Solve simultaneous problems involving an ellipse

Question

I am trying to follow a derivation in a paper. The reason it is of importance is because I want to make a slight change to one of the equations. But in order to do that, I need to understand how they were able to arrive at the original final result. Here are the three equations:

$$\frac{hr}{R^2} = \frac{q^2}{R^2}\sin\beta$$

$$\left(\frac{q}{R}sin\phi\right)^2 + \left(\frac{q}{R}cos\phi - \frac{b}{R}\right)^2sin^2\beta = 1$$

q is a variable length and $\phi$ is a variable angle. R is a constant. The angle, $\beta$, is a constant in this equation as well.

The third equation is:

$$\frac{b}{R} = \cot\beta$$

So since the angle $\beta$ is a constant that means that $\frac{b}{R}$ is also a constant.

The authors claim that they combined these three equations to come up with the final result which is,

$$\frac{hr}{R^2} = \frac{\sin^3\beta}{(1-\cos\phi\cos\beta)^2}$$

Right now Eq. 2 is what is giving me problems. I can't see anyway of solving it for $\frac{q^2}{R^2}$ so that I can plug the result into the first equation. If I multiply out the quadratic in Eqn 2, it leaves me with a term that includes $\frac{2qb}{R^2}$. Because of that term, I can't see how to ultimately solve for $\frac{q^2}{R^2}$.

Edit: Below is the diagram that applies to the 2nd equation. Again, the angle $\beta$ is a constant for this scenario even though it is not shown in the drawing. O is the origin and P is a focus of the ellipse.

Answer 1

The straightforward way continuing how you were with that $\frac{2qb}{R^2}$ term. No cleverness needed.

$$ \frac{hr}{R^2} = \frac{q^2}{R^2} \sin \beta\\ (\frac{q}{R} \sin \phi )^2 + (\frac{q}{R} \cos \phi - \frac{b}{R})^2 \sin^2 \beta = 1\\ \frac{b}{R} = \cot \beta\\ \frac{q^2}{R^2} \sin^2 \phi + \bigg( \frac{q^2}{R^2} \cos^2 \phi + \frac{b^2}{R^2} - 2 \frac{q b \cos \phi}{R^2} \bigg) \sin^2 \beta = 1\\ \frac{q^2}{R^2} (\sin^2 \phi + \cos^2 \phi \sin^2 \beta) + \cot^2 \beta \sin^2 \beta - 2 \frac{q \cos \phi}{R} \cot \beta \sin^2 \beta = 1\\ \frac{q^2}{R^2} (\sin^2 \phi + \cos^2 \phi \sin^2 \beta) + \cos^2 \beta - 2 \frac{q \cos \phi}{R} \cos \beta \sin \beta = 1\\ \frac{q^2}{R^2} (\sin^2 \phi + \cos^2 \phi \sin^2 \beta) - \sin^2 \beta - 2 \frac{q \cos \phi}{R} \cos \beta \sin \beta = 0\\ $$

$$ A = \sin^2 \phi + \cos^2 \phi \sin^2 \beta\\ B = -2 \cos \phi \cos \beta \sin \beta\\ C = - \sin^2 \beta\\ \frac{q}{R} = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\\ B^2 - 4 A C = 4 \cos^2 \phi \cos^2 \beta \sin^2 \beta + 4 \sin^2 \beta \sin^2 \phi + 4 \sin^2 \beta \cos^2 \phi \sin^2 \beta\\ = 4 \sin^2 \beta \bigg( \cos^2 \phi \cos^2 \beta + \sin^2 \phi + \cos^2 \phi \sin^2 \beta \bigg)\\ = 4 \sin^2 \beta \bigg( \cos^2 \phi (\cos^2 \beta + \sin^2 \beta) + \sin^2 \phi \bigg)\\ = 4 \sin^2 \beta\\ \frac{q}{R} = \frac{2 \cos \phi \cos \beta \sin \beta \pm 2 \sin \beta}{2 \sin^2 \phi + 2 \cos^2 \phi \sin^2 \beta} $$

Mathematica can handle this as but we could do by hand too.

Also note that we pick the positive sign because otherwise we would get a negative $\frac{q}{R}$.

The point is that the expression we have for $\frac{q}{R}$ is now only in $\phi$ and $\beta$ so plugging into $\frac{q^2}{R^2} \sin \beta$ will retain that and it will be a matter of trig simplification.