2.05 Transformations of hyperbolas

Transformations - dilation

Use the following applet to explore the effect that $a$a has on the hyperbola $y=\frac{a}{x}$y=ax​. Adjust the values of $a$a and try to summarise the effect.

Summary:

• $a$a dilates (stretches) the graph by a factor of $a$a from the $x$x-axis
• The larger the magnitude of $a$a the further the graph is from the origin
• The point $\left(1,1\right)$(1,1) will be stretched to the point $\left(1,a\right)$(1,a)
• When $a$a is negative the graph lies in the $2$2nd and $4$4th quadrants. This is a reflection of the graph $y=\frac{\left|a\right|}{x}$y=|a|x​ in the $x$x-axis.
• The graphs still has the same symmetry properties of $y=\frac{1}{x}$y=1x​
• The graph has a vertical asymptote of $x=0$x=0 and a horizontal asymptote of $y=0$y=0

Can you find the coordinates of the 'corner' point which is the closest point to the origin? Hint: It lies on the line $y=x$y=x.

 

Transformations - translation

Use the next applet to explore the effect that $h$h and $k$k have on the hyperbola $y=\frac{a}{x-h}+k$y=ax−h​+k. Adjust the values of $h$h and $k$k and try to summarise the effect.

Summary:

• 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
• The centre (intersection of the asymptotes) will move to the point $\left(h,k\right)$(h,k)
• The orientation of the hyperbola will remain unaltered
• Vertical asymptote is translated to $x=h$x=h
• Horizontal asymptote in translated to $y=k$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 at 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.

 

Domain and range

We have seen that the function $y=\frac{a}{x-h}+k$y=ax−h​+k has asymptotes given by $x=h$x=h and $y=k$y=k. Thus $x=h$x=h is the only point excluded from the domain and $y=k$y=k is the only point excluded from the range.

We usually state this formally as, in the case of the domain, $x:x\in\mathbb{R},x\ne h$x:x∈ℝ,x≠h and in the case of the range, $y:y\in\mathbb{R},y\ne k$y:y∈ℝ,y≠k. Alternatively, we can use interval notation, then the domain can be written as $\left(-\infty,h\right)\cup\left(h,\infty\right)$(−∞,h)∪(h,∞). And the range can be written as $\left(-\infty,k\right)\cup\left(k,\infty\right)$(−∞,k)∪(k,∞).

Rather than thinking of translations we can also see from the equation that the domain and range exclude these values. From the form $y=\frac{a}{x-h}+k$y=ax−h​+k, we can see that $x=h$x=h would cause the denominator to be zero and hence, the expression to be undefined. We can rearrange the equation to either $y=\frac{a}{x-k}+h$y=ax−k​+h or $\left(x-h\right)\left(y-k\right)=a$(x−h)(y−k)=a, to see that $y=k$y=k will also cause the equation to be undefined.

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3