Equation of a Hyperbola with Center at the Origin

A hyperbola is a conic section formed by the intersection of a cone by a plane at an angle where both bases are intersected. The hyperbola is composed of two branches that are a reflection of each other. The hyperbola is also defined as the set of all points in the Cartesian plane so that the difference of the distances between any point and the foci is equal to a constant.

elements of a hyperbola

Hyperbolas have two lines of symmetry. The transverse axis is the segment that passes through the center and joins the vertices. The foci are located on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and connects the covertices. The center is the point of intersection of the transverse axis and the conjugate axis. Hyperbolas also have two asymptotes, which also intersect at the center.

PRECALCULUS
hyperbola oriented vertically with center at the origin

Relevant for

Finding the equation of the hyperbola with center at the origin.

See equations

PRECALCULUS
hyperbola oriented vertically with center at the origin

Relevant for

Finding the equation of the hyperbola with center at the origin.

See equations

Standard form of hyperbolas with center at the origin

The standard form of a hyperbola gives us information about the location of the vertices and the foci and from there we can define the hyperbola completely. There are two variations of the hyperbola equations that have the center at the origin depending on their orientation.

We can have hyperbolas oriented horizontally or vertically in the Cartesian plane.

Equation of the horizontal hyperbola

The standard form of a hyperbola that has its center at (0, 0) and whose transversal axis is on the x-axis is:

$latex \frac{{{x}^2}}{{{a}^2}}-\frac{{{y}^2}}{{{b}^2}}=1$

where,

  • $latex 2a$ is the length of the transversal axis (segment that joins the vertices)
  • The coordinates of the vertex are $latex (\pm a, 0)$
  • $latex 2b$ is the length of the conjugate axis (segment that joins the covertices)
  • The coordinates of the covertices are $latex (0, \pm b)$
  • $latex 2c$ is the distance between the foci
  • We find c using $latex {{c}^2}={{a}^2}+{{b}^2}$
  • The coordinates of the foci are $latex (\pm c, 0)$
  • The equations of the asymptotes are $latex y=\pm \frac{b}{a}x$
hyperbola oriented horizontally with center at the origin

Equation of the vertical hyperbola

When the hyperbola has the center at the origin, (0, 0), and its transversal axis is the y axis, its equation is:

$latex \frac{{{y}^2}}{{{a}^2}}-\frac{{{x}^2}}{{{b}^2}}=1$

where,

  • $latex 2a$ is the length of the transverse axis
  • The coordinates of the vertex are $latex (0, \pm a)$
  • $latex 2b$ is the length of the conjugate axis
  • The coordinates of the covertices are $latex (\pm b, 0)$
  • $latex 2c$ is the distance between the foci, where,  $latex {{c}^2}={{a}^2}+{{b}^2}$
  • The coordinates of the foci are $latex (0, \pm c)$
  • The equations of the asymptotes are $latex y=\pm \frac{a}{b}x$
hyperbola oriented vertically with center at the origin

Finding the vertices and foci of a hyperbola centered at the origin

We can find the vertices and the foci using the equation of a hyperbola and following these steps:

Determine the orientation of the hyperbola by finding whether the transverse axis is located on the x-axis or on the y axis:

Case 1. If the equation has the form $latex \frac {{{x}^2}}{{{a}^2}} – \frac{{{y}^2}}{{{b}^2}}=1$, the transverse axis is located on the x-axis. The coordinates of the vertices are $latex (\pm a, 0)$ and the coordinates of the foci are $latex (\pm c, 0)$.

Case 2. If the equation has the form $latex \frac{{{y}^2}}{{{a}^2}}-\frac{{{x}^2}}{{{b}^2}}=1$. The coordinates of the vertices are $latex (0, \pm a)$ and the coordinates of the foci are $latex (0, \pm c)$.

We can find the value of a using the equation $latex a=\sqrt{{{a}^2}}$.

We can find the value of c using the equation $latex {{c}^2}={{a}^2}+{{b}^2}$.


Determine the equation of hyperbolas using vertices and foci

To find the equation of a hyperbola centered at the origin if we know the coordinates of the vertices and the foci, we can follow the following steps:

Step 1: Determine the orientation of the hyperbola. This requires us to find out whether the transverse axis is located on the x-axis or on the y axis.

