Hyperbolas: vertices, foci and asymptotes

Warm-up

Three quick questions from earlier lessons.

Discover

$\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ . Does this curve cross the $y$ -axis?

No
Yes, at $y = \pm 4$

Key idea

A hyperbola is the set of points whose distances to two foci differ by a constant.

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

: vertices

(\pm a, 0)

, foci

(\pm c, 0)

with

c^2 = a^2 + b^2

, asymptotes

y = \pm\dfrac{b}{a}x

. If the

y^2

term is the positive one, the branches open up and down: vertices

(0, \pm a)

.

Worked example

Describe $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ .

Vertices

(\pm 3, 0)

x² is positive: a² = 9, b² = 16.

Foci

(\pm 5, 0)

c² = 9 + 16.

Asymptotes

y = \pm\dfrac{4}{3}x

Slope b/a = 4/3.

Your turn, guided

Your turn: $\dfrac{y^2}{25} - \dfrac{x^2}{144} = 1$ .

Vertices

(0, \pm 5)

y² is positive: vertices on the y-axis, a = 5.

Foci

(0, \pm 13)

c² = 25 + 144 = 169.

Practice ★

The hyperbola $\dfrac{y^2}{ 64 } - \dfrac{x^2}{ 25 } = 1$ has its vertices at $(0, \pm c)$ . What is $c$ ?
The hyperbola $\dfrac{y^2}{ 16 } - \dfrac{x^2}{ 25 } = 1$ has its vertices at $(0, \pm c)$ . What is $c$ ?
The hyperbola $\dfrac{y^2}{ 25 } - \dfrac{x^2}{ 36 } = 1$ has its vertices at $(0, \pm c)$ . What is $c$ ?

Watch out!

Practice ★★

The hyperbola $\dfrac{x^2}{ 144 } - \dfrac{y^2}{ 256 } = 1$ has foci at $(\pm c, 0)$ . Find $c$ .
The hyperbola $\dfrac{x^2}{ 225 } - \dfrac{y^2}{ 64 } = 1$ has foci at $(\pm c, 0)$ . Find $c$ .

The hyperbola $\dfrac{x^2}{ 1 } - \dfrac{y^2}{ 36 } = 1$ has asymptotes $y = \pm mx$ . What is $m$ ?
The hyperbola $\dfrac{x^2}{ 16 } - \dfrac{y^2}{ 64 } = 1$ has asymptotes $y = \pm mx$ . What is $m$ ?
The hyperbola $\dfrac{x^2}{ 81 } - \dfrac{y^2}{ 36 } = 1$ has asymptotes $y = \pm mx$ . What is $m$ ?

Exit ticket

Three questions to check you have got it.

The hyperbola $\dfrac{x^2}{ 1 } - \dfrac{y^2}{ 36 } = 1$ has asymptotes $y = \pm mx$ . What is $m$ ?
The hyperbola $\dfrac{x^2}{ 16 } - \dfrac{y^2}{ 64 } = 1$ has asymptotes $y = \pm mx$ . What is $m$ ?
The hyperbola $\dfrac{x^2}{ 81 } - \dfrac{y^2}{ 36 } = 1$ has asymptotes $y = \pm mx$ . What is $m$ ?

Summary card

Hyperbolas

Vertices on the axis of the positive term, at distance $a$ .
Foci: $c^2 = a^2 + b^2$ .
$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ : asymptotes $y = \pm\dfrac{b}{a}x$ .

Watch out: plus for hyperbolas, minus for ellipses.

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