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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

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  • $\begingroup$ there could be two hyperbolas with those asymptotes $\endgroup$ – J. W. Tanner Nov 3 '19 at 18:29
  • $\begingroup$ Actually, there are infinitely-many hyperbolas with a given pair of asymptotes. Consider: $xy=k$, for any $k$, is a hyperbola with the coordinate axes as asymptotes. $\endgroup$ – Blue Nov 3 '19 at 18:35
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    $\begingroup$ It's worth noting that the product of the distances from a point on a hyperbola to the asymptotes is a constant. So, if you know the formula for the distance from a point to a line ... $\endgroup$ – Blue Nov 3 '19 at 18:46
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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 $P$.

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