I am trying to find the points where two hyperbolas intersect; where one hyperbola, $H1$, is a hyperbola centered at $(0,0)$ and lying horizontal: $$\frac{x^2}{a_1^2} - \frac{y^2}{b_1^2} = 1\tag 1$$ and $H2$ is a hyperbola rotated to $\theta$ and centered at $(h, k)$: $$\frac{[(y-k)\sin\theta - (x-h)\cos\theta]^2}{a_2^2} - \frac{[(y-k)\cos\theta - (x-h)\sin\theta]^2}{b_2^2} = 1\tag 2$$
After solving (1) for $y$ and substituting that into (2), it took a ton of manipulations but I was able to eventually get it into the form of:
$$\alpha_4x^4 + \alpha_3x^3 + \alpha_2x^2 + \alpha_1x + \alpha_0 = 0$$
where:
$$\alpha_4 =(A^2-B^2F)$$ $$\alpha_3 = (2AC - 2BDF)$$ $$\alpha_2 = (C^2 - 2AE - D^2F + B^2G)$$ $$\alpha_1 = (2BDG - 2CE)$$ $$\alpha_0 =E^2 + D^2G$$
and
$$\left[ \frac{b_1^2b_2^2\sin^2\theta}{a_1^2} -\frac{b_1^2a_2^2\cos^2\theta}{a_1^2} + b_2^2\cos^2\theta - a_2^2\sin^2\theta \right]\tag A$$
$$\left[2a_2^2\sin\theta\cos\theta - 2b_2^2\sin\theta\cos\theta \right] \tag B$$
$$\left[2a_2^2j\sin\theta - 2b_2^2i\cos\theta \right]\tag C$$
$$\left[2b_2^2i\sin\theta - 2a_2^2j\cos\theta \right]\tag D$$
$$\left[ a_2^2 b_2^2 - b_2^2i^2 - b_1^2a_2^2\cos^2\theta + b_1^2b_2^2\sin^2\theta + a_2^2j^2\right]\tag E$$
$$\frac{b_1^2}{a_1^2}\tag F$$
$$b_1^2\tag G$$
and
$$h\sin\theta - k\cos\theta\tag j$$
$$h\cos\theta-k\sin\theta\tag i$$
where $\theta, a_1, a_2, b_1, b_2, h,$ and $k$ are all known constants.
At this point my brain hurts, and I am awful at trigonometric substitutions, but it seems like there is an opportunity to simplify each of the $\alpha$ coefficients. For example, B looks like it contains the double angle, $\sin(2\theta)$, but it's not obvious to a non-mathematician like me that it helps simplify this beast.
In order to get to the general equation I am seeking I need help simplifying any and all of the $\alpha$ coefficients, if possible.
-Or- if there is a book or link to some site that will give me the equation(s) for the x values of the (up to) four points that would be even better -- I will gladly throw my work in the dumpster. It seems to me this was likely solved in antiquity, but I couldn't find a solution with Google.
Note: I have put all my work to get to the polynomial in LaTeX form but it is so long winded I didn't want to include it. I am happy to do so if it helps or if there is concern I screwed it up.
