Hyperbola Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Hyperbola.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

The set of all points in a plane where the absolute difference of the distances to two fixed points (foci) is constant. The curve has two separate branches and asymptotes.

While an ellipse keeps the SUM of distances to foci constant, a hyperbola keeps the DIFFERENCE constant. This creates two separate curves that open away from each other, each curving toward (but never reaching) a pair of asymptotic lines.

Read the full concept explanation β†’

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Two branches opening apart, each hugging a pair of asymptotes it never touches.

Common stuck point: The procedure for hyperbola is the easy part; the trap is using a2βˆ’b2a^2-b^2 for the foci. Asking "Is one squared term subtracted from the other (opposite signs) with the result equaling 1?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

Sense of Study hint: Ask: Is one squared term subtracted from the other (opposite signs) with the result equaling 1?

Worked Examples

Example 1

easy
Identify the vertices and the direction of opening for the hyperbola x29βˆ’y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1.

Answer

Vertices:Β (Β±3,0);Β opensΒ leftΒ andΒ right\text{Vertices: } (\pm 3, 0); \text{ opens left and right}

First step

1
The standard form x2a2βˆ’y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 opens left and right (horizontally).

Full solution

  1. 2
    a2=9a^2 = 9, so a=3a = 3. The vertices are at (Β±a,0)=(Β±3,0)(\pm a, 0) = (\pm 3, 0).
  2. 3
    The transverse axis is along the xx-axis with vertices at (βˆ’3,0)(-3, 0) and (3,0)(3, 0).
In the standard form of a hyperbola, the positive fraction determines the direction of opening. When x2x^2 is positive, the hyperbola opens horizontally; when y2y^2 is positive, it opens vertically. The vertices are at distance aa from the center along the transverse axis.

Example 2

medium
Find the equations of the asymptotes for the hyperbola y24βˆ’x29=1\frac{y^2}{4} - \frac{x^2}{9} = 1.

Example 3

medium
Write the equation of a vertical hyperbola with vertices (0,Β±4)(0, \pm 4) and foci (0,Β±5)(0, \pm 5).

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

medium
Find the foci of the hyperbola x225βˆ’y2144=1\frac{x^2}{25} - \frac{y^2}{144} = 1.

Example 2

hard
Write the equation of the hyperbola with foci at (0,Β±5)(0, \pm 5) and vertices at (0,Β±3)(0, \pm 3).

Example 3

easy
Which way does x29βˆ’y216=1\frac{x^2}{9} - \frac{y^2}{16} = 1 open?

Example 4

easy
In x29βˆ’y216=1\frac{x^2}{9}-\frac{y^2}{16}=1, what are aa and bb?

Example 5

easy
Find the vertices of x225βˆ’y24=1\frac{x^2}{25}-\frac{y^2}{4}=1.

Example 6

easy
Which way does y24βˆ’x29=1\frac{y^2}{4} - \frac{x^2}{9} = 1 open?

Example 7

easy
Give the center of (xβˆ’2)216βˆ’(y+3)29=1\frac{(x-2)^2}{16}-\frac{(y+3)^2}{9}=1.

Example 8

easy
What are the asymptote slopes of x29βˆ’y216=1\frac{x^2}{9}-\frac{y^2}{16}=1?

Example 9

easy
Find aa and bb for y236βˆ’x264=1\frac{y^2}{36}-\frac{x^2}{64}=1.

Example 10

easy
Write the equation of a horizontal hyperbola centered at origin with a=2a=2, b=3b=3.

Example 11

medium
Find the foci of x216βˆ’y29=1\frac{x^2}{16}-\frac{y^2}{9}=1.

Example 12

medium
Find the asymptotes of x24βˆ’y29=1\frac{x^2}{4}-\frac{y^2}{9}=1.

Example 13

medium
Find the foci of y29βˆ’x216=1\frac{y^2}{9}-\frac{x^2}{16}=1.

Example 14

medium
Find the eccentricity of x216βˆ’y29=1\frac{x^2}{16}-\frac{y^2}{9}=1.

Example 15

medium
Write the equation of a hyperbola with vertices (Β±3,0)(\pm 3,0) and foci (Β±5,0)(\pm 5,0).

Example 16

medium
Convert 4x2βˆ’9y2=364x^2 - 9y^2 = 36 to standard form.

