3.01 Hyperbolas

Directly versus inversely proportional

Two variables are directly proportional if one can be expressed as a constant multiple of the other. That is, given $y$y is directly proportional to $x$x, which can be written as $y\propto x$y∝x, we know that $y=kx$y=kx, for some non-zero constant $k$k. This constant, unsurprisingly, is called the constant of proportionality. This is a linear relationship with a gradient of $k$k and a $y$y-intercept of $0$0. We saw some examples of this when looking at linear functions.  Now we are going to look an inverse proportion.

Two variables are inversely proportional if one is proportional to the inverse of the other. That is, given $y$y is inversely proportional to $x$x, which can be written as $y\propto\frac{1}{x}$y∝1x​, we know that $y=\frac{k}{x}$y=kx​, for some non-zero constant $k$k. This gives us two features we can look for in determining if a relationship is inversely proportional. Firstly, as one variable increases, the other must decrease. Secondly, rearranging the equation we can see that $x\times y$x×y will always give us the constant of proportionality $k$k. 

 

Worked example

The table below shows the travel time to a destination $240$240 km away given a speed.

Speed ($x$x, km/h)	$40$40	$60$60	$80$80	$120$120
Travel time ($y$y, h)	$6$6	$4$4	$3$3	$2$2

Are the variables $x$x and $y$y inversely proportional? If so, find an equation relating the two variables.

Think: As one variable increases, does the other decrease? Does $x\times y$x×y always give the same answer?

Do: As the speed increases does the travel time decrease? Yes.

For each pair $\left(x,y\right)$(x,y) in the table find $x\times y$x×y. Here we have $40\times6=60\times4=80\times3=120\times2=240$40×6=60×4=80×3=120×2=240. So yes, all pairs give us the constant of proportionality $k=240$k=240, in this situation this represents the distance to the destination.

We have found the two variables are inversely proportional and in doing so, we also found the constant of proportionality. Hence, the equation relating the two variables is $y=\frac{240}{x}$y=240x​.  

Reflect: What happens as $x$x gets very big? Can the travel time ever be zero? What are the limitations of this relationship? What does the graph of this relationship look like?

 

Practice question

Question 1

 

Rectangular hyperbola

When two variables are inversely proportional, the graph we obtain is called a rectangular hyperbola. Have you seen this shape before? Sketch the graph of $y=\frac{1}{x}$y=1x​ in the following question to see the shape and look more closely at key features of the graph.

 

Practice question

Question 2

The graph has two smooth sections of curves, one in the first quadrant and one in the third quadrant. The graph is symmetrical about the lines $y=x$y=x and $y=-x$y=−x. It also has point symmetry about the origin by $180^\circ$180°, that is if we spin the graph $180^\circ$180° about the point $\left(0,0\right)$(0,0) the graph would remain unchanged.

The graph is undefined at $x=0$x=0. If we substitute $x=0$x=0 into the equation, we get $y=\frac{1}{0}$y=10​ which is undefined. This also makes sense if we rearrange the equation: $xy=1$xy=1, if $x=0$x=0 there is no value of $y$y that could give us $1$1. This also tells us that the graph is undefined at $y=0$y=0 as well. Hence, both the domain and range of this function are $\left(-\infty,0\right)\cup\left(0,\infty\right)$(−∞,0)∪(0,∞).

What happens to the graph near the lines $x=0$x=0 and $y=0$y=0? The graph gets closer and closer to these lines but never touches them. These lines are called asymptotes and the 'rectangular' part of the graph's name is because these asymptotes meet at right angles.

We have a vertical asymptote of $x=0$x=0 and a horizontal asymptote of $y=0$y=0. And we have four asymptotic behaviours we can describe:

• As $x$x approaches infinity, $y$y approaches zero from above ($y$y becomes a smaller and smaller positive number)
• As $x$x approaches negative infinity, $y$y approaches zero from below ($y$y becomes a smaller and smaller negative number)
• As $x$x approaches zero from above, $y$y approaches infinity
• As $x$x approaches zero from below, $y$y approaches negative infinity

 

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 3

QUESTION 4

QUESTION 5