ellipse, unit circle, angles

**rainer** · November 28th 2009, 12:19 PM

Please see attached image.

Knowing only theta and x, find theta_2 and theta_3

(the green curve is an ellipse inscribed within the reddish unit circle)

**Opalg** · November 30th 2009, 12:50 AM

Spoiler:

**simplependulum** · November 30th 2009, 01:30 AM

In general , we have

$\tan(\theta_n) = (\frac{1}{b})^n \tan(\theta_0)$

$x_n = (\frac{1}{b})^{n+1} \cdot \frac{ 1 - b }{ \sqrt{ 1 + (\frac{1}{b})^{2n} \tan^2(\theta_0) }}$

Once the value of $b$ is comfirmed ,

we can find out the angle . ( if we keep drawing the lines similarly , a sequence is formed ! )

**rainer** · December 1st 2009, 09:21 AM

These are very unexpected, interesting responses (to me), which is precisely what I hoped to get by posting this problem.

Opalg, I'm hesitant to say you're wrong, but your answer doesn't work when I graph it. Either it's wrong or I'm making some mistake when I graph it.

Simplependulum, I'm still trying to wrap my head around your response. I had never thought of expressing things in that way.

Here is my solution (obviously not the only one):

First, use the given info-- $x, \sin{\theta}$ --to find the semi-minor axis of the ellipse, $b$ :

$x=\sin{\theta}-b\sin{\theta}$

so, $b=\frac{\sin{\theta}-x}{\sin{\theta}}$

Then, the equations for $\theta_2$ and $\theta_3$ are, respectively:

$\tan{\theta_2}=\sqrt{b^2(\sec^2{\theta}-1)}$

and

$\tan{\theta_3}=\arccos{\frac{b}{\sqrt{\tan^2{\thet a}+b^2}}}$

You derive these equations by fooling around with (x, y) intercepts and pythagorus' theorem. I won't go into the details unless you really want.

Happily, these equations turn out to be reducible to:

$\tan{\theta_2}=b\tan{\theta}$

and

$\tan{\theta_3}=\frac{\tan{\theta}}{b}$

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