# Hyperbola: Equation

How to use the hyperbola equation to find the transverse axis and the foci (and vics versa): definition, formula, 6 examples, and their solutions.

## Definition

### Definition

A hyperbola is the set of points

whose difference of the distances from the foci

is constant.

|PF - PF'| = (constant)

## Formula: x^{2}/a^{2} - y^{2}/b^{2} = 1

### Equation

This is the graph of the hyperbola

x^{2}/a^{2} - y^{2}/b^{2} = 1.

The x^{2} term, x^{2}/a^{2}, is (+).

Then this is a horizontal hyperbola.

To show that

the hyperbola is a horizontal hyperbola,

we write x^{2} term first.

### Transverse Axis

The transverse axis is

the distance between the vertices.

The transverse axis is 2a.

### Foci

For x^{2}/a^{2} - y^{2}/b^{2} = 1,

the foci are (c, 0) and (-c, 0).

a, b, and c satisfy

a^{2} + b^{2} = c^{2}.

Unlike the ellipse formula,

a^{2} - b^{2} = c^{2},

the middle sign of the left side is (+).

## Example 1: Transverse Axis

### Example

### Solution

9 = 3^{2}

16 = 4^{2}

x^{2}/3^{2} - y^{2}/4^{2} = 1

The x^{2} term is (+).

Then the transverse axis is

2⋅3.

2⋅3 = 6

So 6 is the answer.

## Example 2: Foci

### Example

### Solution

You just found that

the given hyperbola is

x^{2}/3^{2} - y^{2}/4^{2} = 1.

x^{2}/3^{2} - y^{2}/4^{2} = 1

a = 3

b = 4

Then

c^{2} = 3^{2} + 4^{2}.

3^{2} = 9

+4^{2} = +16

9 + 16 = 25

c^{2} = 25

Then c = √25.

Think the sign of the c plus.

25 = 5^{2}

√5^{2} = 5

Square Root

c = 5

See x^{2}/3^{2} - y^{2}/4^{2} = 1.

The x^{2} term, x^{2}/3^{2}, is (+).

Then the hyperbola is a horizontal hyperbola.

So the foci are

(5, 0) and (-5, 0).

So

(5, 0), (-5, 0)

is the answer.

## Example 3: Equation

### Example

### Solution

The foci are (3, 0) and (-3, 0).

So draw a horizontal hyperbola like this.

And draw the foci (3, 0) and (-3, 0).

Then c = 3.

The transverse axis is

the distance between the vertices.

It's 4.

So 2a = 4.

Divide both sides by 2.

Then a = 2.

a = 2

The foci are (3, 0) and (-3, 0).

So c = 3.

Then

2^{2} + b^{2} = 3^{2}.

2^{2} = 4

3^{2} = 9

Move 4 to the right side.

Then b^{2} = 5.

Instead of finding b,

use b^{2} = 5

to write the hyperbola equation.

The hyperbola is a horizontal hyperbola.

So write the x^{2} term first.

a = 2

b^{2} = 5

Then the hyperbola is

x^{2}/2^{2} - y^{2}/5 = 1.

2^{2} = 4

So

x^{2}/4 - y^{2}/5 = 1

is the answer.

## Formula: y^{2}/a^{2} - x^{2}/b^{2} = 1

### Equation

This is the graph of the hyperbola

y^{2}/a^{2} - x^{2}/b^{2} = 1.

The y^{2} term, y^{2}/a^{2}, is (+).

Then this is a vertical hyperbola.

To show that

the hyperbola is a vertical hyperbola,

we write y^{2} term first.

### Transverse Axis

The transverse axis is 2a.

### Foci

For y^{2}/a^{2} - x^{2}/b^{2} = 1,

the foci are (0, c) and (0, -c).

a, b, and c satisfy

a^{2} + b^{2} = c^{2}.

## Example 4: Transverse Axis

### Example

### Solution

To make the right side 1,

divide both sides by 4.

4 = 2^{2}

-x^{2}

= -x^{2}/1

= -x^{2}/1^{2}

y^{2}/2^{2} - x^{2}/1^{2} = 1

The y^{2} term is (+).

Then the transverse axis is

2⋅2.

2⋅2 = 4

So 4 is the answer.

## Example 5: Foci

### Example

### Solution

You just found that

the given hyperbola is

y^{2}/2^{2} - x^{2}/1^{2} = 1.

y^{2}/2^{2} - x^{2}/1^{2} = 1

a = 2

b = 1

Then

c^{2} = 2^{2} + 1^{2}.

2^{2} = 4

+1^{2} = +1

4 + 1 = 5

c^{2} = 5

Then c = √5.

Think the sign of the c plus.

c = √5

See y^{2}/2^{2} - x^{2}/1^{2} = 1.

The y^{2} term, y^{2}/2^{2}, is (+).

Then the hyperbola is a vertical hyperbola.

So the foci are

(0, √5) and (0, -√5).

So

(0, √5), (0, -√5)

is the answer.

## Example 6: Equation

### Example

### Solution

The foci are (0, 5) and (0, -5).

So draw a vertical hyperbola like this.

And draw the foci (0, 5) and (0, -5).

Then c = 5.

The transverse axis is

the distance between the vertices.

It's 6.

So 2a = 6.

Divide both sides by 2.

Then a = 3.

a = 3

The foci are (0, 5) and (0, -5).

So c = 5.

Then

3^{2} + b^{2} = 5^{2}.

3^{2} = 9

5^{2} = 25

Move 9 to the right side.

Then b^{2} = 16.

Instead of finding b,

use b^{2} = 16

to write the hyperbola equation.

The hyperbola is a vertical hyperbola.

So write the y^{2} term first.

a = 3

b^{2} = 16

Then the hyperbola is

y^{2}/3^{2} - x^{2}/16 = 1.

3^{2} = 9

Then

y^{2}/9 - x^{2}/16 = 1

is the answer.