Parametric equation of ellipse, why is radius not variable?

Question

Taken the equation of an ellipse

$$x^2/a^2 + y^2/b^2 = 1$$

The parametric equation of an ellipse is usually given as

$\begin{array}{c} x = a\cos(t)\\ y = b\sin(t) \end{array}$

Let's rewrite this as the general form (*assuming a "friendly" shape, i.e. only one point for each radial vector at angle T)

$\begin{array}{c} x = r(t)\cos(t)\\ y = r(t)\sin(t) \end{array}$

where $r(t)$ is the radius at angle $t$. This mimic the idea of a sin/cos pair with variable radius (which is, indeed, an ellipse)

So how can $r(t)=a$ and $r(t)=b$ at the same time?

Answer 1

Let's look at the suggested form: $$\left\lbrace \begin{aligned} x &= r(\theta) \cos(\theta) \\ y &= r(\theta) \sin(\theta) \\ \end{aligned} \right.$$ As Achille Hui mentioned in a comment to the question, this can seem paradoxical, because $\theta$ is not the ordinary polar angle here. We all know that the expected form is $$\left\lbrace \begin{aligned} x &= a \cos(\theta) \\ y &= b \sin(\theta) \\ \end{aligned} \right.$$

Note, "seem paradoxical". Whether one finds it paradoxical or not only depends on what one thinks $\theta$ is. The suggested form is written with $\theta$ as the proper polar angle; it is not that in the expected form, because here, $\theta \ne \arctan(y/x)$ (assuming $a \ne b$, and excepting the four points where $x = 0$ or $y = 0$).

In the expected form, $\theta$ is just an angular parameter, eccentric angle, which is related to the actual polar angle $\varphi$ via, $$\theta = \arctan\left(\frac{a y}{b x}\right) = \arctan\left(\frac{a}{b}\tan\varphi\right) \iff \varphi = \arctan\left(\frac{b}{a}\tan\theta\right)$$

In a polar coordinate system, this ellipse is described by $$r(\varphi) = \frac{a b}{\sqrt{(b \cos\varphi)^2 + (a \sin \varphi)^2}} $$

You can always convert from polar to Cartesian coordinates, getting $$\left\lbrace\begin{aligned} x &= \frac{a b}{\sqrt{(b \cos \varphi)^2 + (a \sin \varphi)^2}} \cos\varphi \\ y &= \frac{a b}{\sqrt{(b \cos \varphi)^2 + (a \sin \varphi)^2}} \sin\varphi \\ \end{aligned}\right.$$ where $\varphi$ is the correct polar angle; $\varphi = \arctan(y/x)$.

The difference between polar angle $\varphi$ and eccentric angle $\theta$ is subtle, but important. It is also quite annoying for us non-mathematicians, because we do not always notice the difference, and get bitten by it.

Answer 2

A classical mistake: $t$ is not the polar angle !

From

$$x=a\cos t,\\y=b\sin t\ $$

you draw the parametric polar equation

$$r(t)=\sqrt{a^2\cos^2t+b^2\sin^2t},$$ $$\theta(t)=\arctan\left(\dfrac ba\tan t\right),$$

or after elimination of $t$,

$$r(\theta)=\sqrt{a^2\cos^2\left(\arctan\left(\frac ab\tan\theta\right)\right)+b^2\sin^2\left(\arctan\left(\frac ab\tan\theta\right)\right)}.$$

Ugly, isn't it ?

Answer 3

Points set

$$\left\lbrace \begin{aligned} x &= r(\theta) \cos(\theta) \\ y &= r(\theta) \sin(\theta) \\ \end{aligned} \right.\tag 1$$

is an identity concerning any point for all curves.

$$\left\lbrace \begin{aligned} x &= a \cos(\theta) \\ y &= b \sin(\theta) \\ \end{aligned} \right.\tag 2$$ is parametrization of a particular curve, the ellipse where $ \theta $ is not its polar coordinate but of another point that connects projections of segments between circles of radius $(a,b).$

$$\left\lbrace \begin{aligned} x &= a \cos^2(\theta) \\ y &= b \sin^2(\theta) \\ \end{aligned} \right.\tag 3$$

is polar parametrization of a particular case of straight line.

Identities should not be pulled into particular situations, as the context in which the separate entities are first defined is lost or taken out.

In common parlance ( no insinuation implied ) one would say.. it is not known what is being talking about.

Reading (1) and (3) together throws it out of context:

$$ \frac{y}{x}=\frac{a}{b}\; ! $$

which is not the straight line we started with in (3).

Geometric derivation of a particular case of the straight line in polar parameterized form is mentioned even if elementary. For a line making intercepts $(a,b)$ passing through $P(x,y)$

$$ \frac{x}{OP}=\frac{OP}{a},\quad OP= a \cos \theta $$

$$ \frac{y}{OP}=\frac{OP}{b},\quad OP= b \sin \theta $$

$$ (x,y)= ( a \cos ^2\theta, b \sin ^2\theta). $$