then center to the a vertex on the ellipse is $a$ so you get the coordinates of 2 vertices by this method
$endgroup$
– Fareed AF
Dec 11 '18 at 19:01
1
1
$begingroup$
Yes: the slope of that line is precisely $tanphi$.
$endgroup$
– Aretino
Dec 13 '18 at 17:12
$begingroup$
Yes: the slope of that line is precisely $tanphi$.
$endgroup$
– Aretino
Dec 13 '18 at 17:12
|
show 2 more comments
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1
$begingroup$
You mean that you have $h,k,a,b$ and the foci right? If I am right, then you can know easily enough the 4 vertices of the ellipse so you can substitute $x$ and $y$ and get some equations having only $phi$ as an unknown
$endgroup$
– Fareed AF
Dec 11 '18 at 16:00
$begingroup$
First of all thanks for the reply. Yes, I know $h$, $k$, $a$, $b$ and foci. If you are right how can I derive the vertices from these information?
$endgroup$
– thinker.92
Dec 11 '18 at 16:05
1
$begingroup$
If you know the foci, then you’re basically done: they lie on the semimajor axis.
$endgroup$
– amd
Dec 11 '18 at 18:38
1
$begingroup$
If you know the center and the foci then you can get the equation of this line joining the center with the foci (the major axis) and you know that the distance from then center to the a vertex on the ellipse is $a$ so you get the coordinates of 2 vertices by this method
$endgroup$
– Fareed AF
Dec 11 '18 at 19:01
1
$begingroup$
Yes: the slope of that line is precisely $tanphi$.
$endgroup$
– Aretino
Dec 13 '18 at 17:12