ellipse polar co-ordinate conversion

Question

I have a somewhat trivial question out of interest. Given the equation of an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ why is the substitution $x = \sqrt{a}\cos t$ and $y = \sqrt{b}\sin t$ valid?

For the unit circle $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ it is clear graphically why the polar co-ordinate conversion $x = r \cos t$ and $y = r \sin t$ is valid since the radius $r$ is fixed, but with an ellipse, this is not the case. Why is it taken trivially that this substitution is valid?

Thanks for any assistance.

Answer 1

You don't actually have these square roots there. The line of thought is like this: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \iff \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1,$$so making a change of coordinates $\overline{x} = x/a$ and $\overline{y} = y/b$, we have that this equation reads $\overline{x}^2+\overline{y}^2 = 1$. That ellipse in the $xy$ plane is a circle in the $\overline{x}\overline{y}$ plane, with the axes stretched. We can parametrize it as ${\overline{x}} = \cos t$, $\overline{y}= \sin t$. Going back to the $xy$ coordinates we obtain $x = a \cos t$ and $y = b \sin t$.

Answer 2

Note that

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

are not polar coordinates for ellipse:

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

The polar coordinate equation of central ellipse is

$$ \left(\frac{ \cos t }{a}\right)^2 + \left(\frac{\sin t }{b}\right)^2 = \frac{1}{r^2} $$

And for a circle radius r

$$\frac{x^2}{r^2}+\frac{y^2}{r^2}=1 $$

is valid.

Answer 3

For the ellipse $$\left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1 $$

the expression $[x,y] = [a \cos t,b \sin t]$ is a parametrization that solves the equation of the ellipse. If you plug in $x$ and $y$ you get

$$\left( \frac{a \cos t}{a} \right)^2 + \left( \frac{b \sin t}{b} \right)^2 = \cos^2 t + \sin^2 t = 1 $$

This parametrization does not represent the polar coordinates. To get the polar coordinates $r$ and $\theta$ you need $[x,y] = [r \cos \theta, r \sin \theta]$ which leads to the solution

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

If you plot the above for any constants $a$ and $b$ you will get an ellipse with semi-major and semi-minor radii of $a$ and $b$ respectively.