Equation for Angle Around an Ellipse given semi-major (a), semi-minor (b), and length of radius (r)?

Question

What is the equation to find the angle $\theta$ around an ellipse if given only horizontal semi-axis $a$, vertical semi-axis $b$, and just a length for radius $r$? I've also attached an image that graphically shows the problem.

I'm stuck trying to rearrange and isolate $\theta$ for solving using the equation:

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

Thank you for any help.

Answer 1

$$ r^2(a^2\sin^2\theta+b^2\cos^2\theta)=a^2b^2;$$ $$ r^2(a^2\sin^2\theta+b^2(1-\sin^2\theta))=a^2b^2;$$ Let $\hat a = a/r$ and $\hat b=b/r$ $$ \hat a ^2\sin^2\theta+\hat b^2(1-\sin^2\theta)=\hat a^2 \hat b^2;$$ $$ (\hat a ^2-\hat b^2)\sin^2\theta+\hat b^2=\hat a^2 \hat b^2;$$ $$ (\hat a ^2-\hat b^2)\sin^2\theta=(\hat a^2-1) \hat b^2;$$ $$ \sin^2\theta=\frac{(\hat a-1)(\hat a+1) \hat b^2}{(\hat a+\hat b)(\hat a-\hat b)}$$

Answer 2

Polar equation of the ellipse is $$r=\frac{a b}{\sqrt{a^2 \sin ^2(t)+b^2 \cos ^2(t)}}$$ which can be reduced to $$r =\frac{\sqrt{2} a b}{\sqrt{a^2+b^2+(b^2 -a^2)\cos (2 t)}}$$ Solving wrt $t$ we get $$\cos(2t)=\frac{a^2 r^2+b^2 r^2-2 a^2 b^2}{r^2 \left(a^2-b^2\right)}$$ and then $$t = \frac{1}{2}\,\arccos\left(\frac{a^2 r^2+b^2 r^2-2 a^2 b^2}{r^2 \left(a^2-b^2\right)}\right)$$

The ellipse has four points with the same $r$-value. This formula only gives the angle of the point in the first quadrant, with both coordinates positive.

Answer 3

Let $F_1,F_2$ be the foci of the ellipse, $|OA|=|OC|=|BF_1|=|BF_2|=a$, $|OB|=b$, $\angle DOA=\theta$, $|OD|=r$, $|DF_2|=u$, $|DF_1|=v$.

We know that \begin{align} |OF_1|=|OF_1|&=\sqrt{a^2-b^2} \tag{1}\label{1} ,\\ u+v&=2a \tag{2}\label{2} . \end{align}

By the Stewart’s Theorem

\begin{align} \triangle F_1F_2D:\quad |DF_1|^2\cdot|OF_2|+ |DF_2|^2\cdot|OF_1| &=|F_1F_2|\cdot(|OD|^2+|OF_1|\cdot|OF_2|) \tag{3}\label{3} ,\\ (u^2+v^2)\sqrt{a^2-b^2}&=2\sqrt{a^2-b^2}\,(r^2+a^2-b^2) \tag{4}\label{4} ,\\ u^2+v^2&= 2(r^2+a^2-b^2) \tag{5}\label{5} . \end{align}

Combination of \eqref{2} and \eqref{5} gives

\begin{align} uv&=2a^2-\tfrac12(u^2+v^2) \tag{6}\label{6} ,\\ uv&=a^2+b^2-r^2 \tag{7}\label{7} . \end{align}

A pair of equations \eqref{2}, \eqref{7} results in

\begin{align} u&=a-\sqrt{r^2-b^2} \tag{8}\label{8} ,\\ v&=a+\sqrt{r^2-b^2} \tag{9}\label{9} , \end{align} since $u<v$ by construction.

Given that, we have all the side lengths of $\triangle OF_2D$ in terms of known values $a,b,r$, and the angle $\theta$ now can be found by the cosine rule:

\begin{align} \theta&=\arccos\left(\frac ar\sqrt{\frac{r^2-b^2}{a^2-b^2}}\right) \tag{10}\label{10} . \end{align}

Answer 4

For easy calculation... if reciprocals of $(a,b,r)$, are computed as $(a',b',r')$ then for non-circular ellipses

$$ \frac{\cos2 \theta }{2}= \dfrac { (2 r'^2 - a'^2 - b'^2)}{(a'^2-b'^2)}$$

which solve to four angles.