The hyperbola given by $y=\frac{a}{x-h}+k$y=ax−h+k can be thought of as the basic rectangular hyperbola $y=\frac{a}{x}$y=ax translated horizontally (parallel to the $x$x axis) a distance of $h$h units and translated vertically (parallel to the $y$y axis) a distance of $k$k units. We note that under this type of translation:

The centre will move to the point $\left(h,k\right)$(h,k).

The orientation of the hyperbola will remain unaltered.

The asymptotes will become the straight lines $x=h$x=h and $y=k$y=k.

For example, to sketch the hyperbola $y=\frac{12}{x-3}+7$y=12x−3+7, first place the centre at $\left(3,7\right)$(3,7). Then draw in the two orthogonal asymptotes (orthogonal means at right angles) given by $x=3$x=3 and $y=7$y=7. Finally, draw the hyperbola as if it were the basic hyperbola $y=\frac{12}{x}$y=12x but now centred on the point $\left(3,7\right)$(3,7).

Note that the domain includes all values of $x$x not equal to $3$3 and the range includes all values of $y$y not equal to $7$7. Here is a graph showing how the basic function $y=\frac{12}{x}$y=12x is translated to horizontally and vertically to become the transformed function $y=\frac{12}{x-3}+7$y=12x−3+7.

We can also note that as $x\rightarrow3,y\rightarrow\infty$x→3,y→∞and as $x\rightarrow\infty,y\rightarrow7$x→∞,y→7.

Finding points

Continuing with our example, to find the point where $x=9$x=9, we simply substitute $x=9$x=9 into $y=\frac{12}{x-3}+7$y=12x−3+7 so that $y=\frac{12}{9-3}+7$y=129−3+7 or when simplified $y=9$y=9. Thus the point $\left(9,9\right)$(9,9) lies on the hyperbola.

To find the point where $y=13$y=13, set $13=\frac{12}{x-3}+7$13=12x−3+7 and solve for $x$x, so that:

$13$13	$=$=	$\frac{12}{x-3}+7$12`x`−3+7
$6$6	$=$=	$\frac{12}{x-3}$12`x`−3
$6\left(x-3\right)$6(`x`−3)	$=$=	$12$12
$6x-18$6`x`−18	$=$=	$12$12
$6x$6`x`	$=$=	$30$30
$x$`x`	$=$=	$5$5

Thus another point on the hyperbola is $\left(5,13\right)$(5,13).

Applet

Use the applet to understand how the translation of the basic function works. Try positive and negative values of $a$a.

Created with Geogebra

Worked Examples

QUESTION 1

This is a graph of the hyperbola $y=\frac{1}{x}$y=1x.

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What would be the new equation if the graph of $y=\frac{1}{x}$y=1x was shifted upwards by $4$4 units?
What would be the new equation if the graph of $y=\frac{1}{x}$y=1x was shifted to the right by $7$7 units?

QUESTION 2

This is a graph of $y=\frac{1}{x}$y=1x.

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A cartesian plane is shown with both the x-axis and y-axis ranging from -10 to 10. A hyperbola is plotted with a function $y=\frac{1}{x}$y=1x.

How do we shift the graph of $y=\frac{1}{x}$y=1x to get the graph of $y=\frac{1}{x}+3$y=1x+3?
Move the graph $3$3 units to the left.
A
Move the graph upwards by $3$3 unit(s).
B
Move the graph downwards by $3$3 unit(s).
C
Move the graph $3$3 units to the right.
D
Hence sketch $y=\frac{1}{x}+3$y=1x+3 on the same graph as $y=\frac{1}{x}$y=1x.

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A cartesian plane is shown with both the x-axis and y-axis ranging from -10 to 10. A hyperbola is drawn with a function $y = 1/x$ .

QUESTION 3

Answer the following.

Consider $y=\frac{-1}{x}$y=−1x. What value cannot be substituted for $x$x?
In which two quadrants does the graph of $y=\frac{-1}{x}$y=−1x lie?
$1$
A
$2$
B
$4$
C
$3$
D
Consider $y=\frac{-1}{x-4}$y=−1x−4. What value cannot be substituted for $x$x?
In which three quadrants does the graph of $y=\frac{-1}{x-4}$y=−1x−4 lie?
$4$
A
$1$
B
$3$
C
$2$
D
How can the graph of $y=\frac{-1}{x}$y=−1x be altered to create the graph of $y=\frac{-1}{x-4}$y=−1x−4?
translated $4$4 units $right$
A
reflected about $x$x-axis
B
steepened
C
translated $4$4 units $down$
D

Outcomes

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.

Transformations of Hyperbolas

Principles of translation