4 Ways of Parametrization of a Hyperbola

A hyperbola is a conic section defined as the locus of all points in the plane such that the difference of the distances from two fixed points is a constant. In a rectangular coordinates, a hyperbola centered at origin and with semimajor axis a on the x-axis and semiminor axis b on the y-axis is expressed as

\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2} = 1

This equation is also called standard form equation or general equation for a hyperbola. In some instances, it's more convenient to use parametric equations to represent a hyperbola. In the standard form equation, y is not the function of x for a hyperbola since there are two outputs y for each input x. So to parametrization of a hyperbola is to transform its coordinates x and y to a pair of functions in terms of independent parameter.

Usually there could be more than one way to parameterize a curve. In this article, we are going to discuss 4 ways of parametrization of a hyperbola using trigonometric identities, polynomial identity, geometrical way and hyperbolic functions.

Parametrization of Hyperbola by Trigonometric Identity

The idea of trigonometric method is inspired by the way of parametrization of a circle. The general equation of a unit circle in a rectangular coordinates is

x^2+y^2 = 1

which is identical to the Pythagorean Identity

\sin^2 \alpha+\cos^2 \alpha =1

Comparing (2) and (3) yields the parametric equations for the unit circle.

x = \cos\alpha
y = \sin \alpha

Now for a hyperbola, is it possible to find a trigonometric identity which is difference of squares and equal to 1? Yes, the following identity will fit to our purpose.

\sec^2 \alpha -\tan^2\alpha = 1


x = a\sec t \\y = b\tan t

Substituting x, y into the general form equation (1) shows the LHS is equal to 1. And the parametric equations for a hyperbola could be a pair of trigonometric functions.

Using Polynomial Identity

Similar to using trigonometrical identity, the idea of polynomial method is to find a polynomial identity that has two terms in terms of one independent variable and the difference of them is equal to 1.

Let's consider the following identity.

\Big( t+\dfrac{1}{t} \Big) ^2-\Big( t-\dfrac{1}{t} \Big) ^2 = 4

Dividing both sides by 4 gives similar identity like (4)

\Big[ \dfrac{1}{2} \Big( t+\dfrac{1}{t} \Big)\Big] ^2-\Big[\dfrac{1}{2} \Big( t-\dfrac{1}{t} \Big)\Big] ^2 = 1

Comparison (6) with the general equations of the hyperbola gives a polynomial version parametric equation.

\begin{alignedat}{2} x = \dfrac{1}{2} a( t+\dfrac{1}{t} ) \\ y = \dfrac{1}{2} b( t-\dfrac{1}{t} ) \end{alignedat}

The equations can be also expressed as

\begin{alignedat}{2} x = \dfrac{1+t^2}{2t} a \\ y = \dfrac{t^2-1}{2t} b \end{alignedat}

If we refer to the double angle identities of trigonometric functions, the parametric parts of above equations are identical to the left hand sides of \csc 2\alpha and \cot 2\alpha

\def\arraystretch{3.5} \begin{array}{|c|c|} \hline \tan 2 \alpha = \dfrac{2\tan \alpha }{1-\tan^2\alpha } &\cot 2 \alpha =\dfrac{ 1-\tan^2\alpha }{ 2\tan \alpha } \\ \hline \sin 2\alpha = \dfrac{2\tan \alpha }{1+\tan^2 \alpha } & \csc2\alpha = \dfrac{1+\tan^2\alpha }{2\tan\alpha } \\ \hline \cos 2\alpha =\dfrac{1-\tan^2 \alpha }{1+\tan^2\alpha }& \sec 2\alpha = \dfrac{1+\tan^2\alpha }{1-\tan^2\alpha } \\ \hline \end{array}

The polynomial identity is equivalent to the trigonometric identity,

\csc^2 \alpha -\cot^2\alpha = 1

Find Parametric Equations of a hyperbola using geometrical method

In the figure are a hyperbola and its auxiliary circle centered at origin. P is an arbitrary point on the hyperbola with its perpendicular point to axis of symmetry at point A. AB is the tangent to the circle with tangent point at B, then

BC = a

Let t be the angle formed by radius BC with the positive x axis. Since \triangle ABC is a right triangle,

\cos t = \dfrac{BC}{AC} =\dfrac{a}{AC}
AC = a\cdot \dfrac{1}{\cos t} =a\sec t

which is the abscissa of the point P. Now we get parametric function for x,

x = a\sec t

Next we substitute x into general equation to solve for y.

\dfrac{(a\sec t)^2}{a^2}-\dfrac{y^2}{b^2} = 1
\sec^2 t-\dfrac{y^2}{b^2} = 1

Apply the trigonometric identity

\sec^2 \alpha -1 = \tan^2\alpha


y^2 = b^2 \tan^2 t

Taking square root gives the parametric function for ordinate of point P,

y = b\tan t

Now we have derived the parametric equations for a hyperbola. The parameter t is an angle formed by the radius of the auxiliary circle of the hyperbola width its value dependent on point P.

Represent hyperbola using hyperbolic functions

Hyperbolic functions are a set of six functions analogous to the trigonometric functions but related to the hyperbola rather than the circle. Hyperbolic functions, which are named after trigonometric functions with suffix h, are defined in terms of exponential functions of e^x and e^{-x}.

\def\arraystretch{3.5} \begin{array}{|c|c|} \hline \sinh x= \dfrac{e^x-e^{-x}}{2} & \text{csch}\,x=\dfrac{1}{\sinh x} = \dfrac{2}{e^x-e^{-x}} \\ \hline \cosh x = \dfrac{ e^x+e^{-x} }{2 }& \text{sech}\,x= \dfrac{1}{\cosh x} = \dfrac{2}{e^x+e^{-x}} \\ \hline \tanh x =\dfrac{e^x-e^{-x} }{e^x+e^{-x} }& \coth x =\dfrac{1}{\tanh x} = \dfrac{e^x+e^{-x} }{e^x-e^{-x} } \\ \hline \end{array}

Addition and subtraction of sinh and cosh gives the following two equations.

\cosh x+ \sinh x = e^x \\ \cosh x - \sinh x = e^{-x}

Multiplying the two equations yields the identity.

\cosh^2 x- \sinh^2 x = 1

Comparing the identity (10) with (4) gives the relations between hyperbolic and trigonometric functions.

\cosh x = \sec \alpha \\ \sinh x = \tan \alpha

Substituting (11) into (5), gives hyperbolic version of parametric equations for a hyperbola.

x = a\cosh x \\ y = b \sinh x


  • There are always more than one way to find the parametric equations of a Hyperbola.
  • The process of parametrization of a hyperbola is to utilize the identity that the difference of squares is equalt to 1.
  • The geometric method illustrates the meanings of parameters in the parametric equations.

Collected in the board: Conic sections

Steven Zheng Steven Zheng posted 5 months ago

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