# ellipse polar co-ordinate conversion

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.

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 $y = b \sin t$.

• Thanks for your response. The square roots was a typo on my part. I see what you are saying, so it isn't actually just a polar co-ordinate conversion as I suspected, but rather first a change to (as you call it) the $\bar{xy}$ plane, then a polar co-ordinate conversion. Is this a trivial question that I asked? – user116403 Feb 21 '16 at 20:21
• I wouldn't say that the question was trivial, but the point is that we use the parametrization of the circle of find the parametrization of the ellipse (or at least, we're motivated by it). I just see this change of coordinates as the best way to make it more rigorous. – Ivo Terek Feb 21 '16 at 20:27

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.

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.