# 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: x2/a2 - y2/b2 = 1

### Equation

This is the graph of the hyperbola
x2/a2 - y2/b2 = 1.

The x2 term, x2/a2, is (+).

Then this is a horizontal hyperbola.

To show that
the hyperbola is a horizontal hyperbola,
we write x2 term first.

### Transverse Axis

The transverse axis is
the distance between the vertices.

The transverse axis is 2a.

### Foci

For x2/a2 - y2/b2 = 1,
the foci are (c, 0) and (-c, 0).

a, b, and c satisfy
a2 + b2 = c2.

Unlike the ellipse formula,
a2 - b2 = c2,
the middle sign of the left side is (+).

## Example 1: Transverse Axis

### Solution

9 = 32
16 = 42

x2/32 - y2/42 = 1

The x2 term is (+).

Then the transverse axis is
2⋅3.

2⋅3 = 6

## Example 2: Foci

### Solution

You just found that
the given hyperbola is
x2/32 - y2/42 = 1.

x2/32 - y2/42 = 1

a = 3
b = 4

Then
c2 = 32 + 42.

32 = 9
+42 = +16

9 + 16 = 25

c2 = 25

Then c = √25.

Think the sign of the c plus.

25 = 52

52 = 5

Square Root

c = 5

See x2/32 - y2/42 = 1.

The x2 term, x2/32, is (+).

Then the hyperbola is a horizontal hyperbola.

So the foci are
(5, 0) and (-5, 0).

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

## Example 3: Equation

### 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
22 + b2 = 32.

22 = 4

32 = 9

Move 4 to the right side.

Then b2 = 5.

use b2 = 5
to write the hyperbola equation.

The hyperbola is a horizontal hyperbola.
So write the x2 term first.

a = 2
b2 = 5

Then the hyperbola is
x2/22 - y2/5 = 1.

22 = 4

So
x2/4 - y2/5 = 1

## Formula: y2/a2 - x2/b2 = 1

### Equation

This is the graph of the hyperbola
y2/a2 - x2/b2 = 1.

The y2 term, y2/a2, is (+).

Then this is a vertical hyperbola.

To show that
the hyperbola is a vertical hyperbola,
we write y2 term first.

### Transverse Axis

The transverse axis is 2a.

### Foci

For y2/a2 - x2/b2 = 1,
the foci are (0, c) and (0, -c).

a, b, and c satisfy
a2 + b2 = c2.

## Example 4: Transverse Axis

### Solution

To make the right side 1,
divide both sides by 4.

4 = 22

-x2
= -x2/1
= -x2/12

y2/22 - x2/12 = 1

The y2 term is (+).

Then the transverse axis is
2⋅2.

2⋅2 = 4

## Example 5: Foci

### Solution

You just found that
the given hyperbola is
y2/22 - x2/12 = 1.

y2/22 - x2/12 = 1

a = 2
b = 1

Then
c2 = 22 + 12.

22 = 4
+12 = +1

4 + 1 = 5

c2 = 5

Then c = √5.

Think the sign of the c plus.

c = √5

See y2/22 - x2/12 = 1.

The y2 term, y2/22, is (+).

Then the hyperbola is a vertical hyperbola.

So the foci are
(0, √5) and (0, -√5).

So
(0, √5), (0, -√5)

## Example 6: Equation

### 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
32 + b2 = 52.

32 = 9

52 = 25

Move 9 to the right side.

Then b2 = 16.

use b2 = 16
to write the hyperbola equation.

The hyperbola is a vertical hyperbola.
So write the y2 term first.

a = 3
b2 = 16

Then the hyperbola is
y2/32 - x2/16 = 1.

32 = 9

Then
y2/9 - x2/16 = 1