Equilateral Hyperbola

An equilateral hyperbola is a specific type of hyperbola where the transverse and conjugate axes are of equal length, meaning $ a=b $. The standard equation for this type of hyperbola is $$ \frac{x^2}{a^2} - \frac{y^2}{a^2} = \pm 1 $$ which can also be written as $$ x^2 - y^2 = \pm a^2 $$

This represents an open curve defined by the set of all points that satisfy the equation x² - y² = a².

As with other hyperbolas, every point P on the curve has the same difference in distance from the two foci, F₁ and F₂.

Note: In any hyperbola, including an equilateral one, the difference in distances between a point P and the foci is always equal to 2a. $$ | \overline{PF_1} - \overline{PF_2} | = 2a $$ Here, "a" is the distance from a vertex to the center of the hyperbola. For example, if a=2 and b=2, then the difference in distances is 2a=4.

Characteristics

The equilateral hyperbola is unique in that its axes are symmetric and equal in length.

As a result, the distance of both the real vertices (a) and the conjugate vertices (b) from the center (O) is identical.

$$ a=b $$

The real vertices (a) are the points where the hyperbola intersects the transverse axis, while the conjugate vertices (b) are where it intersects the conjugate axis (or normal axis).

Because $ 2a = 2b $, the transverse and conjugate axes form a square rather than a rectangle, with the asymptotes aligning along the diagonals of this square.

In the case of an equilateral hyperbola, the asymptotes are perpendicular to each other and coincide with the bisectors of the quadrants in the Cartesian plane.

$$ y = x $$

$$ y = -x $$

Additionally, the foci are located on the transverse axis at a distance from the center equal to $ c=\sqrt{a^2+b^2} $, where "a" and "b" are the distances from the center O to the real and conjugate vertices, respectively.

$$ c= \sqrt{a^2+b^2} $$

If the transverse axis is the x-axis, the coordinates of the foci are:

$$ F_1(-c;0) \\ F_2(c;0) $$

If the transverse axis is the y-axis, the coordinates of the foci are:

$$ F_1(0;-c) \\ F_2(0;c) $$

The semi-focal distance $ c $ in the equilateral hyperbola simplifies to

$$ c = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a \sqrt{2} $$

As a result, the eccentricity of the hyperbola becomes:

$$ e = \frac{c}{a} = \frac{a \sqrt{2}}{a} = \sqrt{2} $$

Finally, the canonical equation of the equilateral hyperbola is given by:

• When the foci are on the x-axis: $$ x^2 - y^2 = a^2 $$
• When the foci are on the y-axis: $$ y^2 - x^2 = a^2 $$

Every point on the equilateral hyperbola must satisfy this equation.

A Practical Example

Consider an equilateral hyperbola centered at the origin O(0, 0) of the Cartesian axes:

$$ x^2 - y^2 = 4 $$

In this case, a² = 4

$$ a^2 = 4 $$

Taking the square root of both sides, we get a = 2.

$$ \sqrt{a^2} = \sqrt{4} $$

$$ a = 2 $$

This gives us the distance from the center to the real vertices on the transverse axis A₁A₂.

$$ A_1(-a, 0) = (-2, 0) $$

$$ A_2(a, 0) = (2, 0) $$

Plotting the two real vertices on the plane

Since the hyperbola is equilateral, the variables a = b have the same value, so b = 2 as well.

The coordinates of the vertices on the non-transverse axis are:

$$ B_1(-b, 0) = (-2, 0) $$

$$ B_2(b, 0) = (2, 0) $$

Plotting the non-real vertices on the plane

To find the diagonals, construct a square centered at the origin with sides of length 2a.

The diagonals of the square lie on the asymptotes of the equilateral hyperbola.

Knowing "a" and "b", we can calculate "c", the distance from the foci to the center, using Pythagoras' theorem.

$$ c = \sqrt{a^2 + b^2} $$

$$ c = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} = 2.83 $$

So, the coordinates of the foci are:

$$ F_1(-c, 0) = (-2.83, 0) $$

$$ F_2(c, 0) = (2.83, 0) $$

Plotting the coordinates of the foci on the plane

Next, we can calculate the points of the hyperbola using the canonical equation:

$$ x^2 - y^2 = a^2 $$

In this case, a = 2

$$ x^2 - y^2 = 2^2 $$

$$ x^2 - y^2 = 4 $$

Solving for y in terms of x:

$$ y^2 = x^2 - 4 $$

$$ \sqrt{y^2} = \sqrt{x^2 - 4} $$

$$ y = \pm \sqrt{x^2 - 4} $$

We can now find the coordinates (x, y) of the hyperbola points for various x values in the intervals (-∞, -2.81] ∪ [2.81, ∞).

x	y
-5	±4.58
-4	±3.46
-3	±2.24
3	±2.24
4	±3.46
5	±4.58

This way, we can plot the initial points of the hyperbola on the graph.

