Graphs of Hyperbolas
Theory
The graph of \(y = \dfrac{k}{x}\) is a hyperbola: two separate branches that approach the axes, its asymptotes, without touching them. This Year 10A maths page covers which quadrants a hyperbola lies in, the transformed form \(y = \dfrac{k}{x - b} + c\) with its asymptotes and intercepts, finding the equation, and where lines meet hyperbolas.
A hyperbola of the form \(y = \dfrac{k}{x}\) (or \(xy = k\)), with \(k \ne 0\), has two branches. If \(k \gt 0\) they lie in the 1st and 3rd quadrants; if \(k \lt 0\), in the 2nd and 4th.
\(x = 0\) is not allowed (division by zero), and \(\dfrac{k}{x}\) is never \(0\). So the graph never meets either axis. The lines \(x = 0\) and \(y = 0\) are its asymptotes: as \(x\) gets very large the curve gets closer to \(y = 0\), and as \(x\) gets close to \(0\) the curve shoots up or down beside \(x = 0\).
\(y = \dfrac{k}{x - b} + c\) is the same shape moved \(b\) units right and \(c\) units up. Its asymptotes are \(x = b\) and \(y = c\), and these replace the axes when you decide where the branches go.
A hyperbola describes inverse variation: as one quantity doubles, the other halves, such as time and speed for a fixed distance.
The basic and transformed hyperbola:
| Feature | How to find it |
|---|---|
| vertical asymptote | set the denominator equal to \(0\) |
| horizontal asymptote | \(y = c\), because the fraction is never \(0\) |
| \(y\)-intercept | substitute \(x = 0\) (none if \(b = 0\)) |
| \(x\)-intercept | let \(y = 0\) and solve (none if \(c = 0\)) |
| quadrants / position | sign of \(k\), measured from the asymptotes |
How to sketch a hyperbola
- Asymptotes: draw \(x = b\) and \(y = c\) as dashed lines.
- Branches: \(k \gt 0\) gives top-right and bottom-left of the asymptotes; \(k \lt 0\) gives top-left and bottom-right.
- Intercepts: substitute \(x = 0\), and solve \(y = 0\).
- Sketch each branch through its points, bending towards both asymptotes.
Problem types
- Finding \(k\): substitute a known point. For \(y = \dfrac{k}{x}\) through \((5,\ 3)\), \(k = 15\).
- From asymptotes and a point: asymptotes give \(b\) and \(c\); one point gives \(k\) (Example 2). From a graph, read the asymptotes off the dashed lines and use a marked point.
- From two points: substitute both to get two equations. For \(y = \dfrac{k}{x} + c\), subtract to remove \(c\). For \(y = \dfrac{k}{x - b}\), write \(k = y(x - b)\) for each point and set the two expressions equal.
- Meeting a line: substitute, multiply by the denominator, and solve the quadratic. Two, one or no real solutions mean two points, a tangent, or no meeting (a negative number under the square root in the quadratic formula). Answers may involve surds.
- Meeting horizontal or vertical lines: \(y = d\) gives \(x = \dfrac{k}{d}\); \(x = d\) gives \(y = \dfrac{k}{d}\).
- Transformations: describe \(x \to x - b\) as \(b\) right and \(+c\) as \(c\) up; move the asymptotes with the graph.
- Sharing and rates: a fixed total divided among \(n\) (cost per person, time for a distance) gives \(\dfrac{T}{n}\). Write the difference of two such fractions, multiply by both denominators, and solve the quadratic (Example 4).