1.1. When the coordinates of the vertices have the form $latex (\pm a, 0)$ and the coordinates of the foci have the form $latex (\pm c, 0)$, the transverse axis is on the x axis and we use the equation $latex \frac{{{x}^2}}{{{a}^2}}-\frac{{{y}^2}}{{{b}^2}}=1$.

1.2. When the coordinates of the vertices have the form $latex (0, \pm a)$ and the coordinates of the foci have the form $latex (0, \pm c)$, the transverse axis is on the y axis and we use the equation $latex \frac{{{y}^2}}{{{a}^2}}-\frac{{{x}^2}}{{{b}^2}}=1$.

Step 2: We use the equation $latex {{b}^2}={{c}^2}-{{a}^2}$ to find the value of $latex {{b}^2}$.

Step 3: We use the values of $latex {{a}^2}$ and $latex {{b}^2}$ in the equation obtained in step 1.


Hyperbolas with center at the origin – Examples with answers

The methods and steps used to find the equations of hyperbolas and the coordinates of vertices and foci seen above are applied to solve the following examples. Look at the examples carefully and analyze the process used.

EXAMPLE 1

What are the vertices and foci of the hyperbola that has the equation $latex \frac{{{y}^2}}{16} – \frac{{{x}^2}}{9}=1$

Solution

We can see that the equation has the form $latex \frac{{{y}^2}}{{{a}^2}}-\frac{{{x}^2}}{{{b}^2}}=1$, so the transverse axis is located on the y axis. Since the hyperbola is centered at the origin, the vertices are the y intercepts of the graph. To find the vertices, we use $latex x = 0$ and solve for :

$latex \frac{{{y}^2}}{16}-\frac{{{x}^2}}{9}=1$

$latex \frac{{{y}^2}}{16}-\frac{0}{9}=1$

$latex \frac{{{y}^2}}{16}=1$

$latex {{y}^2}=16$

$latex y=\pm 4$

The vertices are located at $latex (0,\pm 4)$.

Now, we use the equation $latex {{c}^2}={{a}^2}+{{b}^2}$ to get the value of c. Therefore, we have:

$latex {{c}^2}={{a}^2}+{{b}^2}$

$latex c=\sqrt{{{a}^2}+{{b}^2}}$

$latex c=\sqrt{16+9}$

$latex c=\sqrt{25}$

$latex c=\pm 5$

The foci are located at $latex (0, \pm 5)$.

EXAMPLE 2

What is the equation of the hyperbola that has vertices at (±4, 0) and foci at (±5, 0)?

Solution

The foci and vertices are located on the x-axis. This means that the transverse axis is on the x-axis. Therefore, the equation will have the following form:

$latex \frac{{{x}^2}}{{{a}^2}}-\frac{{{y}^2}}{{{b}^2}}=1$

The vertices are $latex (\pm 4, 0 )$, which means that $latex a=4$ and we have $latex {{a}^2}=16$.

The foci are $latex (\pm 5,0)$, which means that $latex c=5$ and we have $latex {{c}^2}=25$.

We determine the value of $latex {{b}^2}$ using the equation $latex {{b}^2}={{c}^2}-{{a}^2}$:

$latex {{b}^2}={{c}^2}-{{a}^2}$

$latex {{b}^2}=25-16$

$latex {{b}^2}=9$

Using these values, we have the following hyperbola equation:

$latex \frac{{{x}^2}}{16}-\frac{{{y}^2}}{9}=1$


Hyperbolas with center at the origin – Practice problems

Use what you have learned to solve the following hyperbola equation problems. See the solved examples above in case you need help with this.

What are the vertices and foci of the hyperbola $latex \frac{{{x}^2}}{9}-\frac{{{y}^2}}{25}=1$?

Choose an answer






What is the equation of the hyperbola that has the vertices $latex (\pm 6, 0)$ and the foci $latex (\pm 2 \sqrt{10}, 0)$?

Choose an answer







See also

Interested in learning more about the equations of a hyperbola? Take a look at these pages:

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