What is the focus of $f(x)=\exp(\frac{1}{\log(x)})?$

Question

Trying hard to find the focus of $f(x)$ but it doesn't fit in great with the standard equation for conics. I tried to algebraically manipulate $f(x)$ but couldn't proceed after getting logarithms in the standard equation. What am I missing? I've been taught to use a rotation matrix to transform the hyperbola to the orientable form, compute the solution and then transform back. I could have made errors in the process, so thanks for the help.

Answer 1

$f(x)=e^{\frac{1}{\log(x)}}$ can be rewritten as $\ln(x)\ln(y)=1$ which is a rectangular hyperbola in log-log space. $f(x)$ is then the image of the hyperbola $xy=1.$ The foci of $xy=1$ are $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2}).$ Transforming the foci of $xy=1$ accordingly, the foci for $\ln(x)\ln(y)=1$ become $(e^{\sqrt{2}},e^\sqrt{2})$ and $(e^{-\sqrt{2}},e^{-\sqrt{2}}).$