Transformations of Hyperbolas
Principles of translation
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$12x−3+7 |
| $6$6 | $=$= | $\frac{12}{x-3}$12x−3 |
| $6\left(x-3\right)$6(x−3) | $=$= | $12$12 |
| $6x-18$6x−18 | $=$= | $12$12 |
| $6x$6x | $=$= | $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.
Worked Examples
QUESTION 1
This is a graph of the hyperbola $y=\frac{1}{x}$y=1x.
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.
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.
AMove the graph upwards by $3$3 unit(s).
BMove the graph downwards by $3$3 unit(s).
CMove the graph $3$3 units to the right.
DHence plot $y=\frac{1}{x}+3$y=1x+3 on the same graph as $y=\frac{1}{x}$y=1x.
Loading Graph...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
A2
B4
C3
DConsider $y=\frac{-1}{x-4}$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
A1
B3
C2
DHow 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
Areflected about $x$x-axis
Bsteepened
Ctranslated $4$4 units down
D