I've been scratching my head over this for the last few days...
I've been doing some maths relating to orbital mechanics and elliptical orbits, and there's one simple thing that I can't get right. I can't figure out how to calculate the angle $\alpha$ between the edge of the ellipse and the two foci.
Let me illustrate with the case of an isosceles triangle:
First of all, it is immediately obvious that $\alpha = 180-2\theta$. However if I calculate it in a different and more general way, I get a different result.
The length between the foci, $A$ is known and is equal to $2ae$ where $a$ is the semi-major axis and $e$ is the eccentricity, and the angle $\theta$ can be arbitrarily set. In this case it is chosen so that it forms an isosceles triangle for the sake of this question. $B$ is unknown but will be calculated, and $\alpha$ is unknown.
The cosine rule: $$ C^2 = A^2 + B^2 -2ABcos(\theta)$$ side $B = C $ so substitute $C$ for $B$ $$B^2 = 4a^2e^2 + B^2 - 4aeBcos(\theta)$$ $$0 = 4a^2e^2-4aeBcos(\theta)$$ $$Bcos(\theta) = ae$$ $$B = \frac{ae}{cos(\theta)} $$ The sine rule can then be used to find $\alpha$: $$ \frac{sin(\alpha)}{A} = \frac{sin(\theta)}{B} $$ $$ \frac{sin(\alpha)}{2ae} = \frac{cos(\theta)sin(\theta)}{ae} $$ $$ sin(\alpha) = 2sin(\theta)cos(\theta) = sin(2\theta)$$ $$ \alpha = sin^{-1}(sin(2\theta)) $$ $$ \alpha = 2\theta $$
Now, for almost all values of $\theta$, $\alpha$ quite clearly cannot both be equal to $2\theta$ and $180-2\theta$. But for the life of me I cannot figure out where my mistake is.