1DParamHyperb


 * Parametric Equations of Hyperbolas**

We can use parametric equations to define a hyperbola with oblique asymptotes.
 * x = a sec(t)
 * y = b tan(t)

Note : sec(t) is defined as the reciprocal of cos(t) (See 2A Reciprocal Trig Functions)

if we use radians, then t Î [0, 2 p ] ... (or equivalent) if we use degrees, then t Î [0, 360°]

Note: If t is set to a domain smaller than 2 p, a partial hyperbola will be obtained. math \\ t \in \left( -\dfrac {\pi}{2}, \, \dfrac {\pi}{2} \right) \text { will give the right branch} \\ \\ t \in \left( \dfrac {\pi}{2}, \, \dfrac {3\pi}{2} \right) \text { will give the left branch} math

We can confirm that these parametric equations create a hyperbola by finding the cartesian equation from the equations. This can be done by isolating the trig functions, squaring each equation, then adding the two equations.

You will need to use: (See 2C Pythagorean Trig Identities) ... ... 1 + tan 2 (t) = sec 2 (t)

which can be rearranged to give: ... ... sec 2 (t) – tan 2 (t) = 1

Isolate the trig functions and square

math \\ . \qquad x=a\sec(t) \qquad \qquad \qquad y=b\tan(t) \\. \\ . \qquad \sec(t)=\dfrac{x}{a} \qquad \qquad \qquad \tan(t)=\dfrac{y}{b} \\. \\ . \qquad \sec^2(t)=\dfrac{x^2}{a^2} \qquad \qquad \quad \; \tan^2(t)=\dfrac{y^2}{b^2} math

__Subtract__ equations math . \qquad \sec^2(t)-\tan^2(t)=\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2} math

Using the trig identity from above, we get math . \qquad \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 math

This is the basic equation of a hyperbola with its centre at the origin.

General Parametric Equations of Hyperbolas

For a hyperbola that has been translated h units to the right and k units up, the parametric equations become:
 * x = a sec(t) + h
 * y = b tan(t) + k

if we use radians, then t Î [0, 2 p ] (or equivalent) if we use degrees, then t Î [0, 360°]

Return to Hyperbolas

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