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Consider an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ such that $a>b$ and its parametrization

$$\vec{r}(\phi)=a\cos\phi\ \hat{x}+b\sin\phi\ \hat{y}$$

where $\phi\in[0,2\pi)$. Also consider a focus at $\vec{f}=c\ \hat{x}$, where $c^2=a^2-b^2$. The vector from the focus to a point on the ellipse is given by

$$\vec{d}=\vec{r}-c\hat{x} = (a\cos\phi\ -c)\hat{x}+b\sin\phi\ \hat{y}$$

ellipse

It follows from the dot product that,

\begin{align} d^2 &= a^2\cos^2\phi + b^2\sin^2\phi +c^2 -2ca\cos\phi \\ &= r^2 +c^2 -2ca\cos\phi \end{align}

But in the triangle formed by the vectors, the Law of Cosines gives:

$$d^2 = r^2 +c^2 -2cr\cos\phi$$

This seems to imply that $r=a$, but that cannot always be true. Where has this argument gone astray?

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2 Answers 2

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The problem is that the $\phi$ angle in the first equation is not the same as the one later. Let's call that $t$ instead. Then, according to https://en.wikipedia.org/wiki/Ellipse#Parametric_form_in_canonical_position, you get $\tan t=\frac{a}{b}\tan \phi$

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  • $\begingroup$ Interesting. That's what I feared... It seems counter-intuitive, though. $\endgroup$
    – zahbaz
    Commented Jan 23, 2017 at 3:06
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    $\begingroup$ But expected. If it would be the same angle, the first equation should be $|r|\cos t \hat{x}+|r|\sin t \hat{y}$ $\endgroup$
    – Andrei
    Commented Jan 23, 2017 at 3:22
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    $\begingroup$ @zahbaz To improve your intuition: take a circle, the angle $\pi/4$ and then an ellipse, for which longer axis is , say, thousand times longer then the shorter one. $\endgroup$
    – Przemysław Scherwentke
    Commented Jan 23, 2017 at 3:24
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    $\begingroup$ @zahbaz There is also an animation in the link. If we use $t$ for the "angle" in the parametrization, that is $\vec{r}(t)=a\cos t\ \hat{x}+b\sin t\ \hat{y}$, then $t$ is the angle from the $\hat{x}$ direction to the direction of the (rotating) blue half-line. See that the running point tracing the ellipse is usually not on that blue line. If you scaled by $a$ both horizontally and vertically, it would work (large circle), or by $b$ in both directions (small circle). But if you stretch one dimension more, the angle is changed. $\endgroup$
    – Jeppe Stig Nielsen
    Commented Jan 23, 2017 at 7:06
  • $\begingroup$ Thank you for your recent comment. $\endgroup$
    – Sebastiano
    Commented Oct 10, 2020 at 20:14
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It is well-known, that the angle from the parametrization is not the angle between the running point and half-axis $x\geq 0$. If you assume the equality, you will obtain a circle, as in your example.

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