Domain and range of hyperbolas

Recall that the domain of a relation is the set of $x$x-values in the relation. Graphically, we can think of the domain as all values of $x$x which correspond to one or more points in the relation.

Correspondingly, the range of a relation is the set of $y$y-values in the relation. Graphically, we can think of the range as all values of $y$y which correspond to one or more points in the relation.

It is often easiest to determine the domain and range of relations by looking at their graphs.

 

Hyperbolas with a horizontal transverse axis

A hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2​−y2b2​=1 has a horizontal transverse axis and is centred at the origin.

We can see that the hyperbola has two branches; one on the right which corresponds to $x\ge a$x≥a, and one on the left which corresponds to $x\le-a$x≤−a. Values of $x$x between $-a$−a and $a$a do not correspond to any part of the hyperbola. So we have that:

Domain$=$=$\left(-\infty,-a\right]\cup\left[a,\infty\right)$(−∞,−a]∪[a,∞)

Looking at the other axis, we can see that each branch of the hyperbola covers all real values of $y$y. That is:

Range$=$=$\left(-\infty,\infty\right)$(−∞,∞)

 

Let's now look at a graph of $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​−(y−k)2b2​=1, which still has a horizontal transverse axis but is no longer centred at the origin.

This hyperbola is centred at the point $\left(h,k\right)$(h,k), with semi-major axis length $a$a. This means that the branch on the right corresponds to $x\ge h+a$x≥h+a, while the branch on the left corresponds to $x\le h-a$x≤h−a. Values of $x$x that are closer to the centre than $a$a units (that is, values of $x$x between $h-a$h−a and $h+a$h+a) do not correspond to any part of the hyperbola. So we have that:

Domain$=$=$\left(-\infty,h-a\right]\cup\left[h+a,\infty\right)$(−∞,h−a]∪[h+a,∞)

Once again, the branches of the hyperbola still cover all real values of $y$y. So, as before, we have that:

Range$=$=$\left(-\infty,\infty\right)$(−∞,∞)

 

Hyperbolas with a vertical transverse axis

A hyperbola of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$y2a2​−x2b2​=1 has a vertical transverse axis instead.

The two branches of this hyperbola are separated vertically, where the top branch corresponds to $y\ge a$y≥a and the bottom one corresponds to $y\le-a$y≤−a. So for this orientation of hyperbola, it is the range that has two parts. On the other axis, we can see that all real values of $x$x correspond to both branches of the hyperbola. So we have that

Domain$=$=$\left(-\infty,\infty\right)$(−∞,∞),

and that

Range$=$=$\left(-\infty,-a\right]\cup\left[a,\infty\right)$(−∞,−a]∪[a,∞).

 

Here is a graph of $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(y−k)2a2​−(x−h)2b2​=1, which has been translated away from the origin.

This hyperbola is centred at the point $\left(h,k\right)$(h,k), with semi-major axis length $a$a. This means that the top branch corresponds to $y\ge k+a$y≥k+a, while the bottom branch corresponds to $y\le k-a$y≤k−a. Values of $y$y that are closer to the centre than $a$a units (that is, values of $y$y between $k-a$k−a and $k+a$k+a) do not correspond to any part of the hyperbola. Meanwhile, all real values of $x$x still correspond to points on the hyperbola. So we have that

Domain$=$=$\left(-\infty,\infty\right)$(−∞,∞),

and that 

Range$=$=$\left(-\infty,k-a\right]\cup\left[k+a,\infty\right)$(−∞,k−a]∪[k+a,∞).

 

 

Practice questions

Question 1

Question 2