Parametric form of ellipse $x^2 + 16y^2 = 4$
The given ellipse is $-$
$x^2 +16y^2 = 4$
I know the standard method of changing the equation of ellipse in parametric form.
Instead of complicated calculation, can we simply put $x= 2 cos theta , y= frac{1}{2} sin theta $ to satisfy the equation. Does this still represent the same ellipse? How do I know if my parametric equation is representing the same ellipse as given ,without using graph?
conic-sections parametrization
asked Dec 26 at 8:14
Mathsaddict
1758
|
show 1 more comment
The given ellipse is $-$
$x^2 +16y^2 = 4$
I know the standard method of changing the equation of ellipse in parametric form.
Instead of complicated calculation, can we simply put $x= 2 cos theta , y= frac{1}{2} sin theta $ to satisfy the equation. Does this still represent the same ellipse? How do I know if my parametric equation is representing the same ellipse as given ,without using graph?
conic-sections parametrization
asked Dec 26 at 8:14
Mathsaddict
1758
2
You can verify this by looking at the graph. Your parametric form is the same as the initial one, assuming the range of $t$ is from $0$ to $2pi$
– roman
Dec 26 at 8:16
@roman is there any other way to verify this than graph?
– Mathsaddict
Dec 26 at 8:21
@roman Just looking at a graph usually isn't enough to rigourously prove that two things are the same.
– Arthur
Dec 26 at 8:21
@Arthur I know, but that helps to gain some intuition
– roman
Dec 26 at 8:22
@Mathsaddict That's parametric form, not polar.
– Arthur
Dec 26 at 8:22
|
show 1 more comment
The given ellipse is $-$
$x^2 +16y^2 = 4$
I know the standard method of changing the equation of ellipse in parametric form.
Instead of complicated calculation, can we simply put $x= 2 cos theta , y= frac{1}{2} sin theta $ to satisfy the equation. Does this still represent the same ellipse? How do I know if my parametric equation is representing the same ellipse as given ,without using graph?
conic-sections parametrization
asked Dec 26 at 8:14
Mathsaddict
1758
The given ellipse is $-$
$x^2 +16y^2 = 4$
I know the standard method of changing the equation of ellipse in parametric form.
Instead of complicated calculation, can we simply put $x= 2 cos theta , y= frac{1}{2} sin theta $ to satisfy the equation. Does this still represent the same ellipse? How do I know if my parametric equation is representing the same ellipse as given ,without using graph?
conic-sections parametrization
conic-sections parametrization
asked Dec 26 at 8:14
Mathsaddict
1758
asked Dec 26 at 8:14
Mathsaddict
1758
edited Dec 26 at 8:25
asked Dec 26 at 8:14
Mathsaddict
1758
asked Dec 26 at 8:14
Mathsaddict
1758
asked Dec 26 at 8:14
Mathsaddict
1758
1758
2
You can verify this by looking at the graph. Your parametric form is the same as the initial one, assuming the range of $t$ is from $0$ to $2pi$
– roman
Dec 26 at 8:16
@roman is there any other way to verify this than graph?
– Mathsaddict
Dec 26 at 8:21
@roman Just looking at a graph usually isn't enough to rigourously prove that two things are the same.
– Arthur
Dec 26 at 8:21
@Arthur I know, but that helps to gain some intuition
– roman
Dec 26 at 8:22
@Mathsaddict That's parametric form, not polar.
– Arthur
Dec 26 at 8:22
|
show 1 more comment
2
You can verify this by looking at the graph. Your parametric form is the same as the initial one, assuming the range of $t$ is from $0$ to $2pi$
– roman
Dec 26 at 8:16
@roman is there any other way to verify this than graph?
– Mathsaddict
Dec 26 at 8:21
@roman Just looking at a graph usually isn't enough to rigourously prove that two things are the same.
– Arthur
Dec 26 at 8:21
@Arthur I know, but that helps to gain some intuition
– roman
Dec 26 at 8:22
@Mathsaddict That's parametric form, not polar.
– Arthur
Dec 26 at 8:22
2
2
You can verify this by looking at the graph. Your parametric form is the same as the initial one, assuming the range of $t$ is from $0$ to $2pi$
– roman
Dec 26 at 8:16
You can verify this by looking at the graph. Your parametric form is the same as the initial one, assuming the range of $t$ is from $0$ to $2pi$
– roman
Dec 26 at 8:16
@roman is there any other way to verify this than graph?
– Mathsaddict
Dec 26 at 8:21
@roman is there any other way to verify this than graph?
– Mathsaddict
Dec 26 at 8:21
@roman Just looking at a graph usually isn't enough to rigourously prove that two things are the same.
– Arthur
Dec 26 at 8:21
@roman Just looking at a graph usually isn't enough to rigourously prove that two things are the same.
