Lesson Explainer: Equation of a Hyperbola Mathematics

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 π‘₯π‘Žβˆ’π‘¦π‘=1,π‘Ž,𝑏>0.

This means the following:

  1. The center of the hyperbola is at the origin, (0,0).
  2. The major axisβ€”the one that contains its vertices and focal pointsβ€”is the π‘₯-axis (this is also expressed by saying that the hyperbola has an east-west opening).
  3. The vertices are (βˆ’π‘Ž,0) and (π‘Ž,0).
  4. The asymptotes are the lines 𝑦=π‘π‘Žπ‘₯ and 𝑦=βˆ’π‘π‘Žπ‘₯.
  5. The foci are the points (𝑐,0) and (βˆ’π‘,0), where 𝑐=βˆšπ‘Ž+π‘οŠ¨οŠ¨.

Note that the geometric definition of the hyperbola is that the difference between the distances to the foci is constant, specifically that, for any (π‘₯,𝑦) on the hyperbola, √(π‘₯βˆ’π‘)+π‘¦βˆ’βˆš(π‘₯+𝑐)+𝑦=Β±2π‘Ž.

This is the distance from (π‘₯,𝑦) to focus (𝑐,0) less its distance to (βˆ’π‘,0). The sign, of course, depends on which (left or right) branch of the hyperbola (π‘₯,𝑦) is on.

The left-hand side above gives the difference between lengths |𝑃𝐹| and |𝑃𝐹|. In defining the locus of the hyperbola from the two foci, this difference is a constant. So, it must be the value when 𝑃=π‘‰οŠ§, where we see that |𝑉𝐹|βˆ’|𝑉𝐹|=|𝑉𝑉|=2π‘ŽοŠ§οŠ¨οŠ§οŠ§οŠ§οŠ¨ if the coordinates of π‘‰οŠ¨ are (π‘Ž,0).

The Cartesian equation involving π‘Ž and 𝑏 above is obtained by thinking of the hyperbola’s two branches as being the locus of pairs (π‘₯,𝑦) satisfying the single equation ο€»βˆš(π‘₯βˆ’π‘)+π‘¦βˆ’βˆš(π‘₯+𝑐)+𝑦+2π‘Žο‡ο€»βˆš(π‘₯βˆ’π‘)+π‘¦βˆ’βˆš(π‘₯+𝑐)+π‘¦βˆ’2π‘Žο‡=0.

The left-hand side is a difference of squares, and everything simplifies initially to βˆšπ‘+π‘₯+π‘¦βˆ’2𝑐π‘₯βˆšπ‘+π‘₯+𝑦+2𝑐π‘₯=2π‘Žβˆ’π‘βˆ’π‘₯βˆ’π‘¦οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨ and eventually to π‘₯π‘Žβˆ’π‘¦(π‘βˆ’π‘Ž)=1.

This defines our 𝑏 in terms of π‘Ž and 𝑐.

The asymptotes are more closely related to 𝑏 (rather than 𝑐) since the hyperbola can also be written as ο€»π‘₯π‘Žβˆ’π‘¦π‘ο‡ο€»π‘₯π‘Ž+𝑦𝑏=1, which is related to the β€œcurve” ο€»π‘₯π‘Žβˆ’π‘¦π‘ο‡ο€»π‘₯π‘Ž+𝑦𝑏=0 in the same way that the coordinate axes π‘₯𝑦=0 are related to the hyperbola π‘₯𝑦=1. Of course, the zero set of ο€»π‘₯π‘Žβˆ’π‘¦π‘ο‡ο€»π‘₯π‘Ž+𝑦𝑏=0 is the two lines through the origin 𝑦=Β±π‘π‘Žπ‘₯.

For example, if we take a point like the 𝑃 in the figure, with π‘₯<0 and 𝑦>0, then ο€»π‘₯π‘Žβˆ’π‘¦π‘ο‡ has a negative value which becomes larger and larger as π‘₯ goes to βˆ’βˆž. On the hyperbola, this means that, with very large and negative values of π‘₯, π‘₯π‘Ž+𝑦𝑏=1ο€»βˆ’ο‡β‰ˆ0π‘¦π‘β‰ˆβˆ’π‘₯π‘Ž,ο—οŒΊο˜οŒ»so which describes the asymptotic line of slope βˆ’π‘π‘Ž.

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 𝑦=34π‘₯ and 𝑦=βˆ’34π‘₯, and its closest distance to the center fountain is 20 yards. Find the equation of the hyperbola.


We will assume that the hyperbola is in standard position with the π‘₯-axis as its major axis. This means we seek the equation π‘₯π‘Žβˆ’π‘¦π‘=1 for suitable values of π‘Ž and 𝑏. 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 (0,0) and are therefore at (βˆ’20,0) and (20,0) because of the 20-yard distance given.

To find 𝑏, we note that the asymptotes of the hyperbola are the lines 𝑦=Β±π‘π‘Žπ‘₯. With 𝑏>0, π‘Ž=20, and π‘π‘Ž=34, we conclude 𝑏=34π‘Ž=34(20)=15.

The equation is π‘₯20βˆ’π‘¦15=1.

If a hyperbola in standard position is translated so that it has center (𝑝,π‘ž), the equation of the new hyperbola is (π‘₯βˆ’π‘)π‘Žβˆ’(π‘¦βˆ’π‘ž)𝑏=1 simply because the point (π‘₯,𝑦) on the curve must be moved to (π‘₯βˆ’π‘,π‘¦βˆ’π‘ž) to give a point on the original hyperbola π‘₯π‘Žβˆ’π‘¦π‘=1. 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 object’s trajectory in the solar system by a hyperbolic path in the coordinate plane. The π‘₯-axis is a line of symmetry of this hyperbola. The object enters in the direction of 𝑦=3π‘₯βˆ’9 and leaves in the direction 𝑦=βˆ’3π‘₯+9. The sun is positioned at the origin and the object passes within 1 AU (astronomical unit) of the sun at its closest. Using the asymptote’s equations, find the equation of the object’s path.


We will take it that the π‘₯-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 𝑦=3π‘₯βˆ’9 and 𝑦=βˆ’3π‘₯+9. Adding these equations gives 2𝑦=0 or 𝑦=0 which occurs when π‘₯=3. Therefore, the intersection (3,0) is the center of the hyperbola.

This means that this is a hyperbola which has been shifted by 3 units horizontally. A β€œunit” in this model will correspond to an AU. The equation should, therefore, be of the form (π‘₯βˆ’3)π‘Žβˆ’π‘¦π‘=1 for suitable values of π‘Ž and 𝑏.

The ratio π‘π‘Ž gives the slope of the asymptote (of positive slope) to the original hyperbola π‘₯π‘Žβˆ’π‘¦π‘=1.

This does not change with a translation in the plane, so π‘π‘Ž=3,𝑏=3π‘Ž.i.e.,

Now the equation is (π‘₯βˆ’3)π‘Žβˆ’π‘¦9π‘Ž=1.

The closest point that the object gets to the sun is when it is at the location (1,0). Setting these coordinates in the equation above allows us to solve for π‘Ž: (1βˆ’3)π‘Žβˆ’09π‘Ž=1(βˆ’2)π‘Ž=1βˆ΄π‘Ž=4.

So π‘Ž=2, since it is positive. The equation is (π‘₯βˆ’3)4βˆ’π‘¦36=1.

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