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.

A hyperbola of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$`x`2`a`2−`y`2`b`2=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`)2`a`2−(`y`−`k`)2`b`2=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$=$=$\left(-\infty,\infty\right)$(−∞,∞)

A hyperbola of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$`y`2`a`2−`x`2`b`2=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`)2`a`2−(`x`−`h`)2`b`2=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`,∞).

Summary

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

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

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

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

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

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

A graph of the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$`x`225−`y`216=1 is shown below.

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State the domain of the hyperbola.

State the range of the hyperbola.

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.

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State the domain of the hyperbola.

State the range of the hyperbola.

Apply the geometry of conic sections

Apply the geometry of conic sections in solving problems