In this explainer, we will learn how to analyze and write equations of hyperbolas.

The Cartesian equation of the hyperbola in its standard position is

This means the following:

- The center of the hyperbola is at the origin, .
- The major axisβthe one that contains its vertices and focal pointsβis the -axis (this is also expressed by saying that the hyperbola has an east-west opening).
- The vertices are and .
- The asymptotes are the lines and .
- The foci are the points and , where .

Note that the geometric definition of the hyperbola is that the difference between the distances to the foci is constant, specifically that, for any on the hyperbola,

This is the distance from to focus less its distance to . The sign, of course, depends on which (left or right) branch of the hyperbola is on.

The left-hand side above gives the difference between lengths and . In defining the locus of the hyperbola from the two foci, this difference is a constant. So, it must be the value when , where we see that if the coordinates of are .

The Cartesian equation involving and above is obtained by thinking of the hyperbolaβs two branches as being the locus of pairs satisfying the single equation

The left-hand side is a difference of squares, and everything simplifies initially to and eventually to

This defines our in terms of and .

The asymptotes are more closely related to (rather than ) since the hyperbola can also be written as which is related to the βcurveβ in the same way that the coordinate axes are related to the hyperbola . Of course, the zero set of is the two lines through the origin .

For example, if we take a point like the in the figure, with and , then has a negative value which becomes larger and larger as goes to . On the hyperbola, this means that, with very large and negative values of , which describes the asymptotic line of slope .

### Example 1: Finding the Equation of a Hyperbola

A hedge is to be constructed in the shape of a hyperbola near a fountain at the center of a yard. The hedge will follow the asymptotes and , and its closest distance to the center fountain is 20 yards. Find the equation of the hyperbola.

### Answer

We will assume that the hyperbola is in standard position with the -axis as its major axis. This means we seek the equation for suitable values of and . We also assume that the unit of measurement of the coordinates is yards. The vertices are then the closest points to the fountain at the origin and are therefore at and because of the 20-yard distance given.

To find , we note that the asymptotes of the hyperbola are the lines . With , , and , we conclude

The equation is

If a hyperbola in standard position is translated so that it has center , the equation of the new hyperbola is simply because the point on the curve must be moved to to give a point on the original hyperbola . This idea is used in the following example.

### Example 2: Finding the Equation of a Hyperbola from Its Asymptotes in a Real-World Context

Suppose that we model an objectβs trajectory in the solar system by a hyperbolic path in the coordinate plane. The -axis is a line of symmetry of this hyperbola. The object enters in the direction of and leaves in the direction . The sun is positioned at the origin and the object passes within 1 AU (astronomical unit) of the sun at its closest. Using the asymptoteβs equations, find the equation of the objectβs path.

### Answer

We will take it that the -axis is also the major axis of the hyperbola: it is the line of symmetry that contains the vertices (and the foci). The asymptotes intersect at the common solution to and . Adding these equations gives or which occurs when . Therefore, the intersection is the center of the hyperbola.

This means that this is a hyperbola which has been shifted by 3 units horizontally. A βunitβ in this model will correspond to an AU. The equation should, therefore, be of the form for suitable values of and .

The ratio gives the slope of the asymptote (of positive slope) to the original hyperbola .

This does not change with a translation in the plane, so

Now the equation is

The closest point that the object gets to the sun is when it is at the location . Setting these coordinates in the equation above allows us to solve for :

So , since it is positive. The equation is