I have searched through math.stackexchange for related posts, but failed to either connect or transfer/map to my following mind-experiment:
Suppose we are given two lines described by: $3x+4y=2$ and further $-3x-2y=2$. They both intersect at $(-2,2)$. How can I compute the hyperbola using these two lines as the hyperbolas asymptotes?
A simple first naive approach with $$\frac{(x-x_0)^2}{a^2}-\frac{(y-y_0)^2}{b^2}=1$$ did not yield the expected result. It seems like the rotation is missing.
Thank you in advance for any hints and With best regards
The equation of a hyperbola whose asymptotes have equations $3x+4y-2=0$ and $-3x-2y-2=0$ can be written as: $$ (3x+4y-2)(-3x-2y-2)=k, $$ where $k$ is a constant. For every value of $k$ you'll get a different hyperbola. Given any point $P$ in the plane, you can choose $k$ such that the hyperbola passes through $P$.