The hyperbola

What you will be able to do

• say what a hyperbola is and how it differs from an ellipse;
• read the axes, foci, directrices, eccentricity, latus rectum and asymptotes from the equation;
• work with the conjugate and the rectangular hyperbola;
• use the parametric form.

A bang heard in two places

A bang happens somewhere, and two listeners at known places hear it at slightly different times. The difference in time tells you the difference of the two distances. All the places with that same difference lie on a hyperbola. Radio navigation systems have used this idea to find ships, and some use it still. The hyperbola is the curve that matches an ellipse, with a difference of distances in place of a sum.

The definition

Let $F_1$ and $F_2$ be two points with $F_1F_2 = 2c$. The hyperbola is the set of points $P$ with

In the focus and directrix form, $PF = e \cdot PM$ with $e > 1$. As for the ellipse, the two descriptions give the same curve with $c = ae$.

The standard equation

With the foci at $(\pm c, 0)$ and the centre at the origin, squaring twice gives

Here $a$ and $b$ can be in any order. The curve has two separate branches, one for $x \ge a$ and one for $x \le -a$.

• Vertices: $(\pm a, 0)$.
• Transverse axis: the segment between the vertices, length $2a$, along the $x$-axis. The foci are on it.
• Conjugate axis: the segment from $(0, -b)$ to $(0, b)$, length $2b$.
• Foci: $(\pm ae, 0)$, which is $(\pm c, 0)$ with $c = \sqrt{a^2 + b^2}$.
• Eccentricity: $e = \dfrac{c}{a} = \sqrt{1 + \dfrac{b^2}{a^2}}$, which is larger than $1$.
• Directrices: $x = \pm\dfrac{a}{e}$.
• Latus rectum: $\dfrac{2b^2}{a}$.
• Focal distances: for $x_1 \ge a$ on the right branch, $PF_1 = ex_1 + a$ and $PF_2 = ex_1 - a$. The difference is $2a$.

For an ellipse $c^2 = a^2 - b^2$. For a hyperbola $c^2 = a^2 + b^2$. That one sign change is the main thing to remember.

The conjugate hyperbola swaps the roles of $x$ and $y$: $\dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1$. It opens upwards and downwards. If $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$, the transverse axis is along the $y$-axis, the foci are $(0, \pm ae)$ and the vertices are $(0, \pm a)$. In both cases the number with the plus sign in front tells you which way the branches open.

Asymptotes

As $x$ becomes very large, the branches of $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ come closer and closer to the two lines

These are the asymptotes. To see why, write $y = \pm\dfrac{b}{a}\sqrt{x^2 - a^2}$. For large $x$ the square root is nearly $x$. The asymptotes are found by putting $0$ in place of $1$: $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0$. This also gives the quick way to remember: the hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, the asymptotes are $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0$, and the conjugate hyperbola is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = -1$.

Draw the rectangle with sides $x = \pm a$ and $y = \pm b$. Its diagonals extended are the asymptotes, and the hyperbola touches the rectangle at the vertices.

A worked example

Take $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$, so $a = 3$ and $b = 4$.

1. Foci. $c = \sqrt{9 + 16} = 5$, so the foci are $(\pm 5, 0)$.
2. Eccentricity. $e = \dfrac{5}{3} \approx 1.667$.
3. Directrices. $x = \pm\dfrac{9}{5}$.
4. Latus rectum. $\dfrac{2b^2}{a} = \dfrac{32}{3} \approx 10.67$.
5. Asymptotes. $y = \pm\dfrac{4}{3}x$.
6. A point. The point $\left(5, \dfrac{16}{3}\right)$ is on the curve, because $\dfrac{25}{9} - \dfrac{256}{9 \times 16} = \dfrac{25}{9} - \dfrac{16}{9} = 1$. It is on the right branch. Its distances from the foci are $\dfrac{34}{3}$ from $F_1(-5, 0)$ and $\dfrac{16}{3}$ from $F_2(5, 0)$, and the difference is $6 = 2a$.

