Ellipse in polar coordinates

I think Wikipedia's polar coordinate elliptical equation isn't correct. Here is my explanation: Imagine constants $a$ and $b$ in this format -

Where $2a$ is the total height of the ellipse and $2b$ being the total width. You can then find the radial length, $r$, at any angle $\theta$ to major axis as...

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

...by just following the Pythagorean theorem. Yet Wikipedia's equation for the polar coordinate ellipse is as follows:

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

Here is the link to the Wikipedia page: Can someone explain this, please? Why divide by the hypotenuse? Why the $ab$? Thank you!

It's easiest to start with the equation for the ellipse in rectangular coordinates:

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

Then substitute $x = r(\theta)\cos\theta$ and $y = r(\theta)\sin\theta$ and solve for $r(\theta)$.

That will give you the equation you found on Wikipedia.

Polar Equation from the Center of the Ellipse

The equation of an ellipse is $$ \left(\frac{x}{a}\right)^2+\left(\frac{y}{a\sqrt{1-e^2}}\right)^2=1\tag1 $$ Using $x=r\cos(\theta)$ and $y=r\sin(\theta)$ in $(1)$, we get $$ r^2\cos^2(\theta)+\frac{r^2\sin^2(\theta)}{1-e^2}=a^2\tag2 $$ and we can solve $(2)$ for $r^2$ to get the polar equation $$ r^2=\frac{\overbrace{a^2\!\left(1-e^2\right)}^{b^2}}{1-e^2\cos^2(\theta)}\tag3 $$

Polar Equation from a Focus of the Ellipse

Centered at the right focus $$ \left(\frac{x+ae}a\right)^2+\left(\frac{y}{a\sqrt{1-e^2}}\right)^2=1\tag4 $$ Using $x=r\cos(\theta)$ and $y=r\sin(\theta)$ in $(4)$, we get $$ r^2\cos^2(\theta)+2aer\cos(\theta)+a^2e^2+\frac{r^2\sin^2(\theta)}{1-e^2}=a^2\tag5 $$ which gives the quadratic equation in $r$: $$ \frac{r^2\left(1-e^2\cos^2(\theta)\right)}{1-e^2}+2aer\cos(\theta)-a^2\!\left(1-e^2\right)=0\tag6 $$ whose solution is $$ r=\frac{a\!\left(1-e^2\right)}{1+e\cos(\theta)}\tag7 $$

Ellipse parametrization is done differently. To more clearly distinguish between them we should note there are two different $\theta$ s. And two different $r\,$s.

The two $r(\theta) $s you mentioned represent different things.

The first angle is used for polar coordinates and is measured from center of the circles with radii $(a,b)$. Call it $ \theta_{polar} $

For an ellipse axes $ (a,b)$ along $(x,y) $ coordinate axes respectively centered at origin given Wiki expression is obtained in polar coordinates thus:

Plug in

$$x=r_{polar}\cos \theta_{polar},\, y=r_{polar}\sin \theta_{polar}$$

to cast the standard equation of an ellipse from the Cartesian form : $$ \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 =1 $$

to get

$$ OE =r_{polar}= \frac{ab}{\sqrt{(b \cos \theta_{polar})^2 + (a \sin \theta_{polar})^2}} \tag1 $$

Next let us denote second angle by $\theta_{ecc}$ and radius by $ r_{ecc}.$

The radius vector in this way represented ( by La Hire originally ) is $r_{ecc}$ depends on $\theta_{ecc}=\theta_{La\, Hire};$ point $E$ is obtained by drawing vertical and horizontal lines from points of intersection of polar/radial line with the two circles radii $(a,b).$

We have

$$ x= a \cos \theta_{ecc},\,y= b \sin \theta_{ecc},\, $$

$$ OE=r_{ecc} = \sqrt{x^2+y^2} =\sqrt{a\cos\theta_{ecc})^2 + (b \sin \theta_ {ecc})^2} \tag2 $$ and

$$ \theta_{ecc} = \tan ^{-1} \dfrac{b \sin \theta_{ecc}}{ a \cos \theta_{ecc}} \tag3 $$

In either case $ \theta = 0 $ or $ \theta= \pi/2$ the same points are reached at ends of major, minor axes respectively. These are plotted, so tha they show that La Hire polar measures less than the Central polar always.

Ellipse dimensions are $( a=5,b=3,e=0.8)$ for this sketch.

Using the same symbols $ (r, \theta) $ to connote different representations may lead to confusion.

You're making the common mistake of using the polar coordinate instead of the eccentric anomaly which is the parameter in the ellipse coordinates.