To complete the curve, we find additional points and connect them.

The final result is the equilateral hyperbola.

Note. As with any other hyperbola, for any point P on the curve, the absolute difference between the distances to the foci is constant and equal to 2a = 4.

Proof

Consider a hyperbola centered at the origin O(0, 0) with its transverse axis on the x-axis and its non-transverse axis on the y-axis, both of equal length.

$$ a = b $$

Since they are equal, the rectangle ABCD with sides of length 2a and 2b, constructed at the center of the hyperbola, becomes a square.

In this case, the diagonals of the square coincide with the bisectors of the quadrants of the Cartesian plane.

$$ y = x $$

$$ y = -x $$

The canonical equation of a hyperbola centered at the origin with foci on the x-axis is:

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

Knowing that a = b, we can replace b² with a²

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

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

By multiplying both sides by a², we get:

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

$$ x^2 - y^2 = a^2 $$

The final result is the canonical equation of the equilateral hyperbola.

Equilateral Hyperbola with Asymptotes Parallel to the Cartesian Axes

The equation for an equilateral hyperbola with asymptotes that align with the Cartesian axes is $$ xy=1 $$. If the asymptotes are parallel to the Cartesian axes, the equation becomes $$ (x-x_0) \cdot (y-y_0) $$ where $ x_0 $ and $ y_0 $ represent the coordinates of the hyperbola's center.

An equilateral hyperbola with its transverse axis on the bisector of the Cartesian plane represents a 45-degree rotation of the standard equilateral hyperbola.

This is a special case of an equilateral hyperbola where the transverse axis coincides with the bisector of the quadrants of the plane instead of the axes.

In other words, the bisectors become the axes of the hyperbola, while the Cartesian plane's axes become the asymptotes of the equilateral hyperbola.

In this system, the equation of the equilateral hyperbola is:

$$ x \cdot y = k $$

Where k is a positive or negative constant.

• If k > 0, the branches of the hyperbola are in the first and third quadrants.
• If k < 0, the branches of the hyperbola are in the second and fourth quadrants.

The axes of symmetry are the bisectors of the quadrants:

$$ y = x $$

$$ y = -x $$

The relationship between $k$ and the length of the transverse semiaxis can be expressed as:

$$ k = \pm \frac{a^2}{2} $$

So, the length of the transverse semi-axis is:

$$ a = \overline{OA_1} = \overline{OA_2} = \sqrt{|k|+|k|} = \sqrt{2|k|} $$

The real vertices are located at the following coordinates:

• For k>0: $$ A_1(- \sqrt{k}, - \sqrt{k} ) $$ and $$ A_2(\sqrt{k}, \sqrt{k} ) $$
• For k<0: $$ A_1(- \sqrt{-k}, \sqrt{-k} ) $$ and $$ A_2(\sqrt{-k}, -\sqrt{-k} ) $$

The semi-focal distance is:

$$ c = a \sqrt{2} = \sqrt{2|k|} \cdot \sqrt{2} = 2 \sqrt{|k|} $$

The coordinates of the foci are:

• For k>0: $$ F_1( - \sqrt{2k}, - \sqrt{2k} ) $$ and $$ F_2( \sqrt{2k}, \sqrt{2k} ) $$
• For k<0: $$ F_1( - \sqrt{-2k}, \sqrt{-2k} ) $$ and $$ F_2( \sqrt{-2k}, - \sqrt{-2k} ) $$

In summary, when k>0, the transverse axis is the bisector passing through the 1st and 3rd quadrants of the Cartesian plane.

When k<0, the transverse axis is the bisector of the 2nd and 4th quadrants of the Cartesian plane.

Proof

To illustrate the formulas, let's consider two coordinate systems: (x,y) and (x',y').

The first system, (x,y), is based on Cartesian axes, while the second system, (x',y'), uses the bisectors of the quadrants.

For instance, in the first coordinate system, point P has coordinates (x,y), while in the second system, it has coordinates (x',y').

