How to get the radius of an ellipse at a specific angle by knowing its semi-major and semi-minor axes?

Please take a look at this picture : enter image description here


The polar form of the equation for an ellipse with "horizontal" semi-axis $a$ and "vertical" semi-axis $b$ is

$$r = \frac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}$$

Here, $\theta$ represents the angle measured from the horizontal axis ($30.5^\circ$ in your case), and $r$ is the distance from the center to the point in question (the radius you seek).

  • $\begingroup$ I am having trouble deriving this formula from the Wikipedia page. Do you use the "Polar form relative to focus" or the "General polar form"? I'm trying to do it using the general polar form, but don't see where $cos^2(\theta)$ comes in. $\endgroup$ – TSJ Mar 14 '15 at 6:55
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    $\begingroup$ @TSJ: This is the polar form with the origin at the center. Just start with standard Cartesian equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \qquad\to\qquad x^2b^2+y^2a^2=a^2b^2$$ and make the substitutions $x=r \cos\theta$, $y=r\sin\theta$. $\endgroup$ – Blue Mar 14 '15 at 7:39

So, assuming the major axis to be the $Y$ axis and centre $(0,0)$

the equation of the ellipse is $$\frac{x^2}{(2.23)^2}+\frac{y^2}{(3.05)^2}=1$$

and the equation of the line of $r$ is $\frac yx=\tan(30.5-90)^\circ$

Now, if $(h,k)$ is one the points of intersection $r=\sqrt{h^2+k^2}$


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