– Arthur
Dec 26 at 8:21
@Arthur I know, but that helps to gain some intuition
– roman
Dec 26 at 8:22
@Arthur I know, but that helps to gain some intuition
– roman
Dec 26 at 8:22
@Mathsaddict That's parametric form, not polar.
– Arthur
Dec 26 at 8:22
@Mathsaddict That's parametric form, not polar.
– Arthur
Dec 26 at 8:22
|
show 1 more comment
3 Answers
3
active
oldest
votes
In terms of 'intuition' - generally the only thing you need to be careful of this technique is that you may not obtain the whole ellipse. But upon checking the domain of $theta$. This particular parameterisation that you have used is okay!
A choice of parameterisation is generally from the knowledge that it will satisfy the cartesian equation and that it is sufficiently variable.
For example if I wanted to find a parameterisation for the parabola $x^2=4ay$ I could simply say
$$x=2t^2 y=frac{t^4}{a}$$
Indeed, it satisfies the equation - however this would not 'capture' the parabola for $x<0$. We need to be aware of such things when we choose a parameterisation.
answered Dec 26 at 8:41
Hugh Entwistle
788217
add a comment |
Well, you can make the substitution in the equation, so you get:
$$(2cos(theta))^2+16(frac 1 2 sin(theta))^2=4.$$
Therefore your parametrization is ok.
answered Dec 26 at 8:27
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
In the case of an ellipse, the simplest solution is to make a change of variables, in order to get the equation of a circle.
For the equation
$$ x^2 +16y^2 = 4 $$
Just consider the change in variables:
$$X = x; quad Y = 4y $$
To get the new equation
$$X^2 + Y^2 = 4 $$
Therefore the equation of a circle of center $(0,0)$ and ray $R=2$.
The corresponding parametric equation is
$$X=Rcos thetaquad Y=Rsin theta $$
For $theta$ varying from $0$ to $2 pi$
Which corresponds to
$$x=2cos thetaquad y=frac{1}{2}sin theta $$
answered Dec 26 at 14:59
Damien
4474
That's easy! Thanks.
– Mathsaddict
Dec 26 at 16:24
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
In terms of 'intuition' - generally the only thing you need to be careful of this technique is that you may not obtain the whole ellipse. But upon checking the domain of $theta$. This particular parameterisation that you have used is okay!
A choice of parameterisation is generally from the knowledge that it will satisfy the cartesian equation and that it is sufficiently variable.
For example if I wanted to find a parameterisation for the parabola $x^2=4ay$ I could simply say
$$x=2t^2 y=frac{t^4}{a}$$
Indeed, it satisfies the equation - however this would not 'capture' the parabola for $x<0$. We need to be aware of such things when we choose a parameterisation.
answered Dec 26 at 8:41
Hugh Entwistle
788217
add a comment |
In terms of 'intuition' - generally the only thing you need to be careful of this technique is that you may not obtain the whole ellipse. But upon checking the domain of $theta$. This particular parameterisation that you have used is okay!
A choice of parameterisation is generally from the knowledge that it will satisfy the cartesian equation and that it is sufficiently variable.
For example if I wanted to find a parameterisation for the parabola $x^2=4ay$ I could simply say
$$x=2t^2 y=frac{t^4}{a}$$
Indeed, it satisfies the equation - however this would not 'capture' the parabola for $x<0$. We need to be aware of such things when we choose a parameterisation.
answered Dec 26 at 8:41
Hugh Entwistle
788217
add a comment |
In terms of 'intuition' - generally the only thing you need to be careful of this technique is that you may not obtain the whole ellipse. But upon checking the domain of $theta$. This particular parameterisation that you have used is okay!
A choice of parameterisation is generally from the knowledge that it will satisfy the cartesian equation and that it is sufficiently variable.
For example if I wanted to find a parameterisation for the parabola $x^2=4ay$ I could simply say
$$x=2t^2 y=frac{t^4}{a}$$
Indeed, it satisfies the equation - however this would not 'capture' the parabola for $x<0$. We need to be aware of such things when we choose a parameterisation.
answered Dec 26 at 8:41
Hugh Entwistle
788217
In terms of 'intuition' - generally the only thing you need to be careful of this technique is that you may not obtain the whole ellipse. But upon checking the domain of $theta$. This particular parameterisation that you have used is okay!
A choice of parameterisation is generally from the knowledge that it will satisfy the cartesian equation and that it is sufficiently variable.
