Graphs of Hyperbolas
Theory
The graph of \(y = \dfrac{k}{x}\) (or \(xy = k\)) is a hyperbola: two separate branches that approach the axes, its asymptotes, without touching them. This Year 10 maths page covers sketching from a table of values, which quadrants a hyperbola lies in, finding \(k\) from a point, symmetry, and inverse variation in real situations.
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 \(x = 0\).
The size of \(k\) sets how far the branches are from the origin: a larger \(|k|\) moves them further out. Every hyperbola \(y = \dfrac{k}{x}\) is symmetric about the lines \(y = x\) and \(y = -x\): swapping \(x\) and \(y\) in \(xy = k\) gives the same equation.
A hyperbola describes inverse variation: as one quantity doubles, the other halves, such as time and speed for a fixed distance.
The hyperbola and its two forms, and how to find \(k\) from a point \((x_1, y_1)\):
| Feature | \(y = \dfrac{k}{x}\) |
|---|---|
| asymptotes | \(x = 0\) and \(y = 0\) (the axes) |
| \(k \gt 0\) | 1st and 3rd quadrants |
| \(k \lt 0\) | 2nd and 4th quadrants |
| intercepts | none |
| domain and range | \(x \ne 0\), \(y \ne 0\) |
How to sketch \(y = \dfrac{k}{x}\)
- Quadrants: \(k \gt 0\) gives the 1st and 3rd; \(k \lt 0\) gives the 2nd and 4th.
- Table of values: choose positive and negative \(x\)-values, including some close to \(0\) and some large. Never use \(x = 0\).
- Plot the points and draw two smooth branches, each bending towards both axes without touching them. Label the curve and a point on each branch.
Problem types
- Finding \(k\): substitute a known point: \(k = xy\). For \(y = \dfrac{k}{x}\) through \((5,\ 3)\), \(k = 15\).
- Finding a missing coordinate: once \(k\) is known, divide: \(x = \dfrac{k}{y}\) or \(y = \dfrac{k}{x}\).
- Two points with an unknown: if \((a,\ p)\) and \((a + d,\ q)\) are both on the hyperbola, their products are equal: \(ap = (a + d)q\). Solve this linear equation for \(a\), then find \(k\).
- Symmetry: \(y = \dfrac{k}{x}\) has lines of symmetry \(y = x\) and \(y = -x\), and rotational symmetry of order \(2\) about the origin.
- Distances and midpoints: use \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) and \(M = \left(\dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2}\right)\); test whether a point is on the hyperbola by checking \(xy = k\).
- In context: a fixed product (area, distance, volume) gives \(xy = k\). Neither quantity can be \(0\), so the graph never meets the axes; usually only the first-quadrant branch makes sense.
(i) complete a table of values for \(x = \pm 1, \pm 2, \pm 5, \pm 10\)
(ii) sketch the graph and state its asymptotes and the quadrants it lies in.
(i) Divide \(-10\) by each \(x\)-value. A negative \(x\) gives a positive \(y\).
| \(x\) | \(-10\) | \(-5\) | \(-2\) | \(-1\) | \(1\) | \(2\) | \(5\) | \(10\) |
|---|---|---|---|---|---|---|---|---|
| \(y\) | \(1\) | \(2\) | \(5\) | \(10\) | \(-10\) | \(-5\) | \(-2\) | \(-1\) |
(ii) The asymptotes are \(\boldsymbol{x = 0}\) and \(\boldsymbol{y = 0}\). Since \(k = -10 \lt 0\), the branches lie in the 2nd and 4th quadrants.
(i) Find \(k\).
(ii) \(B(5,\ b)\) and \(C(c,\ -2)\) are also on the hyperbola. Find \(b\) and \(c\).
(iii) Find the exact length of \(AB\).
| \(k\) | \(=\) | \(xy = 3 \times 10 = \mathbf{30}\) |
(ii) The equation is \(y = \dfrac{30}{x}\), or \(xy = 30\).
| \(b\) | \(=\) | \(\dfrac{30}{5} = \mathbf{6}\) |
| \(c\) | \(=\) | \(\dfrac{30}{-2} = \mathbf{-15}\) |
(iii) Use the distance formula with \(A(3, 10)\) and \(B(5, 6)\).
| \(AB\) | \(=\) | \(\sqrt{(5 - 3)^2 + (6 - 10)^2}\) |
| \(\) | \(=\) | \(\sqrt{4 + 16} = \sqrt{20} = \boldsymbol{2\sqrt{5}}\) |
(ii) State the lines of symmetry of \(y = -\dfrac{7}{x}\) and its order of rotational symmetry.
(i) Every point on the hyperbola has \(xy = k\), so the two products are equal.
| \(14a\) | \(=\) | \(4(a + 5)\) |
| \(14a\) | \(=\) | \(4a + 20\) |
| \(10a\) | \(=\) | \(20\) |
| \(a\) | \(=\) | \(\mathbf{2}\) |
| \(k\) | \(=\) | \(2 \times 14 = \mathbf{28}\) |
Check with the other point: \((2 + 5) \times 4 = 28\). ✓
(ii) Swapping \(x\) and \(y\) in \(xy = -7\) gives the same equation, and so does replacing \(x\) by \(-y\) and \(y\) by \(-x\). The lines of symmetry are \(\boldsymbol{y = x}\) and \(\boldsymbol{y = -x}\). A half-turn about the origin maps each branch onto the other, so the order of rotational symmetry is \(\mathbf{2}\).
(i) Write \(t\) in terms of \(v\).
(ii) How long does the trip take at \(80\) km/h?
(iii) What speed makes the trip in \(2.5\) hours?
(iv) Explain why the graph of \(t\) against \(v\) never meets the \(v\)-axis.
| \(\text{(i) } t\) | \(=\) | \(\boldsymbol{\dfrac{240}{v}}\) |
| \(\text{(ii) } t\) | \(=\) | \(\dfrac{240}{80} = \mathbf{3}\text{ hours}\) |
| \(\text{(iii) } v\) | \(=\) | \(\dfrac{240}{2.5} = \mathbf{96}\text{ km/h}\) |
(iv) On the \(v\)-axis \(t = 0\), but then \(vt = 0\), not \(240\). However fast the car goes, the trip takes some time, so \(t\) gets closer to \(0\) without reaching it. Only the first-quadrant branch makes sense: speed and time are positive.
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 xy = k). It has two separate branches that approach the axes, its asymptotes, without ever touching them.
What are the asymptotes of y = k/x?
The two axes, x = 0 and y = 0. The curve cannot meet x = 0 because dividing by zero is undefined, and it cannot meet y = 0 because k over x is never 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.
What are the lines of symmetry of y = k/x?
The lines y = x and y = negative x. The graph also has rotational symmetry of order 2 about the origin: a half-turn maps each branch onto the other.
Why does a hyperbola describe inverse variation?
If xy = k, doubling x halves y, and the product of the two quantities always stays the same, like speed and time for a fixed distance.