In the second coordinate system, the lengths of segments AP and BP are:

$$ \overline{AP} = | x' | $$

$$ \overline{BP} = | y' | $$

The product of these two lengths is simply x'y':

$$ \overline{AP} \cdot \overline{BP} = |x'| \cdot |y'| = |x'y'| $$

To measure the same segments AP and BP in the first coordinate system (x,y), we need to find the distance of point P from the lines y' and x'.

Here, y' corresponds to the line y=x, or x-y=0, and x' corresponds to the line y=-x, or x+y=0.

The distance from a point P(x₀,y₀) to a line ax+by+c=0 is given by the formula:

$$ d = \frac{|ax_0+by_0+c|}{ \sqrt{a^2+b^2} } $$

The distance AP from point P to the line y', i.e., x-y=0 where a=1 and b=-1, is:

$$ \overline{AP} = \frac{|x-y|}{ \sqrt{1^2+(-1)^2} } = \frac{|x-y|}{\sqrt{2}} $$

The distance BP from point P to the line x', i.e., x+y=0 where a=1 and b=1, is:

$$ \overline{BP} = \frac{|x+y|}{ \sqrt{1^2+1^2} } = \frac{|x+y|}{\sqrt{2}} $$

The product of these two lengths is:

$$ \overline{AP} \cdot \overline{BP} = \frac{|x-y|}{\sqrt{2}} \cdot \frac{|x+y|}{\sqrt{2}} $$

$$ \overline{AP} \cdot \overline{BP} = \frac{|x^2-y^2|}{2} $$

For a rectangular hyperbola, we know that $ x^2 - y^2 = a^2 $, so:

$$ \overline{AP} \cdot \overline{BP} = \frac{a^2}{2} $$

To summarize, in both coordinate systems, the product AP·BP is:

$$ \overline{AP} \cdot \overline{BP} = |x'y'| = \frac{a^2}{2} $$

In other words:

$$ |x'y'| = \frac{a^2}{2} $$

$$ x'y' = \pm \frac{a^2}{2} $$

Since $ k = \frac{a^2}{2} $ is a constant, we denote it by the letter $ k $

$$ x'y' = \pm k $$

Therefore, in the (x',y') coordinate system, x' and y' are inversely proportional.

A Practical Example

Consider the following hyperbola:

$$ xy = - \frac{1}{2} $$

In this case, the constant is k = -1/2, which is a negative value.

Thus, the hyperbola is located in the second and fourth quadrants of the Cartesian plane.

The x and y axes of the plane are the asymptotes of the hyperbola.

Conversely, the bisectors of the plane are the transverse and non-transverse axes of the hyperbola.

In this case, $k<1$, and the transverse axis is $y = -x$.

The vertices are located at the following coordinates:

$$ A_1\left(-\sqrt{-k}, \sqrt{-k}\right) = \left(-\sqrt{-\left(-\frac{1}{2}\right)}, \sqrt{-\left(-\frac{1}{2}\right)}\right) = \left(-\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}}\right) $$

$$ A_2\left(\sqrt{-k}, -\sqrt{-k}\right) = \left(\sqrt{-\left(-\frac{1}{2}\right)}, -\sqrt{-\left(-\frac{1}{2}\right)}\right) = \left(\sqrt{\frac{1}{2}}, -\sqrt{\frac{1}{2}}\right) $$

The length of the transverse semi-axis $a$ is 1:

$$ a = \sqrt{2|k|} = \sqrt{2 \cdot |-\frac{1}{2}|} = 1 $$

The foci are located at the coordinates F₁(-1, 1) and F₂(1, -1):

$$ F_1\left(-\sqrt{-2k}, \sqrt{-2k}\right) = \left(-\sqrt{-2 \cdot \left(-\frac{1}{2}\right)}, \sqrt{-2 \cdot \left(-\frac{1}{2}\right)}\right) = \left(-1, 1\right) $$

$$ F_2\left(\sqrt{-2k}, -\sqrt{-2k}\right) = \left(\sqrt{-2 \cdot \left(-\frac{1}{2}\right)}, -\sqrt{-2 \cdot \left(-\frac{1}{2}\right)}\right) = \left(1, -1\right) $$

Here’s a graphical representation:

