Implicit form of an ellipse - Geometrical interpretation

The general equation of an ellipse is:

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

where $0<b \leq a$. It turns out that:

$$\begin{cases} x = a \cos(\alpha) \\y = b \sin(\alpha)\end{cases}, $$

where $\alpha \in [0, 2\pi]$.

Similarly, it turns out that:

$$\begin{cases} x = r(\theta) \cos(\theta) \\y = r(\theta) \sin(\theta)\end{cases}, $$

where $\theta \in [0, 2\pi]$ and

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

The two formulations are so similar, but at the same time, they are different.

What is the geometrical difference of the angles $\alpha$ and $\theta$?

Solution 1:

The difference is as follows: $\theta$ is the polar angle of a generic point on the ellipse, and $r(\theta)$ the corresponding value of $r$ (which clearly changes with $\theta$ unless the ellipse is a circle). On the other hand, $\alpha$ has the following meaning: draw a vertical line through a point on the ellipse and consider the intersection of that line with the circle of radius $a$ concentric with the ellipse (this circle is tangent to the ellipse at $(a,0)$ and $(-a,0)$). Then, $\alpha$ is the polar angle of that point of intersection. The ellipse is just a shrunk version of the circle of radius $a$ with scaling factor $b/a$. If you start with the circle $x=a\cos\alpha$ $y=a\sin\alpha$ and you multiply $y$ by the scaling factor you get the equation of the ellipse, but of course $\alpha$ keeps its original meaning, the one I explained above. Hope this helps.