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I have a unit circle centered at the origin defined parametrically as $ x = \cos \ t,\ y = \sin \ t $ and an ellipse centered at the origin defined parametrically as $ x = a\cdot \cos \ t^\prime,\ y = b\cdot \sin \ t^\prime $ I need to find a formula to convert from $t$ to $t^\prime$.

I want to be able to draw an arc on the circle given $t_1$ and $t_2$ and draw the corresponding arc (using $t^\prime_1$ and $t^\prime_2$) on the ellipse so that I can draw a straight line from the origin that touches an endpoint of the arc on the circle and the corresponding endpoint on the ellipse.

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    You could use two-argument arctangent for the purpose, I suppose...2012-11-22

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Let

$g(r) := \begin{pmatrix} 0 \\ 0 \end{pmatrix} + r \cdot \begin{pmatrix} \cos t \\ \sin t \end{pmatrix}$

the straight line connecting the origine and the pont $(\cos t,\sin t)$ on the unit circle. There exists exactly one intersecting point of this line and the ellipse, i.e.

$g(r) = \begin{pmatrix} a \cdot \cos x \\ b \cdot \sin x \end{pmatrix}$

for some $r \in \mathbb{R}$, $x \in [0,2\pi]$. Hence by definition of $g$:

$r \cdot \cos t = a \cdot \cos x, r \cdot \sin t = b \cdot \sin x \\ \Rightarrow r=a \cdot \frac{\cos x}{\cos t} = b \cdot \frac{\sin x}{\sin t} \\ \Rightarrow a \cdot \tan t = b \cdot \tan x \\ \Rightarrow x = \arctan \left( \frac{a}{b} \cdot \tan t \right)$

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    If $t=\frac{\pi}{2}$, then $\cos t =0$, thus $\cos x \stackrel{!}{=}0$ and therefore $x= \frac{\pi}{2}$.2012-11-23