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Basically, I want to know the length of the radius of an ellipse, based on the angle this radius makes with either of the 2 main radius.

Is that possible to do?

Shadowxvii
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  • What is the radius of an ellipse? Where are you measuring the angle? – Ross Millikan Oct 01 '13 at 20:42
  • $\sqrt{(a\cos \varphi)^2+(b\sin \varphi)^2}$. – njguliyev Oct 01 '13 at 20:43
  • What do you mean by the radius of an ellipse - a circle has a radius, an ellipse has a major axis and a minor axis - or are you looking for an equation in polar co-ordinates? – Mark Bennet Oct 01 '13 at 20:44
  • I want to know the lenght of a line going from the center of my ellipse to a point on the ellipse. The problem is that I don't know the coordinates of the point, only the angle that line makes with the major and minor axis. – Shadowxvii Oct 01 '13 at 20:50
  • The angle alone does not help. You need the values of the "main radii" as well. – user Feb 28 '20 at 19:26

2 Answers2

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The equation of an ellipse is $(\frac xa)^2+(\frac yb)^2=1$. If you know the angle $\theta$ from the $x$ axis, you have $y= x \tan (\theta)$. Now substitute in to get $x^2(\frac 1{a^2}+\frac {\tan^2 \theta}{b^2})=1$ This gives you $x$, then you can find $y$, then $r=\sqrt {x^2+y^2}$

Ross Millikan
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  • Thanks for the answer. Juste aside, if we add the same amount to both axis, will the radius change by the same amount for every points on the ellipse? As in, if a' = a + c, b' = b + c, will r' = r + c where a,b,c are from the initial ellipse, a',b',c' from the modified ellipse and c is the amount we add to each axis. – Shadowxvii Oct 04 '13 at 15:52
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    No, the ellipses are not parallel in that way. On the $45^\circ$ line, we have $x=y$. For the original ellipse, then $x^2(\frac 1{a^2} +\frac 1{b^2})=1$, so $x=\frac ab{\sqrt a^2+b^2}$. If you substitute $a'=a+c,b'=b+c$, you won't get $x'=x+\frac c{\sqrt 2}$ as would be required. – Ross Millikan Oct 04 '13 at 15:57
  • Ok, good to know. Therefore, is there a way to know r'-r from c ? – Shadowxvii Oct 04 '13 at 17:03
  • You can substitute into the equation and determine it. I don't know if it will be anything clean. – Ross Millikan Oct 04 '13 at 17:09
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$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=r^2$$ $$x=acos(t),y=bsin(t)$$ $$r=\sqrt{x^2+y^2}=\sqrt{a^2cos^2(t)+b^2sin^2(t)}\text{, where "t" is the angle from }0\text{ to }2\pi.$$ If the angle from the minor axis is $\phi$ and the angle from the major axis is $\psi$, then $\phi=\frac{\pi}{2}-t$ and $\psi=\pi-t$. $$r=\sqrt{a^2cos^2(\pi-\psi)+b^2sin^2(\frac{\pi}{2}-\phi)}$$

iamreptar
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