For example if I wanted to find a parameterisation for the parabola $x^2=4ay$ I could simply say
$$x=2t^2 y=frac{t^4}{a}$$
Indeed, it satisfies the equation - however this would not 'capture' the parabola for $x<0$. We need to be aware of such things when we choose a parameterisation.
answered Dec 26 at 8:41
Hugh Entwistle
788217
answered Dec 26 at 8:41
Hugh Entwistle
788217
answered Dec 26 at 8:41
Hugh Entwistle
788217
answered Dec 26 at 8:41
Hugh Entwistle
788217
788217
add a comment |
add a comment |
Well, you can make the substitution in the equation, so you get:
$$(2cos(theta))^2+16(frac 1 2 sin(theta))^2=4.$$
Therefore your parametrization is ok.
answered Dec 26 at 8:27
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Well, you can make the substitution in the equation, so you get:
$$(2cos(theta))^2+16(frac 1 2 sin(theta))^2=4.$$
Therefore your parametrization is ok.
answered Dec 26 at 8:27
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Well, you can make the substitution in the equation, so you get:
$$(2cos(theta))^2+16(frac 1 2 sin(theta))^2=4.$$
Therefore your parametrization is ok.
answered Dec 26 at 8:27
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Well, you can make the substitution in the equation, so you get:
$$(2cos(theta))^2+16(frac 1 2 sin(theta))^2=4.$$
Therefore your parametrization is ok.
answered Dec 26 at 8:27
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 26 at 8:27
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 26 at 8:27
Alonso Quijano
445
answered Dec 26 at 8:27
Alonso Quijano
445
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
In the case of an ellipse, the simplest solution is to make a change of variables, in order to get the equation of a circle.
For the equation
$$ x^2 +16y^2 = 4 $$
Just consider the change in variables:
$$X = x; quad Y = 4y $$
To get the new equation
$$X^2 + Y^2 = 4 $$
Therefore the equation of a circle of center $(0,0)$ and ray $R=2$.
The corresponding parametric equation is
$$X=Rcos thetaquad Y=Rsin theta $$
For $theta$ varying from $0$ to $2 pi$
Which corresponds to
$$x=2cos thetaquad y=frac{1}{2}sin theta $$
answered Dec 26 at 14:59
Damien
4474
That's easy! Thanks.
– Mathsaddict
Dec 26 at 16:24
add a comment |
In the case of an ellipse, the simplest solution is to make a change of variables, in order to get the equation of a circle.
For the equation
$$ x^2 +16y^2 = 4 $$
Just consider the change in variables:
$$X = x; quad Y = 4y $$
To get the new equation
$$X^2 + Y^2 = 4 $$
Therefore the equation of a circle of center $(0,0)$ and ray $R=2$.
The corresponding parametric equation is
$$X=Rcos thetaquad Y=Rsin theta $$
For $theta$ varying from $0$ to $2 pi$
Which corresponds to
$$x=2cos thetaquad y=frac{1}{2}sin theta $$
answered Dec 26 at 14:59
Damien
4474
That's easy! Thanks.
– Mathsaddict
Dec 26 at 16:24
add a comment |
In the case of an ellipse, the simplest solution is to make a change of variables, in order to get the equation of a circle.
For the equation
$$ x^2 +16y^2 = 4 $$
Just consider the change in variables:
$$X = x; quad Y = 4y $$
To get the new equation
$$X^2 + Y^2 = 4 $$
Therefore the equation of a circle of center $(0,0)$ and ray $R=2$.
The corresponding parametric equation is
$$X=Rcos thetaquad Y=Rsin theta $$
For $theta$ varying from $0$ to $2 pi$
Which corresponds to
$$x=2cos thetaquad y=frac{1}{2}sin theta $$
answered Dec 26 at 14:59
Damien
4474
In the case of an ellipse, the simplest solution is to make a change of variables, in order to get the equation of a circle.
For the equation
$$ x^2 +16y^2 = 4 $$
Just consider the change in variables:
$$X = x; quad Y = 4y $$
To get the new equation
$$X^2 + Y^2 = 4 $$
Therefore the equation of a circle of center $(0,0)$ and ray $R=2$.
The corresponding parametric equation is
$$X=Rcos thetaquad Y=Rsin theta $$
For $theta$ varying from $0$ to $2 pi$
Which corresponds to
$$x=2cos thetaquad y=frac{1}{2}sin theta $$
answered Dec 26 at 14:59
Damien
4474
answered Dec 26 at 14:59
Damien
4474
answered Dec 26 at 14:59
Damien
4474
answered Dec 26 at 14:59
Damien
4474
4474
That's easy! Thanks.
– Mathsaddict
Dec 26 at 16:24
add a comment |
That's easy! Thanks.
– Mathsaddict
Dec 26 at 16:24
That's easy! Thanks.
– Mathsaddict
Dec 26 at 16:24
That's easy! Thanks.
– Mathsaddict
Dec 26 at 16:24
add a comment |
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2
You can verify this by looking at the graph. Your parametric form is the same as the initial one, assuming the range of $t$ is from $0$ to $2pi$
– roman
Dec 26 at 8:16
@roman is there any other way to verify this than graph?
– Mathsaddict
Dec 26 at 8:21
@roman Just looking at a graph usually isn't enough to rigourously prove that two things are the same.
– Arthur
Dec 26 at 8:21
@Arthur I know, but that helps to gain some intuition
– roman
Dec 26 at 8:22
@Mathsaddict That's parametric form, not polar.
– Arthur
Dec 26 at 8:22