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Is the graph of the following inverse relation a hyperbola?$$y=\frac kx$$

If yes, is it the only kind of hyperbola whose equation is an explicit function?

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  • $\begingroup$ You mean: an expmicit function? $\endgroup$ – zoli Dec 26 '17 at 7:32
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Yes, $$y=\frac kx$$ is a hyperbola. In fact, it is a type of rectangular hyperbola which you can read more about here.

This means that a function $y(x)$ that takes the form $$y-k=\frac {k}{x-h}$$ is also hyperbola.

I suppose that is the only type of hyperbolas that is a function since any rotation of this type of hyperbolas will immediately cause the plot to fail the vertical line test.

Thanks to @Blue:

Any (non-degenerate) hyperbola with a vertical asymptote is the graph of a function. Rectangularity is not a requirement.

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    $\begingroup$ Any (non-degenerate) hyperbola with a vertical asymptote is the graph of a function. Rectangularity is not a requirement. $\endgroup$ – Blue Dec 26 '17 at 7:59
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    $\begingroup$ @Blue oh yes you are right, my bad. $\endgroup$ – Karn Watcharasupat Dec 26 '17 at 8:01
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It is a hyperbola, and you can even derive it from the original formula of a hyperbola using analytic geometry.

Let $a$ and $b$ be the axes of the $y=1/x$ hyperbola, while $x$ and $y$ are the axes of the $y^2-x^2=k$ hyperbola. If $a=(x+y)/2$ and $b=(y-x)/2$, then graphing the original $y^2-x^2=k$ formula by replacing y with b and x with a respectively we get the exact same hyperbola as $y=k/x$.

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