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

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 7:32

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.

• Any (non-degenerate) hyperbola with a vertical asymptote is the graph of a function. Rectangularity is not a requirement. – Blue Dec 26 '17 at 7:59
• @Blue oh yes you are right, my bad. – Karn Watcharasupat Dec 26 '17 at 8:01

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 $$y=k/x$$.