Given an ellipse centered on the origin in an x-y plane expressed as

$$\bigg(\frac{x}{a} \bigg)^2+\bigg(\frac{y}{b} \bigg)^2 = 1$$

In polar coordinates with radius $R$ and angle = $\theta$, this can be expressed as:

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

If we set a and b to be related as follows:

$a = 1 - \Delta$

$b = 1 + \Delta$

What is the simplest solution for $\Delta$ as a function of $R$ and $\theta$ as determined from the following relationship?

$$R = \sqrt{1 - 2\Delta \cos (2\theta)+ \Delta^2}$$

  • $\begingroup$ Solve it as a quadratic in $\Delta$? $\endgroup$ Dec 11, 2018 at 14:13
  • $\begingroup$ Did you try generalized polar coordinates $x=ra\cos \theta, y=rb\sin \theta?$ $\endgroup$
    – user376343
    Dec 11, 2018 at 14:35

1 Answer 1


$\Delta^2- 2\Delta \cos(2\theta)+1-R^2=0\\\implies\displaystyle \Delta=\cos2\theta\pm\sqrt{\cos^22\theta+R^2-1}=\cos2\theta\pm\sqrt{R^2-\sin^22\theta}$

  • $\begingroup$ sorry that I changed my question on you!! If you have a solution to my expanded question I would be very appreciative! $\endgroup$ Dec 11, 2018 at 14:26
  • 1
    $\begingroup$ @DanBoschen It's regarded poorly here to change the underlying question altogether after an answer has been posted. $\endgroup$ Dec 11, 2018 at 14:28
  • $\begingroup$ @Travis I am so sorry- I will revert it and then repost my expanded question separately $\endgroup$ Dec 11, 2018 at 14:29
  • $\begingroup$ I guess that was a steal... $\endgroup$ Dec 11, 2018 at 14:31
  • 1
    $\begingroup$ @DanBoschen No worries, it's a perhaps nonobvious cultural thing to keep the site maximally usable (and to avoid annoying helpful answerers like Shubham). Welcome to math.se, by the way! $\endgroup$ Dec 11, 2018 at 15:21

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