# Equation of a Hyperbola with Center Outside the Origin

Hyperbolas are defined as conic sections that are obtained when a cone is intersected by a plane. The hyperbola is formed when the plane intersects both faces of the cone. The hyperbola contains two branches that have a parabolic shape and that are reflections of each other. The set of all points in a hyperbola are characterized because the difference of the distances between any point on the hyperbola and the foci is equal to a constant.

A hyperbola has two lines of symmetry. The transverse axis extends from one vertex to the other vertex and passes through the center. The foci define the hyperbola and are located on the line that contains the transverse axis. The conjugate axis extends from one co-vertice to the other and is perpendicular to the transverse axis. The point of intersection of the transverse axis and the conjugate axis is the center of the hyperbola. The center is also the point of intersection of the two asymptotes.

##### PRECALCULUS

Relevant for

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

See equation

##### PRECALCULUS

Relevant for

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

See equation

## Standard form of hyperbolas with center outside the origin

The standard form of hyperbolas centered outside the origin is found by applying a translation of h units on the x-axis and k units on the y axis. This results in the center of the hyperbola being located at $latex (h, k)$. We have two variations of this equation depending on the orientation of the hyperbola.

### Equation of the horizontal hyperbola with center outside the origin

A hyperbola that has its center at $latex (h, k)$ and in which its transversal axis is parallel to the x axis is:

where,

• h is the coordinate of the center and is the y coordinate of the center
• The length of the transverse axis is $latex 2a$ (segment joining vertices)
• The vertices are located at $latex (h\pm a, k)$
• The length of the conjugate axis is $latex 2b$ (segment joining covertices)
• The covertices are located at $latex (h, k\pm b)$
• The distance between the foci is $latex 2c$
• We find c using $latex {{c}^2}={{a}^2}+{{b}^2}$
• The foci are located at $latex (h \pm c, 0)$
• The equations of the asymptotes are $latex y=\pm \frac{b}{a}(x-h)+k$

### Equation of the vertical hyperbola with center outside the origin

When the hyperbola is centered at the point $latex (h, k)$ and its transversal axis is parallel to the y axis, its equation is:

where,

• h is the x coordinate of the center and k is the y coordinate of the center
• The length of the transverse axis is $latex 2a$ (segment joining vertices)
• The vertices have the coordinates $latex (h, k\pm a)$
• The length of the conjugate axis is $latex 2b$ (segment joining covertices)
• The covertices have the coordinates $latex (h\pm b, k)$
• The distance between the foci is $latex 2c$, where, $latex {{c}^2}={{a}^2}+{{b}^2}$
• The foci have the coordinates $latex (h, k\pm c)$
• The asymptotes have the equations $latex y=\pm \frac{a}{b}(x-h)+k$

## Determine the equation of hyperbolas centered outside the origin using vertices and foci

We can use the coordinates of the vertices and the foci to find the equation of a hyperbola centered outside the origin by following these steps:

Step 1:Determine the orientation of the hyperbola. We have to determine if the transverse axis is parallel to the x-axis or parallel to the y axis.

1.1.When the y coordinates of the vertices are the same as the y coordinates of the foci, the transverse axis is parallel to the x-axis, and we use the equation $latex \frac{{{(x-h)}^2}}{{{a}^2}}-\frac{{{(y-k)}^2}}{{{b}^2}}=1$.

1.2. When the x coordinates of the vertices are the same as the x coordinates of the foci, the transverse axis is parallel to the y axis, and we use the equation $latex \frac{{{(y-k)}^2}}{{{a}^2}}-\frac{{{(x-h)}^2}}{{{b}^2}}=1$.

Step 2: We use the coordinates of the vertices and the midpoint formula to find the center $latex (h, k)$.

Step 3: The distance between the vertices is $latex 2a$. We use this to find $latex {{a}^2}$.

Step 4: We use the coordinates of the foci and the values of h and k to find the value of c and $latex {{c}^2}$.

