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I want to transform cartesian space to polar space to draw an ellipse.

For circle the polar space is $(r,\theta)$ and $(x,y)$ being cartesian space. How do we represent polar space for an ellipse? $$x=i \cos (t) \cos (\theta)-j \sin(t) \sin (\theta)$$

$$y=i \sin (t) \cos (\theta)+j \cos (t) \sin (\theta) $$

in the formula above is ellipse parametrized by $(i,j,t,\theta)$ as this doesnt look like polar space?

Ng Chung Tak
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user0193
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    It would help if you would write your rectangle equation(s) for an ellipse. Then a matching discussion can follow. It is a different discussion if you start with an algebraic equation $$ \frac{(x-h)^2}{a^2} + \cdots $$ or if you start with a trigonometric equation (see https://en.wikipedia.org/wiki/Ellipse#Polar_forms for various examples). – Eric Towers Sep 29 '20 at 21:44
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    For en ellipse centered at the origin, $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ $$r^2\left(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right)=1$$ ......Can you fill in the rest? – K.defaoite Sep 29 '20 at 22:00
  • @EricTowers, yes but then the rectangle equation for an ellipse will still be in cartesian coordinates. The formulation I wrote above is transformation from cartesian to polar, but what I dont see is how is that polar coordinates because in polar coordinates we only have $(r,\theta)$ but here we have more parameters and no $r$ – user0193 Sep 29 '20 at 22:01
  • @K.defaoite I understood your equations. What is confusing me what is the form of the equations that I wrote then? They are neither cartesian nor polar but if we program them we can draw the ellipse. So what is the form of the equations that I worte? – user0193 Sep 29 '20 at 22:04
  • @JhonnyS Are $i,j$ supposed to represent the Cartesian unit vectors $\hat{\mathbf{i}},\hat{\mathbf{j}}$ in $\mathbb{R}^2$? If so, we still have two parameters, $t,\theta$ and since an ellipse is a one dimensional manifold there's no way we need two parameters. What are the ranges on $t,\theta$? – K.defaoite Sep 29 '20 at 22:07
  • @K.defaoite $i,j$ define major-minor axis values of an ellipse (if they are the same then it shrinks to circular equation) and not Cartesian unit vectors, appologies for the confusion. Parameter $t$ likewise is used if ellipse is tilted . The range for both $t$, $\theta$ is $(0,2\pi)$. – user0193 Sep 29 '20 at 22:18

1 Answers1

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Note that $\theta$ is not the polar angle $\phi$.

Since $i,j$ can be regarded as unit vectors, quaternion and so on, use $a,b$ instead.

As $t$ varies,

$$x^2+y^2=a^2\cos^2 \theta+b^2\sin^2 \theta \tag{1a}$$

which is a circle of with radius $$r(\phi)=\sqrt{a^2\cos^2 \theta+b^2\sin^2 \theta} \tag{1b}$$

enter image description here

As $\theta$ varies,

$$\frac{(x\cos t+y\sin t)^2}{a^2}+\frac{(x\sin t-y\cos t)^2}{b^2}=1 \tag{2a}$$

which is a rotated ellipse with radius $$r(\phi)=\frac{ab}{\sqrt{a^2\sin^2 (\phi-t)+b^2\cos^2(\phi-t)}} \tag{2b}$$

For both $t$ and $\theta$ vary, we get annular region $b \le r \le a$.

enter image description here

Ng Chung Tak
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  • Also the equations I wrote, what kind of equations are those? Are they still considered cartesian to polar coordinates equation just like the case of a circle? $$x=r cos \theta$$, $$y=r sin \theta$$ – user0193 Sep 29 '20 at 22:28
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    No, $$(x,y)=(r\cos \phi,r\sin \phi)$$ instead. – Ng Chung Tak Sep 29 '20 at 22:29
  • I am bit confused, you essentially wrote what I wrote in a compact form. so for an ellipse if I write $$(x,y)=(a \cos t \cos \phi - b \sin t \sin \phi, a \sin t \cos \phi + b \cos t \sin \phi)$$ then what form will it be called? Note that the equations I wrote can be programmed to draw an ellipse, I just dont know what form is it polar or some other form? certainly not cartesian form. – user0193 Sep 29 '20 at 22:34
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    No, your $\theta$ is one of the parameters only, not the polar angle $$\phi=\tan^{-1} \frac{y}{x}$$ By the way, some texts using $(x,y,z)=(\rho \cos \phi, \rho \sin \phi, z)$ for cylindrical coordinates and $(x,y,z)=r(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$ in spherical coordinates. – Ng Chung Tak Sep 29 '20 at 22:37
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    Equations $(1a)$ and $(2a)$ are called implicit equations (for families of conics) in Cartesian coordinates. Equations $(1b)$ and $(2b)$ are called polar equations. – Ng Chung Tak Sep 29 '20 at 22:51
  • I have one tiny question, you mentioned that $\theta \neq \phi$ but isnt the parameter $\theta$ also defined as $\tan^{-1} \frac{y}{x}$ whats the difference between $\theta$ being parameter and $\phi$ being polar angle (which i though also means a parameter) – user0193 Sep 30 '20 at 12:40
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    Notation is for denotation. To be or not to be. We should make use of notations with care and wisedom. Different resources have different contexts, don't blindly apply individual convention frome one to another. The reason for me in using $\phi$ to represent polar angle is avoiding ambiguity, because $\theta$ has used in the post and, in this case, is nothing related to the polar anlge. – Ng Chung Tak Oct 01 '20 at 05:06
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    Think about an ellipse with an implicit equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we can use $\theta$ (or $\phi$, $t$ whatsoever) as for $(x,y)=(a\cos \theta,b\sin \theta)$, the curves can be plotted as $\theta$ varying (but with specific $a$, $b$ values). An ellipse is distored, the eccentric anlge $\theta$ is not a polar angle unless $a=b$. Compare with a standard circle, $x^2+y^2=c^2$, we can write it as polar equation namely $r=c$ or parametric equation $(c\cos \theta,c\sin \theta)$ and in this case $\theta$ is a polar angle. – Ng Chung Tak Oct 01 '20 at 05:08
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    Alternatively we may have rational parametrization $$c\left( \frac{1-t^2}{1+t^2},\frac{2t}{1+t^2} \right)$$ or parametrization by arc length $$c\left(\cos \frac{s}{c},\sin \frac{s}{c} \right)$$ For instance, a polar equation $$r=2a\cos 2\theta$$ which is a circle with $(x-a)^2+y^2=a^2$ – Ng Chung Tak Oct 01 '20 at 05:15
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    See another answers of mine about convention and symbol usage in post 1, post 2 & post 3. – Ng Chung Tak Oct 01 '20 at 05:25
  • This is massive help! Thanks a lot professor! – user0193 Oct 01 '20 at 13:04