Is the graph of $y=\frac{k}{x}$ a hyperbola?

Question

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?

You mean: an expmicit function? – zoli Dec 26 '17 at 07:32 — zoli, Dec 26 '17 at 07:32

Karn Watcharasupat · Accepted Answer · 2017-12-26T08:02:32.757

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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.

edited Dec 26 '17 at 08:02

answered Dec 26 '17 at 07:34

Karn Watcharasupat

4,323

1

Any (non-degenerate) hyperbola with a vertical asymptote is the graph of a function. Rectangularity is not a requirement. – Blue Dec 26 '17 at 07:59
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@Blue oh yes you are right, my bad. – Karn Watcharasupat Dec 26 '17 at 08:01

score 2 · Answer 2 · edited Dec 29 '18 at 21:23

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 $y=k/x$.

Is the graph of $y=\frac{k}{x}$ a hyperbola?

2 Answers2