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India
Class XI

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.

Loading Graph...

  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.

Loading Graph...

  1. State the domain of the hyperbola.

  2. State the range of the hyperbola.

Outcomes

11.SF.RF.2

Definition of relation, pictorial diagrams, domain, co-domain and range of a relation. Function as a special kind of relation from one set to another. Pictorial representation of a function, domain, co-domain and range of a function. Real valued function of the real variable, domain and range of these functions, constant, identity, polynomial, rational, modulus, signum and greatest integer functions with their graphs. Sum, difference, product and quotients of functions.

11.CG.CS.1

Sections of a cone: Circles, ellipse, parabola, hyperbola, a point, a straight line and pair of intersecting lines as a degenerated case of a conic section. Standard equations and simple properties of parabola, ellipse and hyperbola. Standard equation of a circle.

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