NZ Level 8 (NZC) Level 3 (NCEA) [In development] Domain and range of hyperbolas
Lesson

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 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2y2b2=1 has a horizontal transverse axis and is centred at the origin. A graph of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$x2a2y2b2=1

We can see that the hyperbola has two branches; one on the right which corresponds to $x\ge a$xa, and one on the left which corresponds to $x\le-a$xa. 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 $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2(yk)2b2=1, which still has a horizontal transverse axis but is no longer centred at the origin. A graph of the hyperbola $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2(yk)2b2=1

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 $x\ge h+a$xh+a, while the branch on the left corresponds to $x\le h-a$xha. Values of $x$x that are closer to the centre than $a$a units (that is, values of $x$x between $h-a$ha and $h+a$h+a) do not correspond to any part of the hyperbola. So we have that:

Domain$=$=$\left(-\infty,h-a\right]\cup\left[h+a,\infty\right)$(,ha][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 $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$y2a2x2b2=1 has a vertical transverse axis instead. A graph of the hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$y2a2x2b2=1

The two branches of this hyperbola are separated vertically, where the top branch corresponds to $y\ge a$ya and the bottom one corresponds to $y\le-a$ya. 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 $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(yk)2a2(xh)2b2=1, which has been translated away from the origin. A graph of the hyperbola $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(yk)2a2(xh)2b2=1

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 $y\ge k+a$yk+a, while the bottom branch corresponds to $y\le k-a$yka. Values of $y$y that are closer to the centre than $a$a units (that is, values of $y$y between $k-a$ka 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$=$=$\left(-\infty,k-a\right]\cup\left[k+a,\infty\right)$(,ka][k+a,).

Summary

A hyperbola of the form $\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2(yk)2b2=1, which has a horizontal transverse axis, has:

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

Range$=$=$\left(-\infty,\infty\right)$(,).

A hyperbola of the form $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(yk)2a2(xh)2b2=1, which has a vertical transverse axis, has:

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

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

#### Practice questions

##### Question 1

A graph of the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$x225y216=1 is shown below.

1. State the domain of the hyperbola.

2. State the range of the hyperbola.

##### Question 2

A graph of the hyperbola $\frac{\left(y+1\right)^2}{16}-\frac{\left(x+3\right)^2}{9}=1$(y+1)216(x+3)29=1 is shown below.

1. State the domain of the hyperbola.

2. State the range of the hyperbola.

### Outcomes

#### M8-1

Apply the geometry of conic sections

#### 91573

Apply the geometry of conic sections in solving problems