The hyperbola: standard form, foci and asymptotes

Covers AS91573

What a hyperbola is

A hyperbola is the set of points for which the difference of the distances to two fixed points is constant.
The two fixed points are the foci.
It has two separate branches that open away from each other.

A hyperbola with vertices, foci and asymptotes

Standard form opening left and right

A hyperbola centred at the origin opening left–right has equation:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

$a$ — the distance from the centre to each vertex.
The vertices are at $(\pm a, 0)$ .
The minus sign is what distinguishes the hyperbola from the ellipse.
Swapping the terms, $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ opens up–down with vertices $(0, \pm a)$ .

Finding the foci

The foci are at $(\pm c, 0)$ , found from:

$c^2 = a^2 + b^2$

Note the plus sign — for a hyperbola $c$ is larger than $a$ , so the foci lie beyond the vertices.

The asymptotes

As the branches stretch out, they approach two straight lines called asymptotes:

$y = \pm\frac{b}{a}\,x$

The asymptotes pass through the centre and set the steepness of the branches.
A quick way to draw them: sketch the rectangle with corners $(\pm a, \pm b)$ ; the asymptotes are its diagonals extended.

The rectangular hyperbola

If $a = b$ the asymptotes are $y = \pm x$ (perpendicular), giving a rectangular hyperbola.
The curve $xy = k$ is also a rectangular hyperbola, with the coordinate axes as its asymptotes.

Worked ExampleVertices, foci and asymptotes

For the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ , find the vertices, the foci, and the equations of the asymptotes.

Step 1 — Identify $a$ and $b$

The positive term is $x^2$ , so $a^2 = 9$ ( $a = 3$ ) and $b^2 = 16$ ( $b = 4$ ). The curve opens left–right.

Step 2 — Write the vertices

$\text{Vertices } (\pm 3, 0)$

Step 3 — Find the foci with $c^2 = a^2 + b^2$

$c^2 = 9 + 16 = 25 \;\Rightarrow\; c = 5$

$\text{Foci } (\pm 5, 0)$

Step 4 — Asymptotes

$y = \pm\frac{b}{a}\,x = \pm\frac{4}{3}\,x$

$\boxed{\text{Vertices } (\pm 3, 0), \; \text{foci } (\pm 5, 0), \; y = \pm\tfrac{4}{3}x}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Write down the coordinates of the vertices of the hyperbola $\dfrac{x^2}{36} - \dfrac{y^2}{13} = 1$ .

Merit

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

Excellence

A hyperbola centred at the origin, opening left–right, has one focus at $(13, 0)$ and passes through the point $(5, 0)$ . Find its equation, and state its eccentricity.

The hyperbola: standard form, foci and asymptotes

Covers AS91573

What a hyperbola is

A hyperbola is the set of points for which the difference of the distances to two fixed points is constant.
The two fixed points are the foci.
It has two separate branches that open away from each other.

A hyperbola with vertices, foci and asymptotes

Standard form opening left and right

A hyperbola centred at the origin opening left–right has equation:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

$a$ — the distance from the centre to each vertex.
The vertices are at $(\pm a, 0)$ .
The minus sign is what distinguishes the hyperbola from the ellipse.
Swapping the terms, $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ opens up–down with vertices $(0, \pm a)$ .

Finding the foci

The foci are at $(\pm c, 0)$ , found from:

$c^2 = a^2 + b^2$

Note the plus sign — for a hyperbola $c$ is larger than $a$ , so the foci lie beyond the vertices.

The asymptotes

As the branches stretch out, they approach two straight lines called asymptotes:

$y = \pm\frac{b}{a}\,x$

The asymptotes pass through the centre and set the steepness of the branches.
A quick way to draw them: sketch the rectangle with corners $(\pm a, \pm b)$ ; the asymptotes are its diagonals extended.

The rectangular hyperbola

If $a = b$ the asymptotes are $y = \pm x$ (perpendicular), giving a rectangular hyperbola.
The curve $xy = k$ is also a rectangular hyperbola, with the coordinate axes as its asymptotes.

Worked ExampleVertices, foci and asymptotes

For the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ , find the vertices, the foci, and the equations of the asymptotes.

Step 1 — Identify $a$ and $b$

The positive term is $x^2$ , so $a^2 = 9$ ( $a = 3$ ) and $b^2 = 16$ ( $b = 4$ ). The curve opens left–right.

Step 2 — Write the vertices

$\text{Vertices } (\pm 3, 0)$

Step 3 — Find the foci with $c^2 = a^2 + b^2$

$c^2 = 9 + 16 = 25 \;\Rightarrow\; c = 5$

$\text{Foci } (\pm 5, 0)$

Step 4 — Asymptotes

$y = \pm\frac{b}{a}\,x = \pm\frac{4}{3}\,x$

$\boxed{\text{Vertices } (\pm 3, 0), \; \text{foci } (\pm 5, 0), \; y = \pm\tfrac{4}{3}x}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Write down the coordinates of the vertices of the hyperbola $\dfrac{x^2}{36} - \dfrac{y^2}{13} = 1$ .

Merit

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

Excellence

A hyperbola centred at the origin, opening left–right, has one focus at $(13, 0)$ and passes through the point $(5, 0)$ . Find its equation, and state its eccentricity.

The hyperbola: standard form, foci and asymptotes

What a hyperbola is

Standard form opening left and right

Finding the foci

The asymptotes

The rectangular hyperbola

Step 1 — Identify aaa and bbb

Step 2 — Write the vertices

Step 3 — Find the foci with c2=a2+b2c^2 = a^2 + b^2c2=a2+b2

Step 4 — Asymptotes

Test yourself

Practice by grade

The hyperbola: standard form, foci and asymptotes

What a hyperbola is

Standard form opening left and right

Finding the foci

The asymptotes

The rectangular hyperbola

Step 1 — Identify aaa and bbb

Step 2 — Write the vertices

Step 3 — Find the foci with c2=a2+b2c^2 = a^2 + b^2c2=a2+b2

Step 4 — Asymptotes

Test yourself

Practice by grade

Step 1 — Identify $a$ and $b$

Step 3 — Find the foci with $c^2 = a^2 + b^2$

Step 1 — Identify $a$ and $b$

Step 3 — Find the foci with $c^2 = a^2 + b^2$