This is \(y = \dfrac{3}{x}\) moved \(1\) right and \(2\) up. Asymptotes: \(x = 1\) (the denominator is zero) and \(y = 2\) (the fraction is never zero). Since \(k = 3 \gt 0\), the branches sit top-right and bottom-left of the asymptotes.
| \(y\text{-intercept: } y\) | \(=\) | \(\dfrac{3}{0 - 1} + 2 = -1\) |
| \(x\text{-intercept: } \dfrac{3}{x - 1}\) | \(=\) | \(-2\) |
| \(x - 1\) | \(=\) | \(-\dfrac{3}{2}\) |
| \(x\) | \(=\) | \(-\dfrac{1}{2}\) |
The asymptotes give \(b = -2\) and \(c = 3\), so \(y = \dfrac{k}{x + 2} + 3\). Substitute \((0,\ 1)\):
| \(1\) | \(=\) | \(\dfrac{k}{2} + 3\) |
| \(\dfrac{k}{2}\) | \(=\) | \(-2\) |
| \(k\) | \(=\) | \(-4\) |
So \(y = \boldsymbol{-\dfrac{4}{x + 2} + 3}\). For the \(x\)-intercept, let \(y = 0\):
| \(\dfrac{4}{x + 2}\) | \(=\) | \(3\) |
| \(x + 2\) | \(=\) | \(\dfrac{4}{3}\) |
| \(x\) | \(=\) | \(\mathbf{-\dfrac{2}{3}}\) |
Since \(k \lt 0\), the branches are top-left and bottom-right of the asymptotes.
(i) Find where the line \(y = x + 1\) meets the hyperbola \(y = \dfrac{2}{x}\).
(ii) Show that the line \(y = -2x + 4\) meets it at only one point.
(i) Equate and multiply both sides by \(x\) (\(x \ne 0\)):
| \(x + 1\) | \(=\) | \(\dfrac{2}{x}\) |
| \(x^2 + x - 2\) | \(=\) | \(0\) |
| \((x + 2)(x - 1)\) | \(=\) | \(0\) |
\(x = 1\) gives \(y = 2\); \(x = -2\) gives \(y = -1\). Points \(\mathbf{(1,\ 2)}\) and \(\mathbf{(-2,\ -1)}\).
(ii)
| \(-2x + 4\) | \(=\) | \(\dfrac{2}{x}\) |
| \(-2x^2 + 4x - 2\) | \(=\) | \(0\) |
| \(x^2 - 2x + 1\) | \(=\) | \(0\) |
| \((x - 1)^2\) | \(=\) | \(0\) |
The only solution is \(x = 1\), so the line meets the hyperbola only at \((1,\ 2)\): it is a tangent there.
The time at the faster speed is \(\dfrac{240}{v + 20}\), and it is \(1\) hour less:
| \(\dfrac{240}{v} - \dfrac{240}{v + 20}\) | \(=\) | \(1\) |
| \(240(v + 20) - 240v\) | \(=\) | \(v(v + 20)\) |
| \(4800\) | \(=\) | \(v^2 + 20v\) |
| \(v^2 + 20v - 4800\) | \(=\) | \(0\) |
| \((v + 80)(v - 60)\) | \(=\) | \(0\) |
A speed must be positive, so \(v = \mathbf{60}\) km/h, and \(t = \dfrac{240}{60} = \mathbf{4}\) hours.
Check: at \(80\) km/h the trip takes \(3\) hours, \(1\) hour less. As \(v\) grows, \(t\) shrinks towards \(0\) but never reaches it: the graph of \(t\) against \(v\) is one branch of a hyperbola.
Common pitfalls
Frequently asked questions
What is a hyperbola in maths?
In Year 10, a hyperbola is the graph of y = k over x (or a transformed version). It has two separate branches that approach two asymptotes without ever touching them.
How do you find the asymptotes of a hyperbola?
For y = k over (x - b) plus c, the vertical asymptote is x = b, where the denominator is zero, and the horizontal asymptote is y = c, because the fraction can never equal zero.
Which quadrants is y = k/x in?
If k is positive, the branches are in the first and third quadrants. If k is negative, they are in the second and fourth quadrants.
How do you find where a line meets a hyperbola?
Set the two expressions for y equal, multiply both sides by the denominator, and solve the resulting quadratic. Substitute each solution into the line to find the y-values.
Does a hyperbola have intercepts?
y = k over x has none, because it never meets the axes. A transformed hyperbola usually crosses both axes; find them by substituting x = 0 and solving y = 0.