Proof. Consider an equilateral hyperbola with its transverse axis on the x-axis, with the canonical equation $$ x^2 - y^2 = 1 $$ Rotating the hyperbola by 45° using the rotation matrix, we get the new coordinates x' and y'. $$ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix} $$ The new coordinates are $$ x' = x \cdot \frac{\sqrt{2}}{2} - y \cdot \frac{\sqrt{2}}{2} $$ $$ y' = x \cdot \frac{\sqrt{2}}{2} + y \cdot \frac{\sqrt{2}}{2} $$ Substituting these new coordinates into the canonical equation of the equilateral hyperbola: $$ x'^2 - y'^2 = 1 $$ $$ (x \cdot \frac{\sqrt{2}}{2} - y \cdot \frac{\sqrt{2}}{2})^2 - (x \cdot \frac{\sqrt{2}}{2} + y \cdot \frac{\sqrt{2}}{2})^2 = 1 $$ $$ ( x \cdot \frac{\sqrt{2}}{2} )^2 - 2 xy \frac{\sqrt{2}}{2} \frac{\sqrt{2}}{2} + ( y \cdot \frac{\sqrt{2}}{2} )^2 - [ ( x \cdot \frac{\sqrt{2}}{2} )^2 + 2 xy \frac{\sqrt{2}}{2} \frac{\sqrt{2}}{2} + ( y \cdot \frac{\sqrt{2}}{2} )^2 ] = 1 $$ $$ \frac{1}{2}x^2 - xy + \frac{1}{2}y^2 - [ \frac{1}{2}x^2 + xy + \frac{1}{2} y^2 ] = 1 $$ $$ -xy - xy = 1 $$ $$ -2xy = 1 $$ $$ 2xy = -1 $$ $$ xy = - \frac{1}{2} $$ In this case, the constant is k = -1/2. Since k is negative, the equilateral hyperbola is located in the second and fourth quadrants.

Notes

Here are some observations and additional notes on equilateral hyperbolas:

• The equation of an equilateral hyperbola with asymptotes parallel to the axes is $ (x-x_0)(y-y_0)= \pm 1 $, or when the asymptotes align with the Cartesian axes, $ xy= \pm 1 $. This is actually a special case of a homographic function, expressed as $$ y = \frac{ax+b}{cx+d} $$ where $ c \ne 0 $ and $ ad-bc \ne 0 $. The center of this hyperbola is located at $ C \left( - \frac{d}{c}; \frac{a}{c} \right) $, with asymptotes at $ x = - \frac{d}{c} $ and $ y = \frac{a}{c} $.

  It's important to note that if $ c=0 $ or $ ad-bc=0 $, the function no longer defines a hyperbola and instead degenerates into a straight line.
  

  Example: Let's consider the equation of an equilateral hyperbola centered at the origin with asymptotes aligned with the Cartesian axes: $$ xy=1 $$ Now, let's translate this hyperbola from its original center $ O(0;0) $ to a new center at $ C(2;3) $.
  
  Each point on the hyperbola is shifted by the same amount: $$ \begin{cases} x'=x+2 \\ \\ y'=y+3 \end{cases} $$ Solving for $ x $ and $ y $ in terms of $ x' $ and $ y' $, we get: $$ \begin{cases} x=x'-2 \\ \\ y=y'-3 \end{cases} $$ Substituting $ x=x'-2 $ and $ y=y'-3 $ back into the original hyperbola equation: $$ xy=1 $$ $$ (x'-2) \cdot (y'-3)=1 $$ For simplicity, we'll drop the primes: $$ (x-2) \cdot (y-3)=1 $$ This is the equation of an equilateral hyperbola centered at $ C(2;3) $, with asymptotes parallel to the Cartesian axes at $ x_0 =2 $ and $ y_0=3 $. To find $ y $, we proceed as follows: $$ (x-2) \cdot (y-3)=1 $$ $$ xy-3x-2y+6=1 $$ $$ y(x-2)=1-6+3x $$ $$ y=\frac{ 3x-5 }{x-2} $$ This confirms that an equilateral hyperbola with asymptotes either coinciding with or parallel to the Cartesian axes can be described by a homographic function of the form $ y = \frac{ax+b}{cx+d} $, where $ a= 3 $, $ b=-5 $, $ c=1 $, $ d=-2 $. The condition $ c \ne 0 $ and $ ad-bc \ne 0 $ holds true, confirming that this is indeed a hyperbola centered at $ C(2;3) $ with asymptotes located at $$ x = - \frac{d}{c} = - \frac{-2}{1} = 2 $$ $$ y = \frac{a}{c} = \frac{3}{1} = 3 $$

And so on.