Basic Properties of Conjugate Hyperbola of a Hyperbola

A hyperbola is a type of conic section that can be defined as the locus of points where the difference of the distances from two fixed points (foci) is constant. When we talk about the conjugate hyperbola, we are referring to a hyperbola that is associated with a given hyperbola, sharing the same asymptotes but oriented perpendicularly to the original hyperbola.

Definition

Given a hyperbola with the equation:

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

The conjugate hyperbola has the equation:

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

Table of Differences and Important Points

Property	Hyperbola	Conjugate Hyperbola
Equation	$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$	$\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$
Axes	Transverse axis along x-axis	Transverse axis along y-axis
Asymptotes	Lines $y = \pm \frac{b}{a}x$	Lines $y = \pm \frac{b}{a}x$ (same as hyperbola)
Foci	$(\pm c, 0)$ where $c^2 = a^2 + b^2$	$(0, \pm c)$ where $c^2 = a^2 + b^2$
Vertices	$(\pm a, 0)$	$(0, \pm b)$
Eccentricity	$e = \frac{c}{a}$	$e = \frac{c}{b}$
Orientation	Opens left and right	Opens up and down

Formulas

For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the following formulas are applicable:

• Length of transverse axis: $2a$
• Length of conjugate axis: $2b$
• Distance between foci: $2c$ where $c^2 = a^2 + b^2$
• Eccentricity: $e = \frac{c}{a}$

For the conjugate hyperbola $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$, the formulas are:

• Length of transverse axis: $2b$
• Length of conjugate axis: $2a$
• Distance between foci: $2c$ where $c^2 = a^2 + b^2$
• Eccentricity: $e = \frac{c}{b}$

Examples

Example 1: Identifying the Conjugate Hyperbola

Given the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$, find the equation of its conjugate hyperbola.

Solution:

The given hyperbola has $a^2 = 16$ and $b^2 = 9$. The conjugate hyperbola will swap the positions of $a^2$ and $b^2$ in the equation. Therefore, the equation of the conjugate hyperbola is:

$$ \frac{y^2}{9} - \frac{x^2}{16} = 1 $$

Example 2: Asymptotes of the Conjugate Hyperbola

Find the equations of the asymptotes for the conjugate hyperbola of $\frac{x^2}{25} - \frac{y^2}{16} = 1$.

Solution:

The asymptotes of the original hyperbola are given by $y = \pm \frac{b}{a}x$. Here, $a^2 = 25$ and $b^2 = 16$, so $a = 5$ and $b = 4$. The equations of the asymptotes are:

$$ y = \pm \frac{4}{5}x $$

Since the conjugate hyperbola shares the same asymptotes, these are also the asymptotes for the conjugate hyperbola.

Example 3: Foci of the Conjugate Hyperbola

Given the hyperbola $\frac{x^2}{36} - \frac{y^2}{64} = 1$, determine the foci of its conjugate hyperbola.

Solution:

First, we find $c^2$ for the original hyperbola:

$$ c^2 = a^2 + b^2 = 36 + 64 = 100 $$

So $c = 10$. The foci of the conjugate hyperbola are located along the y-axis at $(0, \pm c)$, which gives us:

$$ (0, \pm 10) $$

Conclusion

The conjugate hyperbola shares many properties with the original hyperbola, including the same asymptotes and the same foci distance. However, it is oriented perpendicularly to the original hyperbola, with its transverse and conjugate axes swapped. Understanding these properties is crucial for solving problems related to hyperbolas and their conjugates, especially in the context of exams.