The hyperbola $\dfrac{y^2}{9} - \dfrac{x^2}{16} = 1$ opens upward and downward. The number $9$ is under $y^2$, so $a = 3$ and $b = 4$, the foci are $(0, \pm 5)$ and the asymptotes are $y = \pm\dfrac{3}{4}x$.

The rectangular hyperbola

When $a = b$ the asymptotes $y = \pm x$ are at right angles, and the hyperbola is rectangular. Its eccentricity is $e = \sqrt{1 + 1} = \sqrt{2}$. Its equation is $x^2 - y^2 = a^2$.

Turn the picture by $45^\circ$ so that the asymptotes become the coordinate axes. The equation becomes

In this form the curve lies in the first and third quadrants, and the axes are the asymptotes. Its parametric form is $\left(ct, \dfrac{c}{t}\right)$. The transverse axis is along the line $y = x$. The vertices are $(c, c)$ and $(-c, -c)$, with $a = c\sqrt{2}$, and the foci are $(c\sqrt{2}, c\sqrt{2})$ and $(-c\sqrt{2}, -c\sqrt{2})$.

For $xy = 18$ we have $c = 3\sqrt{2}$, so the foci are $(6, 6)$ and $(-6, -6)$ and $2a = 12$. A check at $P(3, 6)$: the distance to $(6, 6)$ is $3$ and the distance to $(-6, -6)$ is $\sqrt{81 + 144} = 15$. The difference is $12 = 2a$.

The relation $y = \dfrac{k}{x}$ from algebra, the graph of inverse variation, is a rectangular hyperbola. It is the same curve.

Parametric form and the conjugate pair

A point on $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is $(a\sec\theta, b\tan\theta)$, because $\sec^2\theta - \tan^2\theta = 1$. A second form uses hyperbolic functions, $(a\cosh u, b\sinh u)$, which is what the simulation draws. The hyperbola and its conjugate have eccentricities $e$ and $e'$ with

For $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$ we have $e = \dfrac{5}{4}$, so $\dfrac{1}{e'^2} = 1 - \dfrac{16}{25} = \dfrac{9}{25}$ and $e' = \dfrac{5}{3}$.

Practise with new numbers

Try it in Python

The program takes a hyperbola and finds its foci, eccentricity, directrices, latus rectum and asymptotes. It checks the difference of the focal distances at several points and shows how a branch approaches its asymptote.

Common mistakes

• Using $c^2 = a^2 - b^2$. For a hyperbola it is $c^2 = a^2 + b^2$.
• Taking $a$ from the larger number. The number $a^2$ is the one under the positive term, whether or not it is the larger.
• Writing the asymptotes as $y = \pm\dfrac{a}{b}x$. For $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ the slope is $\dfrac{b}{a}$.
• Forgetting the second branch. The hyperbola has two branches.

Remember these

• $|PF_1 - PF_2| = 2a$, or $PF = e \cdot PM$ with $e > 1$.
• $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, $b^2 = a^2(e^2 - 1)$. Foci $(\pm ae, 0)$. Directrices $x = \pm\dfrac{a}{e}$. Latus rectum $\dfrac{2b^2}{a}$.
• Asymptotes $y = \pm\dfrac{b}{a}x$. Rectangular: $a = b$, $e = \sqrt{2}$, or $xy = c^2$.
• Parametric form $(a\sec\theta, b\tan\theta)$. Conjugate pair: $\dfrac{1}{e^2} + \dfrac{1}{e'^2} = 1$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the eccentricity of $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$.

2. Find the foci of $\dfrac{y^2}{9} - \dfrac{x^2}{16} = 1$.

3. Find the asymptotes of $\dfrac{x^2}{25} - \dfrac{y^2}{144} = 1$.

4. What is the eccentricity of a rectangular hyperbola?

5. Find the latus rectum of $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.

References

• NCERT, Mathematics for Class XI, Chapter 10 (Conic Sections), the section on the hyperbola.
• OpenStax, Precalculus 2e, Chapter 12 (Analytic Geometry), the section on the hyperbola.