# Hyperbola

## Definition and Properties of a Hyperbola

A hyperbola is a plane curve such that the difference of the distances from any point of the curve to two other fixed points (called the foci of the hyperbola) is constant. The midpoint of the line segment joining the foci is called the center of the hyperbola. The distance from the center to a focus is called the focal distance and denoted by c.

Any hyperbola consists of two distinct branches. The points on the two branches that are closest to each other are called the vertices. The distance of a vertex to the center is denoted by a.

The line passing through the vertices of a hyperbola is called the transverse axis. (It contains the segment of length 2a between the vertices). The transverse axis of a hyperbola is its line of symmetry. Another line of symmetry is perpendicular to the transverse axis and is called the conjugate axis. (It contains the segment of length 2b with midpoint at the center of the hyperbola).

The standard (or canonical) equation of the hyperbola is

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

The absolute value of the difference of the distances from any point of a hyperbola to its foci is constant:

$\left|{r_1 - r_2}\right| = 2a$

where $$r_1,$$ $$r_2$$ are the distances from an arbitrary point $$P\left({x,y}\right)$$ of the hyperbola to the foci $$F_1$$ and $$F_2,$$ $$a$$ is the transverse semi-axis of the hyperbola.

## Equations of the Asymptotes of a Hyperbola

$y = \pm \frac{b}{a}x$

## Relationship between Semi-Axes of a Hyperbola and its Focal Distance

$c^2 = a^2 + b^2$

where $$c$$ is the focal distance, $$a$$ is the transverse semi-axis of the hyperbola, $$b$$ is the conjugate semi-axis.

## Eccentricity of a Hyperbola

$e = \frac{c}{a} \gt 1$

## Equations of the Directrices of a Hyperbola

The directrix of a hyperbola is a straight line perpendicular to the transverse axis of the hyperbola and intersecting it at the distance $${\frac{a}{e}}$$ from the center. A hyperbola has two directrices spaced on opposite sides of the center. The equations of the directrices are given by

$x = \pm \frac{a}{e} = \pm \frac{a^2}{c}$

## Parametric Equations of a Hyperbola (Right Branch)

\left\{ \begin{aligned} x &= a\cosh t \\ y &= b\sinh t \end{aligned} \right.,\;\; 0 \le t \le 2\pi

where $$a, b$$ are the semi-axes of the hyperbola, $$t$$ is a parameter.

## General Equation of a Hyperbola

$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$

where $$A,$$ $$B,$$ $$C,$$ $$D,$$ $$E,$$ $$F$$ are real numbers and $$B^2 - 4AC \gt 0.$$

## General Equation of a Hyperbola with Axes Parallel to the Coordinate Axes

$Ax^2 + Cy^2 + Dx + Ey + F = 0$

where $$AC \lt 0.$$

## Equilateral Hyperbola

A hyperbola is called equilateral if its semi-axes are equal to each other: $$a = b.$$ Such a hyperbola has mutually perpendicular asymptotes and its canonical equation looks like

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

If the asymptotes are taken to be the horizontal and vertical coordinate axes (respectively, $$y = 0$$ and $$x = 0),$$ then the equation of the equilateral hyperbola can be reduced to this form:

$xy = \pm\frac{e^2}{4} \;\text{ or }\;y = \pm\frac{k}{x},\;\text{ where }\;k=\frac{e^2}{4}$

## Solved Problems

### Example 1.

Find the transverse and conjugate semi-axes, foci and eccentricity of the hyperbola $${16x^2 - 9y^2 = 144.}$$

Solution.

Let's write the equation of the hyperbola in canonical form:

$16x^2 - 9y^2 = 144, \Rightarrow \frac{16x^2}{144} - \frac{9y^2}{144} = 1, \Rightarrow \frac{x^2}{\frac{225}{9}} + \frac{y^2}{\frac{225}{25}} = 1,$
$\Rightarrow \frac{x^2}{9} -\frac{y^2}{16} = 1, \Rightarrow \frac{x^2}{3^2} - \frac{y^2}{4^2} = 1.$

Thus the length of the transverse semi-axis is $$a = 3$$ and the length of the conjugate semi-axis is $$b = 4.$$

Calculate the focal distance of the hyperbola

$c = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5.$

The two foci of the hyperbola have coordinates

$F_1 = \left({-5,0}\right),\; F_2 = \left({5,0}\right).$

Finally the eccentricity is equal

$e = \frac{c}{a} = \frac{5}{3}.$

### Example 2.

Find the equation of an equilateral hyperbola if it passes through the point $$\left({\sqrt{15},\sqrt{6}}\right).$$

Solution.

The canonical equation of an equilateral hyperbola has the form

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

Let's substitute the coordinates of the given point and find the value of $$a^2:$$

$\frac{\left({\sqrt{15}}\right)^2}{a^2} - \frac{\left({\sqrt{6}}\right)^2}{a^2} = 1, \Rightarrow \frac{15}{a^2} - \frac{6}{a^2} = 1, \Rightarrow \frac{9}{a^2} = 1, \Rightarrow a^2 = 9.$

Therefore, the equation of the hyperbola is written as

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

We can also express it like this

$x^2 - y^2 = 9.$

### Example 3.

Find the equation of a hyperbola if the distance between the foci is 6 and its eccentricity is $$\frac{3}{2}.$$

Solution.

