How To Write The Equation Of A Hyperbola

The Equation Of A Hyperbola

A hyperbola consists of two curves that appear as two symmetric parabolas except that the shapes are less rounded.  The curves of a hyperbola each contain a vertex just like a parabola and are bounded by asymptotes.  These asymptotes are a set of symmetric lines that cross at the geometric center between the curves (midpoint of the vertices) and set boundaries for the curves which aids in the graphing process. 


Like the circle, ellipse and parabola, a hyperbola is a slice of a conic section.  In fact, to understand the appearance of a hyperbola, one could imagine that with two cones joined together at the narrow end, somewhat like an hourglass, a slice is made (offset from the joining point) down the side of the cones creating the characteristic shapes.  Hyperbolas can be positioned such that the axis drawn through their vertices is parallel to the x or y-axis, we call these standard, or such cases where this axis is not parallel to the x or y-axis, we call these nonstandard.


The standard equations of hyperbolas and asymptotes are shown below.


Standard Form:


Vertices lie along vertical axis: (y2 / b2) – (x2 / a2) = 1


Vertices lie along horizontal axis: (x2 / a2) – (y2 / b2) = 1


Asymptotes: y = (b/a)x; y = -(b/a)x


Nonstandard Form:


xy = d; d is a constant that cannot be zero.




With some examples these points can be further illustrated.


1.) The hyperbola, has the given equation: 24y2 – 54x2 = 216 and is centered at the origin.  Find the location of the vertices and the equations for the asymptotes.


We recognize by the form of the equation that the vertices lie along the vertical axis.  We can rewrite the equation in to the form outlined above.


(24y2 – 54x2)(1 / 216) = (216)(1 / 216)

(24y2 / 216) – (54x2 / 216) = 1

(y2 / 9) – (x2 / 4) = 1

(y2 / 32) – (x2 / 22) = 1


Since the vertices lie along the x-axis, they are the x-intercepts: (-2, 0) and (2, 0).


The asymptotes are calculated as follows.


y = (b/a)x; -(b/a)x

y = (3 / 2)x; -(3 / 2)x



2.) Find the vertices for the following hyperbola center at (0, 0), (x2 / 49) = 1 + (y2 / 25).


If we rewrite this equation to be (x2 / 49) – (y2 / 25) = 1, we see that the vertices will be the x-intercepts.  We have a2 = 49, where a = 7, so the vertices are located at: (-7, 0) and (7, 0).





Bittinger, Marvin L., Ellenbogen, David, Keedy, Mervin L. (1994).  Intermediate Algebra: Concepts and Application (4th ed.).  USA: Addison-Wesley Publishing Company, Inc.