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Given the ellipse $(x-1)^2 + \frac{y^2}{4}= 1$, parametrize the curve in polar coordinates.

I've forgotten something very basic here. Can someone help get me started?

Martín-Blas Pérez Pinilla
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Alec
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  • $x=r\cos \theta$, $y=r\sin \theta$? – Henry Nov 13 '14 at 09:31
  • @Henry, this is the parametrization for a circle of radius r. Isn't it ? – creative Nov 13 '14 at 09:40
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    @Abstraction Take $r_1$ and $r_2$ instead and you'll get an ellipse. – Andrei Rykhalski Nov 13 '14 at 09:42
  • @Andrei Rykhalski, thats true !! Thats what I was wondering – creative Nov 13 '14 at 09:46
  • $(x_0+a\cos\phi,y_0+b\sin\phi)$ is what you would do, and what a physicist would just call polar... but it's not strictly speaking the $(r,\phi)$ parametrization that is usually called "polar". To express $r(\phi)$ you need something quite ugly. It only becomes elegant, if the coordinate origin is in the focus. Then you have the famous polar formula with eccentricity. – orion Nov 13 '14 at 10:04

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HINT:

$$(x-1)^2 + \frac{y^2}{4}= 1$$

$$(x-1)^2 + (\frac{y}{2})^2= 1 \tag 1$$

$$x=r\cos \theta \tag 2$$

$$y=r\sin \theta \tag 3$$

Put in Equation 1

$$(x-1)^2 + (\frac{y}{2})^2= 1$$

$$(r\cos \theta-1)^2 + (\frac{r\sin \theta}{2})^2= 1$$

Find r as function of $\theta$

Mathlover
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