Let's denote the focal distance of the hyperbola by $$c.$$ Then the distance between the foci is equal to $$2c.$$ By definition, the eccentricity of a hyperbola is given by the formula

$e = \frac{c}{a},$

where $$a$$ is the length of the transverse semi-axis. Thus, we have a system of two equations with unknowns $$a$$ and $$c:$$

$\left\{ \begin{array}{l} 2c = 6\\ \frac{c}{a} = \frac{3}{2} \end{array} \right..$

Let's solve this system and determine the hyperbola parameters $$a$$ and $$c:$$

$\left\{ \begin{array}{l} c = 3\\ \frac{c}{a} = \frac{3}{2} \end{array} \right., \Rightarrow \left\{ \begin{array}{l} c = 3\\ 2c = 3a \end{array} \right., \Rightarrow \left\{ \begin{array}{l} c = 3\\ a = 2 \end{array} \right..$

Now we can easily find the conjugate semi-axis $$b:$$

$c^2 = a^2+b^2, \Rightarrow b = \sqrt{c^2-a^2} = \sqrt{3^2 - 2^2} = \sqrt{9-4} = \sqrt{5}.$

So the canonical equation of the hyperbola has the form

${\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \Rightarrow \frac{x^2}{2^2} - \frac{y^2}{\left({\sqrt{5}}\right)^2} = 1, \Rightarrow \frac{x^2}{4} - \frac{y^2}{5} = 1}.$

### Example 4.

Find the equation of a hyperbola if its asymptotes are the lines $$y = \pm\frac{2}{3}x$$ and the distance between its foci is $$4\sqrt{13}.$$

Solution.

The asymptotes of a hyperbola are given by the equations

$y = \pm\frac{b}{a}x.$

Hence, we have the first equation

$\frac{b}{a} = \frac{2}{3}.$

The second equation determines the distance between the foci, that is

$2c = 4\sqrt{13}, \Rightarrow c = 2\sqrt{13},$

where $$c$$ is the focal distance.

We also know that

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

So we have a system of two equations with unknowns $$a$$ and $$b:$$

$\left\{ \begin{array}{l} \frac{b}{a} = \frac{2}{3}\\ \sqrt{a^2 + b^2} = 2\sqrt{13} \end{array} \right..$

Let's solve this system and determine the parameters of the hyperbola.

$\left\{ \begin{array}{l} 3b = 2a\\ a^2 + b^2 = \left({2\sqrt{13}}\right)^2 = 52 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} a = \frac{3b}{2}\\ \left({\frac{3b}{2}}\right)^2 + b^2 = 52 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} a = \frac{3b}{2}\\ \frac{9b^2}{4} + b^2 = 52 \end{array} \right.,$
$\Rightarrow \left\{ \begin{array}{l} a = \frac{3b}{2}\\ \frac{13b^2}{4} = 52 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} a = \frac{3b}{2}\\ b^2 = 16 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} a = 6\\ b = 4 \end{array} \right..$

Then the equation of the hyperbola is written as

$\frac{x^2}{6^2} - \frac{y^2}{4^2} = 1\;\text{ or }\;\frac{x^2}{36} - \frac{y^2}{16} = 1.$

### Example 5.

A point with the $$y-$$coordinate equal to 1 is taken on the hyperbola $$x^2-4y^2=16.$$ Find the distance from the point to the foci.

Solution.

Let's convert the hyperbola equation into canonical form and define its semi-axes:

$x^2 - 4y^2 = 16, \Rightarrow \frac{x^2}{16} - \frac{y^2}{4} = 1, \Rightarrow \frac{x^2}{4^2} - \frac{y^2}{2^2} = 1.$

Therefore the semi-axes are equal: $$a = 4,$$ $$b = 2.$$

Find the $$x-$$coordinate of the point $$M:$$

$x^2 - 4y^2 = 16, \Rightarrow x^2 = 16 + 4y^2, \Rightarrow x = \sqrt{16 + 4y^2} = \sqrt{16+4\cdot 1^2} = \sqrt{20}.$

That is, the coordinates of point M are equal $$\left({\sqrt{20},1}\right).$$

Now let's find the coordinates of the foci $$F_1$$ and $$F_2:$$

$c^2 = a^2 + b^2, \Rightarrow c = \sqrt{a^2 + b^2} = \sqrt{4^2 + 2^2} = \sqrt{20}.$

So, the foci have the following coordinates:

$F_1 = \left({-c,0}\right) = \left({-\sqrt{20},0}\right);\;\;F_2 = \left({c,0}\right) = \left({\sqrt{20},0}\right).$

Calculate the distances $$r_1$$ and $$r_2$$ from point $$M$$ to the foci $$F_1$$ and $$F_2,$$ respectively:

$r_1 = MF_1 = \sqrt{\left({x_{F1} - x_M}\right)^2 + \left({y_{F1} - y_M}\right)^2} = \sqrt{\left({-\sqrt{20}-\left({-\sqrt{20}}\right)}\right)^2 + \left({0 - 1}\right)^2} = \sqrt{\left({2\sqrt{20}}\right)^2 + 1^2} = \sqrt{81} = 9.$
$r_2 = MF_2 = \sqrt{\left({x_{F2} - x_M}\right)^2 + \left({y_{F2} - y_M}\right)^2} = \sqrt{\left({\sqrt{20}-\sqrt{20}}\right)^2 + \left({0 - 1}\right)^2} = \sqrt{0^2 + 1^2} = \sqrt{1} = 1.$

So the answer is $$r_1 = 9,$$ $$r_2 = 1.$$