Ellipses: axes, vertices and foci

Warm-up

Three quick questions from earlier lessons.

Discover

$\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ . Where does this ellipse cross the $x$ -axis?

At $x = \pm 5$
At $x = \pm 25$

Key idea

An ellipse is the set of points whose distances to two foci add up to a constant.

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

with

a > b

: vertices

(\pm a, 0)

, co-vertices

(0, \pm b)

, foci

(\pm c, 0)

with

c^2 = a^2 - b^2

. The larger denominator shows the major axis: under

y^2

, the ellipse is vertical.

Worked example

Describe $\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$ .

Vertical major axis,

a = 5

,

b = 4

25 > 16, under y².

Vertices

(0, \pm 5)

Vertices on the y-axis.

Foci

(0, \pm 3)

c² = 25 − 16.

Your turn, guided

Your turn: $\dfrac{x^2}{100} + \dfrac{y^2}{36} = 1$ .

Vertices

(\pm 10, 0)

100 > 36, under x²: a = 10, b = 6.

Foci

(\pm 8, 0)

c² = 100 − 36 = 64.

Practice ★

The ellipse $\dfrac{x^2}{ 36 } + \dfrac{y^2}{ 1 } = 1$ has its vertices at $(\pm c, 0)$ . What is $c$ ?
The ellipse $\dfrac{x^2}{ 25 } + \dfrac{y^2}{ 1 } = 1$ has its vertices at $(\pm c, 0)$ . What is $c$ ?
The ellipse $\dfrac{x^2}{ 9 } + \dfrac{y^2}{ 4 } = 1$ has its vertices at $(\pm c, 0)$ . What is $c$ ?

Watch out!

Practice ★★

The ellipse $\dfrac{x^2}{ 25 } + \dfrac{y^2}{ 16 } = 1$ has foci at $(\pm c, 0)$ . Find $c$ .
The ellipse $\dfrac{x^2}{ 169 } + \dfrac{y^2}{ 144 } = 1$ has foci at $(\pm c, 0)$ . Find $c$ .
The ellipse $\dfrac{x^2}{ 289 } + \dfrac{y^2}{ 64 } = 1$ has foci at $(\pm c, 0)$ . Find $c$ .

Is the major axis of $\dfrac{x^2}{ 1 } + \dfrac{y^2}{ 36 } = 1$ horizontal or vertical?
1. Horizontal
2. Vertical
3. Neither: it is a circle
Is the major axis of $\dfrac{x^2}{ 4 } + \dfrac{y^2}{ 64 } = 1$ horizontal or vertical?
1. Horizontal
2. Vertical
3. Neither: it is a circle

Exit ticket

Three questions to check you have got it.

The ellipse $\dfrac{x^2}{ 25 } + \dfrac{y^2}{ 16 } = 1$ has foci at $(\pm c, 0)$ . Find $c$ .
The ellipse $\dfrac{x^2}{ 169 } + \dfrac{y^2}{ 144 } = 1$ has foci at $(\pm c, 0)$ . Find $c$ .
The ellipse $\dfrac{x^2}{ 289 } + \dfrac{y^2}{ 64 } = 1$ has foci at $(\pm c, 0)$ . Find $c$ .

Summary card

Ellipses

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ : vertices at distance $a$ , co-vertices at $b$ .
Foci: $c^2 = a^2 - b^2$ (larger minus smaller).
The larger denominator gives the major axis.

Watch out: minus for ellipses, plus for hyperbolas.

About this lesson