Step 5: We use the equation $latex {{b}^2}={{c}^2}-{{a}^2}$ to find the value of $latex {{b}^2}$.

Step 6: We substitute the values of $latex {{a}^2}$ and $latex {{b}^2}$ into the equation obtained in step 1.

## Hyperbolas with center outside the origin – Examples with answers

The following examples apply what you have learned about hyperbola equations. Vertices and foci are used to determine the equations. Analyze the examples to learn the process used.

### EXAMPLE 1

What is the equation of the hyperbola that has vertices at (-1, 1), (3, 1) and foci at (-2, 1), (4, 1)?

##### Solution

The y coordinates of the foci and the vertices are the same, so we know that the transverse axis is parallel to the x-axis and the hyperbola equation has the form:

$latex \frac{{{(x-h)}^2}}{{{a}^2}}-\frac{{{(y-k)}^2}}{{{b}^2}}=1$

To find the center, we observe that the center is in the middle of the vertices (-1, 1) and (3, 1). Therefore, we apply the midpoint formula:

$latex (h, k)=(\frac{-1+3}{2}, \frac{1+1}{2})$

$latex =(1, 1)$

To find $latex {{a}^2}$, we determine the length of the transverse axis, 2a, which is bounded by the vertices. Therefore, we find the difference in the x coordinates of the vertices:

$latex 2a=|3-1|$

$latex 2a=2$

$latex a=2$

$latex {{a}^2}=4$

Now, we find $latex {{c}^2}$. The coordinates of the foci are $latex (h\pm c, k)$. Therefore, we have $latex (h-c, k)=(-2, 1)$ and $latex (h+c, k)=(4, 1)$. We can use any of these points to find the value of c:

$latex h+c=4$

$latex 1+c=4$

$latex c=3$

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

Now, we use the equation $latex {{b}^2}={{c}^2}-{{a}^2}$ to find the value of $latex {{b}^2}$:

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

$latex =9-4$

$latex =5$

Substituting the values of h, k, $latex {{a}^2}$ and $latex {{b}^2}$, we have:

$latex \frac{{{(x-1)}^2}}{4}-\frac{{{(y-1)}^2}}{5}=1$

### EXAMPLE 2

If a hyperbola has foci at (2, 0), (2, 6) and vertices at (2, 1), (2, 5), what is its equation?

##### Solution

In this case, we see that the x coordinates of the foci and the vertices are the same. This means that the transverse axis is parallel to the y axis and the hyperbola equation has the form:

$latex \frac{{{(y-k)}^2}}{{{a}^2}}-\frac{{{(x-h)}^2}}{{{b}^2}}=1$

We use the vertices (2, 1) and (2, 5) along with the midpoint formula to find the center:

$latex (h, k)=(\frac{2+2}{2}, \frac{1+5}{2})$

$latex =(2, 3)$

Now, we find the length of the transverse axis, 2a. Therefore, we find the difference in the y coordinates of the vertices:

$latex 2a=|5-1|$

$latex 2a=4$

$latex a=2$

$latex {{a}^2}=4$

We use the coordinates of the foci, $latex (h, k \pm c)$, to find the value of $latex {{c}^2}$. Therefore, we have $latex (h, k-c) = (2, 0)$ and $latex (h, k+c)=(2, 6)$. We can use any of these points to find the value of c:

$latex k+c=6$

$latex 3+c=6$

$latex c=3$

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

Now, we use the equation $latex {{b}^2}={{c}^2}-{{a}^2}$ to find the value of $latex {{b}^2}$:

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

$latex =9-4$

$latex =5$

Substituting the values of h, k, $latex {{a}^2}$ and $latex {{b}^2}$, we have:

$latex \frac{{{(y-3)}^2}}{4}-\frac{{{(y-2)}^2}}{5}=1$

## Hyperbolas with center outside the origin – Practice problems

Apply the methods and steps seen above to solve the following problems and find the equations of the hyperbolas using the coordinates of the vertices and the foci.