Example 17

medium
Find the center and foci of (xβˆ’1)29βˆ’(y+2)216=1\frac{(x-1)^2}{9}-\frac{(y+2)^2}{16}=1.

Example 18

challenge
Write the equation of a hyperbola with asymptotes y=Β±23xy=\pm\frac{2}{3}x and a vertex at (3,0)(3,0).

Example 19

challenge
The difference of distances from a point to foci (Β±5,0)(\pm 5,0) is 6. Find the equation.

Example 20

challenge
Find the eccentricity of a hyperbola whose asymptotes are y=Β±xy=\pm x.

Example 21

medium
Find the eccentricity of y29βˆ’x216=1\frac{y^2}{9}-\frac{x^2}{16}=1.

Example 22

medium
Find the asymptotes of y216βˆ’x29=1\frac{y^2}{16}-\frac{x^2}{9}=1.

Example 23

easy
Find the vertices of x236βˆ’y249=1\frac{x^2}{36} - \frac{y^2}{49} = 1.

Example 24

easy
Identify the opening direction of y216βˆ’x225=1\frac{y^2}{16} - \frac{x^2}{25} = 1.

Example 25

easy
Find the center of (x+4)29βˆ’(yβˆ’1)216=1\frac{(x+4)^2}{9} - \frac{(y-1)^2}{16} = 1.

Example 26

easy
Find the asymptote slopes of x24βˆ’y21=1\frac{x^2}{4} - \frac{y^2}{1} = 1.

Example 27

easy
Write a horizontal hyperbola at origin with a=5a = 5, b=12b = 12.

Example 28

easy
Find aa and bb for y249βˆ’x216=1\frac{y^2}{49} - \frac{x^2}{16} = 1.

Example 29

medium
Find the foci of x29βˆ’y240=1\frac{x^2}{9} - \frac{y^2}{40} = 1.

Example 30

medium
Find the eccentricity of x225βˆ’y211=1\frac{x^2}{25} - \frac{y^2}{11} = 1.

Example 31

medium
Find the asymptotes of (xβˆ’2)29βˆ’(y+1)216=1\frac{(x-2)^2}{9} - \frac{(y+1)^2}{16} = 1.

Example 32

medium
Find the length of the transverse axis of x249βˆ’y216=1\frac{x^2}{49} - \frac{y^2}{16} = 1.

Example 33

medium
Find the length of the conjugate axis of x249βˆ’y216=1\frac{x^2}{49} - \frac{y^2}{16} = 1.

Example 34

medium
Convert 25y2βˆ’16x2=40025y^2 - 16x^2 = 400 to standard form.

Example 35

medium
Find the center, vertices, and foci of (xβˆ’3)216βˆ’(y+2)29=1\frac{(x-3)^2}{16} - \frac{(y+2)^2}{9} = 1.

Example 36

medium
Find the asymptotes of 9x2βˆ’4y2=369x^2 - 4y^2 = 36.

Example 37

medium
Find the foci of y24βˆ’x221=1\frac{y^2}{4} - \frac{x^2}{21} = 1.

Example 38

hard
Find the equation of the hyperbola with foci (Β±10,0)(\pm 10, 0) and asymptotes y=Β±43xy = \pm \frac{4}{3} x.

Example 39

hard
A hyperbola has center at the origin, a vertex at (0,6)(0, 6), and passes through (8,10)(8, 10). Find its equation.

Example 40

hard
Find the eccentricity of x216βˆ’y220=1\frac{x^2}{16} - \frac{y^2}{20} = 1.

Example 41

hard
Identify and convert 4x2βˆ’y2βˆ’16xβˆ’4y+16=04x^2 - y^2 - 16x - 4y + 16 = 0 to standard form.

Example 42

hard
For x2a2βˆ’y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, derive the asymptote equations.

Example 43

hard
Find the equation of a hyperbola with eccentricity 2\sqrt{2} and vertex at (3,0)(3, 0) centered at origin.

Example 44

challenge
Show that a point on the hyperbola x2a2βˆ’y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 satisfies ∣d1βˆ’d2∣=2a|d_1 - d_2| = 2a where d1,d2d_1, d_2 are distances to the foci.

Example 45

challenge
Find the points of intersection between x216βˆ’y29=1\frac{x^2}{16} - \frac{y^2}{9} = 1 and the line y=xy = x.

Background Knowledge

These ideas may be useful before you work through the harder examples.

equation of circleasymptote